MH1810 Math

Chapter 5a: Differentiation

Imron Rosyadi

Chapter 5: Differentiation

What is Differentiation?

Differentiation is a fundamental concept in calculus that deals with the instantaneous rate of change of a function.

It allows us to understand:

• How quantities change at a specific moment.
• The slope of a curve at any point.
• Optimization problems (finding maximums and minimums).

Figure: A smooth curve illustrating a tangent line at a point P.

Key Concepts Today:

• Derivatives & Tangent Lines
• Differentiation Rules
• Applications in Engineering and Science
• Optimization & Curve Sketching

Welcome to Chapter 5 on Differentiation! Today, we’ll explore one of the most powerful tools in mathematics for understanding how things change. Differentiation isn’t just a theoretical concept; it’s the backbone of many engineering and scientific disciplines.

We’ll start by defining what a derivative is, connecting it to the intuitive idea of a tangent line. Then, we’ll build up our arsenal of differentiation rules, which are essential for computing derivatives efficiently. Finally, we’ll dive into practical applications, seeing how derivatives help us solve real-world problems, from optimizing designs to understanding physical motion.

5.1 Derivatives: Tangents and Slopes

Consider a curve and a fixed point \(P\) on it. How do we define a line that “just touches” the curve at \(P\)?

Secant vs. Tangent

• A chord or secant is a straight line passing through any two points of a curve.
• The tangent to a curve at a point \(P\) is a straight line that ‘touches’ the curve at \(P\).
  • In a small neighborhood around \(P\), it cuts the curve in one and only one point.

Note

The tangent at \(P\) is the limiting position of the secant line \(PQ\) as \(Q\) approaches \(P\). \[ \text {slope} = \lim _ {Q \rightarrow P} \text {slope} \]

Figure: As point Q approaches P, the secant line PQ approaches the tangent line at P.

Imagine zooming in on a curve at a specific point. What does it mean for a line to “touch” the curve at that point?

We start with a secant line, which connects two points, P and Q, on the curve. As we slide point Q closer and closer to P, the secant line starts to look more and more like the tangent line at P. The tangent line is, in essence, the limit of these secant lines.

It’s crucial to understand that a tangent line is locally unique. While it might intersect the curve at other points far away from P, near P, it’s the only line that behaves in this “touching” manner.

When a Tangent Doesn’t Exist

Sometimes, a unique tangent line at a point does not exist.

Sharp Corners (Kinks)

• If a curve has a “kink” or a sharp corner at point \(P\), there isn’t a single, well-defined tangent line.
• The slope changes abruptly, and the limit of the secant slopes from different sides won’t match.

Figure: A curve with a sharp corner (kink) at point P, where a unique tangent does not exist.

Not all points on a curve have a well-defined tangent. Consider the image: at point P, the curve makes a sharp turn, forming a “kink.” If we try to draw a tangent here, we’d find multiple lines that seem to “touch” the curve, but none uniquely defines the direction. This is a point where the concept of a unique tangent breaks down, and as we’ll see, the derivative will not exist.

Mathematically Describing the Tangent

To find the equation of a tangent, we need its gradient (slope).

Slope of the Secant Line \(PQ\)

Let \(P = (x_P, y_P)\) and \(Q = (x_Q, y_Q)\) be two points on the curve.

The slope of the chord \(PQ\) is: \[ \mathrm {S l o p e o f c h o r d P Q} = \frac {y _ {Q} - y _ {P}}{x _ {Q} - x _ {P}}. \]

Slope of the Tangent at \(P\)

The slope of the tangent at \(P\) is the limit of the secant slopes as \(Q\) approaches \(P\): \[ \mathrm {S l o p e o f t a n g e n t a t P} = \lim _ {Q \to P} \frac {y _ {Q} - y _ {P}}{x _ {Q} - x _ {P}}. \]

Figure: Illustrating points P and Q on a curve with a secant line, used to derive the tangent’s slope.

Now, let’s put this into mathematical terms. If we have two points P and Q on a curve, the slope of the line connecting them (the secant line) is simply the “rise over run”.

To get the tangent line’s slope, we let Q get infinitesimally close to P. This “getting infinitesimally close” is where the concept of a limit comes into play. The slope of the tangent is precisely this limit. This forms the very definition of a derivative.

Introducing Increments and the Derivative

Let \(\Delta x\) denote the change in \(x\), and \(\Delta y\) the corresponding change in \(y\).

• \(\Delta x = x_Q - x_P\)
• \(\Delta y = y_Q - y_P\)

The slope of the tangent at \(P\) can then be written as: \[ \mathrm {S l o p e o f t a n g e n t a t P} = \lim _ {\Delta x \to 0} \frac {\Delta y}{\Delta x}. \]

If the curve is the graph of a function \(f(x)\), we have \(y_P = f(x_P)\) and \(y_Q = f(x_Q)\). So the slope of the tangent becomes: \[ \mathrm {S l o p e o f t a n g e n t a t P} = \lim _ {x \rightarrow x _ {P}} \frac {f (x) - f \left(x _ {P}\right)}{x - x _ {P}}. \] This limit, if it exists, is called the derivative of \(f\) at \(x_P\).

To make this more general, we introduce the notation of increments. \(\Delta x\) represents a small change in x, and \(\Delta y\) is the resulting change in y. As Q approaches P, \(\Delta x\) approaches 0.

If our curve is defined by a function \(f(x)\), then \(y_Q = f(x_Q)\) and \(y_P = f(x_P)\). Substituting this into our limit definition gives us the formal definition of the derivative. This derivative, \(f'(x_P)\), represents the instantaneous rate of change of \(f(x)\) with respect to \(x\) at the point \(x_P\). It’s a fundamental concept that underlies all of differentiation.

Instantaneous Rate of Change

Beyond geometry, derivatives describe how things change in the real world.

Increments (\(\Delta x\), \(\Delta f\))

• \(\Delta x = x_2 - x_1\): Change in an independent variable.
• \(\Delta f = f(x_2) - f(x_1)\): Corresponding change in the function value.
• \(\Delta x\) can be positive, negative, or zero.

Average Rate of Change

The ratio \(\frac{\Delta f}{\Delta x}\) is the average change in \(f\) as \(x\) changes from \(x_1\) to \(x_2\).

Instantaneous Rate of Change

The limit \(\lim_{\Delta x \to 0} \frac{\Delta f}{\Delta x}\) is the instantaneous rate of change in \(f\) with respect to \(x\).

Examples in Science & Engineering:

• Physics: Velocity (\(dx/dt\)), acceleration (\(d^2x/dt^2\))
• Chemistry: Reaction rates
• Biology: Population growth rates
• Economics: Marginal costs and revenues

Tip

The derivative \(f'(x)\) is equivalent to the instantaneous rate of change!

The concept of rate of change is not just about slopes; it’s about how one quantity responds to a change in another. Think about a sprinter’s position \(x\) at time \(t\). The average speed over an interval is \(\Delta x / \Delta t\). But what about their speed at an exact moment? That’s the instantaneous rate of change, or the derivative.

This idea is incredibly powerful. Whether you’re in physics calculating velocities and accelerations, chemistry studying reaction kinetics, or economics analyzing marginal costs, the derivative provides the mathematical framework to understand these instantaneous changes. It’s the core of how we model dynamic systems.

5.1.1 Derivatives & Differentiable Functions

Formal definitions of the derivative and differentiability.

Definition 5.1.2: Derivative at a Number

The derivative of a function \(f\) at a number \(c\), denoted by \(f'(c)\), is: \[ f ^ {\prime} (c) = \lim _ {x \to c} \frac {f (x) - f (c)}{x - c}, \] if this limit exists.

If \(f'(c)\) exists, we say \(f\) is differentiable at \(c\).

Definition 5.1.3: Derivative Function

1. A function \(f\) is differentiable on an open interval \((a, b)\) if it is differentiable at every number \(c\) in \((a, b)\).
2. The derivative function (or simply derivative) \(f'\) is defined by: \[ f ^ {\prime} (x) = \lim _ {y \rightarrow x} \frac {f (y) - f (x)}{y - x}. \] The domain of \(f'\) is the set of real numbers \(x\) such that \(f'(x)\) exists.

Important

Alternative Notations for \(f'(c)\):

• Using \(\Delta x = x - c\): \[ f ^ {\prime} (c) = \lim _ {\Delta x \to 0} \frac {f (c + \Delta x) - f (c)}{\Delta x}. \]
• Using \(h = x - c\): \[ f ^ {\prime} (c) = \lim _ {h \to 0} \frac {f (c + h) - f (c)}{h}. \]

These forms are often easier to use in calculations.

Here we formalize the definitions. The derivative at a single point \(c\), denoted \(f'(c)\), is simply the limit we discussed – the slope of the tangent at that specific point. If this limit produces a finite number, then the function is “differentiable” at \(c\).

Extending this, if a function is differentiable at every point within an interval, then it’s differentiable on that interval. The derivative function, \(f'(x)\), gives us a new function whose output is the slope of the original function at any given x.

Notice the alternative notations using \(\Delta x\) or \(h\). These are algebraically equivalent but can sometimes simplify the process of evaluating the limit, especially when dealing with polynomials or complex expressions.

Leibniz Notation and Example

Remarks on Derivative Notation

1. Leibniz Notation: For \(y = f(x)\), the derivative can be written as:
   • \(\frac{dy}{dx}\)
   • \(\frac{df}{dx}\)
   • \(\left.\frac{df}{dx}\right|_{x=a}\) or \(\frac{df}{dx}(a)\) for the derivative at a specific point \(a\).
2. “To differentiate” means to determine the derivative \(f'(x)\).

Leibniz Notation and Example

Example 5.1.4: Differentiating \(f(x) = x^2\)

1. Is \(f(x) = x^2\) differentiable at \(x = 1\)? \[ \lim _ {x \rightarrow 1} \frac {f (x) - f (1)}{x - 1} = \lim _ {x \rightarrow 1} \frac {x^2 - 1^2}{x - 1} = \lim _ {x \rightarrow 1} \frac {(x - 1)(x + 1)}{x - 1} = \lim _ {x \rightarrow 1} (x + 1) = 2. \]

   Since the limit exists, \(f\) is differentiable at \(x = 1\) and \(f'(1) = 2\).

2. Find \(f'(x)\) and the domain of \(f'\).

   \[ f ^ {\prime} (x) = \lim _ {h \rightarrow 0} \frac {f (x + h) - f (x)}{h} = \lim _ {h \rightarrow 0} \frac {(x + h)^2 - x^2}{h} \\ = \lim _ {h \rightarrow 0} \frac {x ^ {2} + 2 x h + h ^ {2} - x ^ {2}}{h} = \lim _ {h \rightarrow 0} \frac {2 x h + h ^ {2}}{h} \\ = \lim _ {h \rightarrow 0} (2 x + h) = 2 x. \]

   Thus, \(f'(x) = 2x\) for all \(x \in \mathbb{R}\). The domain of \(f'\) is \(\mathbb{R}\).

The Leibniz notation \(\frac{dy}{dx}\) is very intuitive, suggesting a “differential change in y divided by a differential change in x.” This notation is particularly useful when thinking about units and the chain rule.

Let’s work through an example: \(f(x) = x^2\). We can first check differentiability at a specific point, say \(x=1\). Using the limit definition, we can simplify the expression and evaluate the limit. Since we get a finite value, \(f\) is differentiable at \(x=1\), and its derivative at that point is 2.

To find the general derivative function, \(f'(x)\), we use the \(h \to 0\) definition. This involves expanding \(f(x+h)\), subtracting \(f(x)\), dividing by \(h\), and then taking the limit. For \(x^2\), this process reveals that \(f'(x) = 2x\). This means that the slope of the tangent line to \(y=x^2\) at any point \(x\) is \(2x\). For example, at \(x=1\), the slope is \(2(1)=2\), which matches our earlier calculation. The domain of \(f'(x)\) is all real numbers because the limit exists for all \(x\).

Example: The Modulus Function

Example 5.1.5: \(f(x) = |x|\)

1. Is \(f(x) = |x|\) differentiable at \(0\)? Is there a tangent to the graph of \(y = |x|\) at \(x = 0\)?

Solution: We evaluate \(\lim_{x\to 0}\frac{f(x) - f(0)}{x - 0} = \lim_{x\to 0}\frac{|x|}{x}\).

Since \(f(x) = \left\{ \begin{array}{l l} x & \text {i f} x \geq 0 \\ - x & \text {i f} x < 0 \end{array} \right.\), we check left and right limits:

• \(\lim_{x\to 0^{+}}\frac{|x|}{x} = \lim_{x\to 0^{+}}\frac{x}{x} = 1\).
• \(\lim_{x\to 0^{-}}\frac{|x|}{x} = \lim_{x\to 0^{-}}\frac{-x}{x} = -1\).

Since \(\lim_{x\to 0^{+}}\frac{|x|}{x}\neq \lim_{x\to 0^{-}}\frac{|x|}{x}\), the limit \(\lim_{x\to 0}\frac{|x|}{x}\) does not exist.

Therefore, \(f(x) = |x|\) is not differentiable at 0.

Graphically, this corresponds to the “kink” at \(x=0\), where no unique tangent exists.

Example: The Modulus Function

Example 5.1.5: \(f(x) = |x|\)

2. Find \(f'(x)\) and the domain of \(f'\).

• For \(x > 0\): \(f(x) = x \Rightarrow f'(x) = 1\).
• For \(x < 0\): \(f(x) = -x \Rightarrow f'(x) = -1\).

In conclusion:

\[ f ^ {\prime} (x) = \left\{ \begin{array}{l l} 1 & \text {i f} x > 0 \\ - 1 & \text {i f} x < 0 \end{array} \right. \]

The domain of \(f'\) is \(\mathbb{R} \setminus \{0\}\).

This example highlights a crucial point: differentiability often fails at points where the function has a sharp corner or a discontinuity.

For \(f(x)=|x|\), the graph has a distinct “V” shape with a sharp point at the origin. If we try to compute the derivative at \(x=0\), we find that the left-hand limit and the right-hand limit of the difference quotient are different (1 and -1, respectively). Since these limits don’t match, the overall limit does not exist, and thus \(f(x)=|x|\) is not differentiable at \(x=0\).

This directly ties back to our earlier discussion about “kinks” on a curve. A function with a kink is not differentiable at that point. Away from the origin, \(f(x)=|x|\) is a straight line, so its derivative is easily found as 1 for \(x>0\) and -1 for \(x<0\).

Tangents and Differentiability: A Deeper Look

Remarks on Tangents and Derivatives

1. If \(f'(c)\) exists, the graph of \(f\) has a tangent at \(x = c\), and \(f'(c)\) is its slope.

   The equation of the tangent is given by:

   \[ \frac {y - f (c)}{x - c} = f ^ {\prime} (c), \quad \text{or} \quad y = f(c) + f'(c)(x-c). \]

2. It is possible for a graph to have a tangent even when the derivative does not exist (e.g., a vertical tangent).

Tangents and Differentiability: A Deeper Look

Example 5.1.6: \(f(x) = x^{1/3}\)

1. Is \(f(x) = x^{1/3}\) differentiable at \(x = 0\)? Is there a tangent to the graph at \(x = 0\)?

Solution: At \(x = 0\), we look at the limit:

\[ \lim _ {h \to 0} \frac {f (0 + h) - f (0)}{h} = \lim _ {h \to 0} \frac {h ^ {1 / 3} - 0}{h} = \lim _ {h \to 0} \frac {1}{h ^ {2 / 3}} = \infty. \]

Since the limit is not a finite number, \(f\) is not differentiable at 0.

However, the graph of \(f\) has a vertical tangent at \(x = 0\). (The slope is infinite).

2. Find the derivative of \(f(x) = x^{1/3}\).

For \(x \neq 0\):

\[ \lim _ {y \to x} \frac {y ^ {1 / 3} - x ^ {1 / 3}}{y - x} = \frac {1}{3 x ^ {2 / 3}}. \]

Thus, \(f'(x) = \frac{1}{3x^{2/3}}\) for \(x \in \mathbb{R} \setminus \{0\}\).

The existence of a derivative guarantees a unique, finite-sloped tangent. However, the converse isn’t always true. A function can have a tangent even if its derivative doesn’t exist as a finite number. This happens with vertical tangents.

Consider \(f(x) = x^{1/3}\) (the cube root function). At \(x=0\), if we compute the limit for the derivative, we find it goes to infinity. This means the function is not differentiable at \(x=0\) in the usual sense (it doesn’t have a finite slope). Yet, if you visualize the graph, there’s clearly a vertical tangent line at the origin. The slope is infinite, which isn’t a finite number, so the derivative doesn’t “exist” as a number.

This example illustrates an important nuance. Differentiability requires a finite limit for the slope.

Basic Differentiation: Constants and Power Rule

Example 5.1.7: Derivative of a Constant

Use the definition of derivative to prove \(\frac{dC}{dx} = 0\) for any constant \(C\).

Solution: Let \(f(x) = C\).

\[ f ^ {\prime} (x) = \lim _ {h \to 0} \frac {f (x + h) - f (x)}{h} = \lim _ {h \to 0} \frac {C - C}{h} = \lim _ {h \to 0} \frac {0}{h} = \lim _ {h \to 0} 0 = 0. \]

Thus, the derivative of any constant is 0.

Theorem 5.1.8: Power Rule for Positive Integers

Let \(f(x) = x^n\), where \(n\) is a positive integer. Then,

\[ f ^ {\prime} (x) = n x ^ {n - 1}, \quad \text{i.e.,} \quad \frac {d x ^ {n}}{d x} = n x ^ {n - 1}. \]

Now we’ll start building our differentiation toolkit with some fundamental rules.

First, the derivative of a constant function. Intuitively, if a function’s value never changes, its rate of change is zero. The formal definition confirms this: the difference \(f(x+h) - f(x)\) will always be zero, so the limit is zero.

Next, the power rule for positive integer exponents. This is incredibly useful! The proof involves the binomial expansion, where after cancelling the \(x^n\) terms, every remaining term in the numerator has an \(h\). Dividing by \(h\) leaves us with \(nx^{n-1}\) plus terms that still have \(h\) and go to zero as \(h \to 0\). This gives us the elegant result: bring the power down and reduce the power by one. We already saw this in action with \(x^2 \to 2x\).

Derivative of a Constant Multiple

Theorem 5.1.9: Constant Multiple Rule

Let \(\alpha\) be a real constant. Consider the function \(\alpha f\), defined by \((\alpha f)(x) = \alpha \cdot f(x)\).

If \(f\) is differentiable at \(x = c\), then \(\alpha f\) is differentiable at \(x = c\) and its derivative at \(c\) is:

\[ \left(\alpha f\right) ^ {\prime} (c) = \alpha \cdot f ^ {\prime} (c). \]

Example 5.1.10:

We know \(\frac{d}{dx} (x^{1/3}) = \frac{1}{3x^{2/3}}\), \(x \neq 0\).

By the Constant Multiple Rule:

\[ \frac {d}{d x} (1 7 9 x ^ {1 / 3}) = 1 7 9 \frac {d}{d x} (x ^ {1 / 3}) = \frac {1 7 9}{3 x ^ {2 / 3}}. \]

The constant multiple rule is another intuitive and powerful rule. It states that if you multiply a function by a constant, you simply multiply its derivative by that same constant.

The proof is quite straightforward, relying on the limit properties we’ve already covered. You can factor out the constant from the limit expression, and what remains is the definition of the derivative of \(f\).

This rule makes differentiating expressions like \(5x^3\) or \(1/2 \sin x\) very easy, once you know the derivative of the base function. For instance, if you know the derivative of \(x^{1/3}\), finding the derivative of \(179x^{1/3}\) is just a matter of multiplying by 179.

5.1.2 Higher Derivatives

Derivatives don’t stop at the first one! We can differentiate a function multiple times.

• The second derivative of \(f\) is the derivative of \(f'\), denoted \(f''(x)\) or \(\frac{d^2y}{dx^2}\).
• The third derivative is \(f'''(x)\) or \(\frac{d^3y}{dx^3}\).
• In general, the \(n^{th}\) derivative is \(f^{(n)}(x)\) or \(\frac{d^ny}{dx^n}\).

Example 5.1.11: Higher Derivatives of \(f(x) = x^5\)

• First derivative: \(f'(x) = \frac{dy}{dx} = 5x^4\).
• Second derivative: \(f''(x) = (f')'(x) = \frac{d}{dx}(5x^4) = 5 \cdot (4x^3) = 20x^3\).
• Third derivative: \(f'''(x) = (f'')'(x) = \frac{d}{dx}(20x^3) = 20 \cdot (3x^2) = 60x^2\).
• Fourth derivative: \(f^{(4)}(x) = \frac{d}{dx}(60x^2) = 60 \cdot (2x) = 120x\).
• Fifth derivative: \(f^{(5)}(x) = \frac{d}{dx}(120x) = 120\).
• Sixth derivative: \(f^{(6)}(x) = \frac{d}{dx}(120) = 0\).

(All subsequent derivatives will also be zero.)

Just as we can find the rate of change of a function, we can also find the rate of change of that rate of change, and so on. These are called higher derivatives.

The second derivative tells us about the concavity of a function’s graph, which we’ll explore later. In physics, if the first derivative is velocity, the second derivative is acceleration. The third derivative is sometimes called jerk.

The process is simply to apply the differentiation rules repeatedly. As seen with \(x^5\), you keep reducing the power and multiplying by the new coefficient until you eventually reach a constant, and then zero. Higher derivatives become very important in advanced topics like Taylor series, which allow us to approximate functions using polynomials.

Remarks on Higher Derivatives

Higher derivatives provide deeper insights into a function’s behavior.

Applications:

1. Kinematics:

   • If \(x(t)\) is position at time \(t\).
   • \(x'(t)\) is the velocity of the object.
   • \(x''(t)\) is the acceleration at time \(t\).

2. Curve Shape: The first and second derivatives help determine the shape of a function’s graph (e.g., increasing/decreasing, concavity). We’ll cover this in more detail in the next sections.

3. Approximation: Higher derivatives are crucial for approximating functions using Taylor series, which allow us to represent complex functions as polynomials.

Higher derivatives are not just mathematical curiosities; they have profound applications.

In physics, the first derivative of position is velocity, telling us how fast an object is moving and in what direction. The second derivative of position is acceleration, telling us how quickly the velocity is changing. This is fundamental to understanding motion.

Beyond physics, higher derivatives are used to analyze the shape of curves, identifying where they bend upwards or downwards (concavity). This is vital for curve sketching and understanding the overall behavior of a function.

And in numerical methods and approximation, Taylor series rely heavily on higher derivatives to create polynomial approximations of functions, which can be much easier to work with than the original function itself. This highlights the practical utility of understanding higher derivatives.

5.2 Differentiation Rules: Differentiability and Continuity

There’s a strong relationship between differentiability and continuity.

Theorem 5.2.1: Differentiability Implies Continuity

If a function \(f\) is differentiable at \(x = c\), then \(f\) is continuous at \(x = c\).

Figure: A smooth, continuous curve, illustrating how differentiability implies continuity (no breaks or jumps).

Caution

The converse is FALSE: A function can be continuous at \(x=c\) but NOT differentiable at \(x=c\). (Recall \(f(x) = |x|\) at \(x=0\)).

Corollary 5.2.2:

If \(f\) is not continuous at \(x = c\), then \(f\) is not differentiable at \(x = c\).

(This is the contrapositive of Theorem 5.2.1)

This is a foundational theorem connecting two important calculus concepts. Differentiability is a stronger condition than continuity. If a function is differentiable at a point, it must be continuous at that point. Think of it: if you can draw a unique, non-vertical tangent line, the function cannot have any breaks or jumps.

The proof uses algebraic manipulation of limits. We start with the definition of differentiability and work towards the definition of continuity.

However, it’s crucial to remember that the reverse is not true! A function can be continuous but not differentiable. Our example of \(f(x)=|x|\) at \(x=0\) is a perfect illustration: it’s continuous (you can draw it without lifting your pen), but it’s not differentiable because of the sharp corner.

The corollary simply states that if a function has a jump, hole, or asymptote (i.e., it’s not continuous), then it definitely can’t be differentiable there. This gives us a quick check: if it’s discontinuous, don’t even bother checking for differentiability.

Differentiability of Piecewise Functions

Theorem 5.2.3: Differentiability of Piecewise Functions

Suppose \(f\) is continuous at \(x = c\). If

\[ \lim _ {x \to c ^ {+}} f ^ {\prime} (x) = \lim _ {x \to c ^ {-}} f ^ {\prime} (x) = L, \]

then \(f\) is differentiable at \(x = c\) and \(f'(c) = L\).

This theorem is very useful for functions defined by different expressions on different intervals.

Differentiability of Piecewise Functions

Example 5.2.4: Differentiating a Piecewise Function

Let \(f(x) = \left\{ \begin{array}{ll} x^3 + 2 & \text{if } x > 1 \\ 3x & \text{if } x \leq 1 \end{array} \right.\), find \(f'(x)\).

Solution:

1. Check Continuity at \(x=1\):
   • \(f(1) = 3(1) = 3\).
   • \(\lim_{x\to 1^{+}}f(x) = \lim_{x\to 1^{+}}(x^{3} + 2) = 1^3 + 2 = 3\).
   • \(\lim_{x\to 1^{-}}f(x) = \lim_{x\to 1^{-}}(3x) = 3(1) = 3\). Since \(\lim_{x\to 1}f(x) = 3 = f(1)\), \(f\) is continuous at \(x = 1\).
2. Find Derivatives for Each Piece:
   • For \(x > 1\): \(f^{\prime}(x) = \frac{d}{dx}(x^3 + 2) = 3x^{2}\).
   • For \(x < 1\): \(f^{\prime}(x) = \frac{d}{dx}(3x) = 3\).
3. Check Limits of Derivatives at \(x=1\):
   • \(\lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} (3x^2) = 3(1)^2 = 3\).
   • \(\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} (3) = 3\).
   Since \(\lim_{x \to 1^+} f'(x) = \lim_{x \to 1^-} f'(x) = 3\), and \(f\) is continuous at \(x=1\), we conclude \(f\) is differentiable at \(x=1\) and \(f'(1) = 3\).

Final Derivative Function:

\[ f ^ {\prime} (x) = \left\{ \begin{array}{l l} 3 x ^ {2} & \text {i f} x > 1 \\ 3 & \text {i f} x \leq 1 \end{array} \right. \]

This theorem gives us a practical method to determine differentiability for piecewise functions. The key is that not only must the function be continuous at the “split point,” but the derivatives of each piece must also match at that point. Geometrically, this means the two pieces connect smoothly, without a kink.

Let’s apply this to the example. First, we confirm continuity at the split point \(x=1\). Since the left-hand limit, right-hand limit, and function value all agree, \(f\) is continuous.

Next, we find the derivatives of each piece for \(x>1\) and \(x<1\). These are straightforward power rule applications.

Finally, we check if the limits of these derivatives match at \(x=1\). Since they both approach 3, and the function is continuous, the theorem tells us that the derivative exists at \(x=1\) and is equal to 3. So, we can include \(x=1\) in the domain of the \(3x^2\) part of the derivative, or specifically state \(f'(1)=3\). The way it’s written now, it means \(f'(1)=3\) is taken from the second line of the definition (for \(x \le 1\)).

5.2.2 Differentiation Rules: Sum and Difference Rules

These rules allow us to differentiate sums and differences of functions easily.

Theorem 5.2.5 (Sum and Difference Rules):

Suppose \(f\) and \(g\) are differentiable at \(x = c\). Then:

1. Sum Rule: \(f + g\) is differentiable at \(x = c\) and \[ (f + g) ^ {\prime} (c) = f ^ {\prime} (c) + g ^ {\prime} (c). \]
2. Difference Rule: \(f - g\) is differentiable at \(x = c\) and \[ (f - g) ^ {\prime} (c) = f ^ {\prime} (c) - g ^ {\prime} (c). \]

We’re continuing to build our set of differentiation rules. The sum and difference rules are very intuitive: the derivative of a sum (or difference) of functions is simply the sum (or difference) of their individual derivatives.

The proof for the sum rule demonstrates this by breaking down the limit definition of the derivative into two separate limits, each corresponding to the derivative of one of the functions. This relies on the property that the limit of a sum is the sum of the limits, provided those individual limits exist.

These rules, combined with the power rule and constant multiple rule, allow us to differentiate any polynomial function very quickly. For example, to differentiate \(3x^2 + 5x - 7\), you differentiate each term separately: \(3(2x) + 5(1) - 0 = 6x + 5\).

Differentiation Rules: Product and Quotient Rules

These rules handle derivatives of products and quotients of functions.

Theorem 5.2.6 (Product and Quotient Rules):

Suppose \(f\) and \(g\) are differentiable at \(x = c\). Then:

1. Product Rule: \(fg\) is differentiable at \(x = c\) and \[ (f g) ^ {\prime} (c) = f ^ {\prime} (c) g (c) + f (c) g ^ {\prime} (c). \]

   (Memorize as: (derivative of first) times (second) plus (first) times (derivative of second))

2. Quotient Rule: \(f / g\) is differentiable at \(x = c\), provided \(g(c) \neq 0\), and \[ \left(\frac {f}{g}\right) ^ {\prime} (c) = \frac {f ^ {\prime} (c) g (c) - f (c) g ^ {\prime} (c)}{(g (c)) ^ {2}}. \]

   (Memorize as: (derivative of top) times (bottom) minus (top) times (derivative of bottom), all over (bottom squared))

3. Reciprocal Rule: If \(g(x) \neq 0\) near \(c\), then \(\frac{1}{g}\) is differentiable at \(x = c\), and \[ \left(\frac {1}{g}\right) ^ {\prime} (c) = \frac {- g ^ {\prime} (c)}{(g (c)) ^ {2}}. \]

   (This is a special case of the Quotient Rule where \(f(x)=1\).)

The product and quotient rules are essential for differentiating functions that are multiplied or divided. Unlike the sum/difference rules, you cannot simply multiply or divide the derivatives. These rules are more complex but follow a specific pattern.

The product rule has a symmetric form, adding two terms where each function is differentiated once. The proof involves a clever trick of adding and subtracting a term, which allows us to regroup the expression into limits that represent the individual derivatives.

The quotient rule is a bit more involved to memorize due to the subtraction and the denominator being squared. A common mnemonic for the quotient rule is “low dee high minus high dee low, over low squared” (where ‘dee’ means ‘derivative of’). The reciprocal rule is a simplified version where the numerator is a constant 1.

Generalized Power Rule for All Integers

Combining previous rules, we get a general power rule for all integers.

Theorem 5.2.7: Linear Combination Rule

The function \(\alpha f + \beta g\) is differentiable at \(x = c\), where \(\alpha\) and \(\beta\) are real constants, and \[ (\alpha f + \beta g) ^ {\prime} (c) = \alpha f ^ {\prime} (c) + \beta g ^ {\prime} (c). \]

(This follows from the Constant Multiple Rule and the Sum Rule.)

Proposition 5.2.8: The Power Rule for Negative Integers

Suppose \(m\) is a negative integer, say \(m = -n\), where \(n \in \mathbb{Z}^+\). If \(f(x) = x^m\) where \(x \neq 0\), then \(f'(x) = mx^{m-1}\).

Conclusion:

\[ \frac {d}{d x} \left(x ^ {n}\right) = n x ^ {n - 1}, \text {for every (positive and negative) i n t e g e r} n. \]

The linear combination rule is a convenient summary of the constant multiple and sum/difference rules. It states that the derivative of any linear combination of functions is the linear combination of their derivatives.

Building on the power rule for positive integers, we can now extend it to negative integers using the reciprocal rule. If \(m\) is a negative integer, we can write \(x^m\) as \(1/x^n\) for some positive integer \(n\). Applying the reciprocal rule and the positive integer power rule, we arrive at the same form: \(mx^{m-1}\).

This means the power rule, \(\frac{d}{dx}(x^n) = nx^{n-1}\), holds true for all integer values of \(n\) (provided \(x \neq 0\) for negative \(n\)). This is a very powerful and widely used rule in calculus.

The Chain Rule: Differentiating Composite Functions

The Chain Rule is used to differentiate composite functions (functions within functions).

Theorem 5.2.9 (The Chain Rule):

Let \(f\) and \(g\) be functions such that \(g(x)\) is differentiable at \(x = c\) and \(f(u)\) is differentiable at \(u = g(c)\). Then \(f \circ g\) (defined as \(f(g(x))\)) is differentiable at \(x = c\) and

\[ (f \circ g) ^ {\prime} (c) = f ^ {\prime} (g (c)) g ^ {\prime} (c). \]

Or with Leibniz notation (very intuitive!): \[ \frac {d f}{d x} = \frac {d f}{d u} \cdot \frac {d u}{d x}. \]

The Chain Rule: Differentiating Composite Functions

Example 5.2.10: Differentiating \(h(x) = \sqrt{3x^5 - 6x^2 + 7x + \pi}\)

Solution:

Let \(f(u) = \sqrt{u}\) (the “outer” function) and \(g(x) = 3x^5 - 6x^2 + 7x + \pi\) (the “inner” function).

Then \(h(x) = f(g(x))\).

• \(f'(u) = \frac{1}{2\sqrt{u}}\)
• \(g'(x) = 15x^4 - 12x + 7\)

Applying the Chain Rule, \(h'(x) = f'(g(x))g'(x)\):

\[ h ^ {\prime} (x) = \frac {1}{2 \sqrt {3 x ^ {5} - 6 x ^ {2} + 7 x + \pi}} \cdot (1 5 x ^ {4} - 1 2 x + 7). \]

The chain rule is arguably one of the most important differentiation rules, as it allows us to differentiate composite functions – functions that are “nested” inside each other.

The formula states that the derivative of \(f(g(x))\) is the derivative of the outer function \(f\) evaluated at the inner function \(g(x)\), multiplied by the derivative of the inner function \(g(x)\). The Leibniz notation, \(df/dx = (df/du) \cdot (du/dx)\), is incredibly helpful here because it visually suggests that the ‘du’ terms cancel out, leaving \(df/dx\).

Let’s apply it to the example. We identify the outer function (square root) and the inner function (the polynomial inside the square root). We differentiate each separately. Then, we plug the inner function back into the derivative of the outer function, and multiply by the derivative of the inner function. It’s a systematic process that becomes second nature with practice.

Corollaries of the Chain Rule

The Chain Rule leads to convenient forms for common function types.

Corollary 5.2.11:

1. Generalized Power Rule: For \(n \in \mathbb{Z}\), \[ \frac{d}{dx} (f(x))^n = n(f(x))^{n - 1}f'(x). \]

   (This combines the Power Rule and Chain Rule for a function raised to an integer power.)

2. Derivative of Square Root of a Function: For \(f(x) > 0\), \[ \frac{d}{dx}\sqrt{f(x)} = \frac{1}{2}\frac{f'(x)}{\sqrt{f(x)}}. \]

   (This is a special case of (a) where \(n=1/2\), often seen enough to be remembered.)

These corollaries are direct applications of the Chain Rule that are so common they’re worth highlighting.

The generalized power rule is simply the power rule applied via the chain rule. Instead of \(x^n\), we have \((f(x))^n\). We still “bring down the power and reduce by one,” but then we must multiply by the derivative of the inner function, \(f'(x)\). This is incredibly common in differentiating polynomial terms, trigonometric terms, and other expressions raised to a power.

The derivative of the square root of a function is an even more specific application, where \(n=1/2\). It’s so frequently encountered that many students memorize this specific form. It saves a step from explicitly writing \((f(x))^{1/2}\) and applying the generalized power rule. Both these corollaries are powerful shortcuts derived from the fundamental Chain Rule.

5.2.3 Derivatives of Trigonometric Functions

Key derivatives for trigonometric functions.

Theorem 5.2.12:

1. \(\frac{d}{dx} (\sin x) = \cos x\)
2. \(\frac{d}{dx} (\cos x) = -\sin x\)
3. \(\frac{d}{dx} (\tan x) = \sec^2 x\)
4. \(\frac{d}{dx} (\sec x) = \sec x \tan x\)
5. \(\frac{d}{dx} (\csc x) = -\csc x \cot x\)
6. \(\frac{d}{dx} (\cot x) = -\csc^2 x\)

Corollary 5.2.13 (Chain Rule for Trigonometric Functions):

1. \(\frac{d}{dx}\sin (Ax + B) = A\cos (Ax + B)\)
2. \(\frac{d}{dx} \cos (Ax + B) = -A \sin (Ax + B)\)
3. \(\frac{d}{dx} \tan (Ax + B) = A \sec^2 (Ax + B)\)
4. \(\frac{d}{dx}\sin (f(x)) = f'(x)\cos (f(x))\)

Now we move on to the derivatives of trigonometric functions. These are fundamental and must be memorized. Notice the pattern: sine goes to cosine, cosine goes to negative sine. The derivatives of tangent, secant, cosecant, and cotangent are also standard results.

The corollary simply shows these derivatives combined with the chain rule. If the argument of the trigonometric function is a linear expression like \(Ax+B\), we multiply by the derivative of that inner function, which is \(A\). More generally, if the argument is any function \(f(x)\), we multiply by \(f'(x)\). This is a direct application of the chain rule and makes differentiating complex trigonometric expressions much easier.

5.2.4 Derivatives of Exponential and Logarithmic Functions

Another crucial set of derivatives.

Theorem 5.2.15:

1. \(\frac{d}{dx} (e^x) = e^x\)
2. \(\frac{d}{dx} (a^x) = a^x\ln a\), for \(a > 0\) and \(a \neq 1\).
3. \(\frac{d}{dx} (\ln x) = \frac{1}{x}\), for \(x > 0\).
4. \(\frac{d}{dx} (\log_a x) = \frac{1}{x\ln a}\), for \(a > 0\) and \(a \neq 1\).

Note

The exponential function \(e^x\) is unique because its derivative is itself! (\(f'(x) = f(x)\)).

Proposition 5.2.17 (Chain Rule for Exp/Log):

1. \(\frac{d}{dx} e^{f(x)} = e^{f(x)} \cdot f'(x)\)
2. \(\frac{d}{dx}\ln f(x) = \frac{f'(x)}{f(x)}\), for \(f(x) > 0\).

5.2.4 Derivatives of Exponential and Logarithmic Functions

Example 5.2.16:

1. \(\frac{d}{dx}\left(x^5 + \sin(4x) - xe^x\right) = 5x^4 + 4\cos(4x) - (e^x + xe^x)\).
2. \(\frac{d}{dx}\left(\frac{e^x}{3x^3 - x + 1}\right) = \frac{e^x(3x^3-x+1) - e^x(9x^2-1)}{(3x^3-x+1)^2}\).
3. \(\frac{d}{dx}\left(\tan (\sqrt{x})\right) = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\).
4. \(\frac{d}{dx}\left(\ln (\cos^2 x + 1)\right) = \frac{1}{\cos^2 x + 1} \cdot (2\cos x (-\sin x)) = \frac{-2\sin x \cos x}{\cos^2 x + 1}\).

These derivatives are fundamental to calculus and have wide applications in modeling natural growth and decay processes.

The derivative of \(e^x\) being \(e^x\) is a remarkable property that makes it a natural choice for many mathematical models. For other bases \(a^x\), the derivative includes a factor of \(\ln a\).

Similarly, the derivative of \(\ln x\) is \(1/x\). For other bases \(\log_a x\), the derivative includes \(1/\ln a\).

The propositions show these derivatives combined with the Chain Rule, which are incredibly common. For instance, differentiating \(e^{f(x)}\) results in \(e^{f(x)}\) multiplied by the derivative of the exponent, \(f'(x)\). Differentiating \(\ln(f(x))\) results in \(1/f(x)\) multiplied by \(f'(x)\).

The examples demonstrate how to combine all the rules we’ve learned so far: sum/difference, product, quotient, power, trigonometric, exponential, logarithmic, and especially the chain rule. Each part of example 5.2.16 requires identifying the appropriate rule and applying it systematically.

5.2.5 Parametric Differentiation

When \(x\) and \(y\) are both functions of a third variable (a parameter \(t\)), we use parametric differentiation.

Let \(y = u(t)\) and \(x = v(t)\).

Then, to find \(\frac{dy}{dx}\), we use the formula:

\[ \frac {d y}{d x} = \frac {\frac {d y}{d t}}{\frac {d x}{d t}} = \frac {u ^ {\prime} (t)}{v ^ {\prime} (t)}, \quad \text{provided } \frac{dx}{dt} \neq 0. \]

To find the second derivative \(\frac{d^2y}{dx^2}\):

\[ \frac {d ^ {2} y}{d x ^ {2}} = \frac {d}{d x} \left(\frac {d y}{d x}\right) = \frac {\frac {d}{d t} \left(\frac {d y}{d x}\right)}{\frac {d x}{d t}}. \]

(Note: \(\frac{d^2y}{dx^2} \neq \frac{d^2y/dt^2}{d^2x/dt^2}\))

5.2.5 Parametric Differentiation

Example 5.2.18: Cycloid Curve

Let \(x = 9(t - \sin t)\) and \(y = 9(1 - \cos t)\).

1. Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\):

   • \(\frac {d y}{d t} = \frac {d}{d t} (9(1 - \cos t)) = 9 \sin t\).
   • \(\frac {d x}{d t} = \frac {d}{d t} (9(t - \sin t)) = 9 (1 - \cos t)\).

2. Find \(\frac{dy}{dx}\): \[ \frac {d y}{d x} = \frac {9 \sin t}{9 (1 - \cos t)} = \frac {2 \sin \frac {t}{2} \cos \frac {t}{2}}{2 \sin^ {2} \frac {t}{2}} = \cot \frac {t}{2}. \]

3. Find \(\frac{d^2y}{dx^2}\):

   First, differentiate \(\frac{dy}{dx}\) with respect to \(t\):

   \(\frac{d}{dt}\left(\cot \frac{t}{2}\right) = -\csc^2\left(\frac{t}{2}\right) \cdot \frac{1}{2}\). Then, divide by \(\frac{dx}{dt}\):

   \[ \frac {d ^ {2} y}{d x ^ {2}} = \frac {- \frac {1}{2} \csc^2 \left(\frac{t}{2}\right)}{9 (1 - \cos t)} = \frac {- \frac {1}{2 \sin^2 \left(\frac{t}{2}\right)}}{9 \cdot 2 \sin^2 \left(\frac{t}{2}\right)} = - \frac {1}{3 6 \sin^ {4} \left(\frac {t}{2}\right)}. \]

Parametric differentiation is a powerful technique for finding derivatives when a curve is defined by equations for \(x\) and \(y\) in terms of a third variable, called a parameter (often \(t\) for time or an angle).

The core idea for finding \(\frac{dy}{dx}\) is to use the chain rule implicitly: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). This is very intuitive, almost like cancelling the ‘dt’ terms.

Calculating the second derivative, \(\frac{d^2y}{dx^2}\), is a common source of error. You do not simply divide the second derivatives with respect to \(t\). Instead, you treat \(\frac{dy}{dx}\) (which is itself a function of \(t\)) as a new function that you differentiate with respect to \(t\), and then divide that result by \(\frac{dx}{dt}\).

The example of a cycloid demonstrates this process step-by-step. After finding the first derivative \(\frac{dy}{dx}\) as a function of \(t\), we then differentiate that function with respect to \(t\) and apply the formula for the second derivative. Pay close attention to the algebraic simplifications using trigonometric identities, which often makes the final expression much cleaner.

5.3 Implicit Differentiation

When \(x\) and \(y\) are related by an equation \(F(x, y) = 0\) that isn’t easily solved for \(y\), we use implicit differentiation.

We differentiate both sides of the equation with respect to \(x\), treating \(y\) as an unknown function of \(x\) (i.e., \(y(x)\)) and applying the Chain Rule whenever differentiating a term involving \(y\).

Example 5.3.1: Find \(\frac{dy}{dx}\) if \(3x^4 y^2 - 7xy^3 = 4 - 8y\).

Solution: Differentiating each term with respect to \(x\):

\[ \frac {d}{d x} (3x^4 y^2) - \frac {d}{d x} (7xy^3) = \frac {d}{d x} (4) - \frac {d}{d x} (8y). \]

Applying Product Rule and Chain Rule:

• \(\frac{d}{dx} (3x^4 y^2) = 3(4x^3)y^2 + 3x^4(2y \frac{dy}{dx}) = 12x^3 y^2 + 6x^4 y \frac{dy}{dx}\).
• \(\frac{d}{dx} (7xy^3) = 7(1)y^3 + 7x(3y^2 \frac{dy}{dx}) = 7y^3 + 21xy^2 \frac{dy}{dx}\).
• \(\frac{d}{dx} (4) = 0\).
• \(\frac{d}{dx} (8y) = 8 \frac{dy}{dx}\).

5.3 Implicit Differentiation

Example 5.3.1: Find \(\frac{dy}{dx}\) if \(3x^4 y^2 - 7xy^3 = 4 - 8y\). (cont.)

Substituting these back into the equation: \[ 12x^3 y^2 + 6x^4 y \frac{dy}{dx} - (7y^3 + 21xy^2 \frac{dy}{dx}) = 0 - 8 \frac{dy}{dx}. \]

Rearranging terms to isolate \(\frac{dy}{dx}\):

\[ 6x^4 y \frac{dy}{dx} - 21xy^2 \frac{dy}{dx} + 8 \frac{dy}{dx} = 7y^3 - 12x^3 y^2. \] \[ \frac {d y}{d x} (6x^4 y - 21xy^2 + 8) = 7y^3 - 12x^3 y^2. \] Therefore: \[ {\frac {d y}{d x}} = \frac {7y^3 - 12x^3 y^2}{6x^4 y - 21xy^2 + 8}. \]

Implicit differentiation is a technique used when \(y\) is defined as a function of \(x\) indirectly, through an equation that mixes \(x\) and \(y\) terms, rather than explicitly as \(y = f(x)\).

The key idea is to differentiate both sides of the equation with respect to \(x\). When you encounter a term involving \(y\), you treat \(y\) as a function of \(x\) and apply the chain rule. So, the derivative of \(y^2\) with respect to \(x\) is \(2y \frac{dy}{dx}\), and the derivative of \(\sin y\) is \(\cos y \frac{dy}{dx}\).

In the example, we systematically differentiate each term. Notice how the product rule is used for \(3x^4y^2\) and \(7xy^3\), and the chain rule is applied to \(y^2\) and \(y^3\) terms when differentiating with respect to \(x\). After differentiating, the goal is to algebraically solve for \(\frac{dy}{dx}\). This involves collecting all terms with \(\frac{dy}{dx}\) on one side and factoring it out, then dividing. This results in \(\frac{dy}{dx}\) being expressed in terms of both \(x\) and \(y\).

Power Rule for Rational Exponents

We can now differentiate \(x^n\) for all integers \(n\). What about rational exponents like \(x^{3/2}\)?

For a rational number \(r = \frac{m}{n}\), where \(m \in \mathbb{Z}\) and \(n \in \mathbb{Z}^+\), \(x^{\frac{m}{n}} = (x^{\frac{1}{n}})^m\).

Theorem 5.3.2: Power Rule for Rational Powers

Suppose \(r = \frac{m}{n}\), where \(n\) is a positive integer, and \(m \in \mathbb{Z}\). Then \[ \frac {d}{d x} \left(x ^ {r}\right) = r x ^ {r - 1}. \]

Example 5.3.3:

\[ \frac {d}{d x} \left(x ^ {3 / 2} - \pi x ^ {- 9 / 5} + x ^ {1 / 3}\right) = \frac{3}{2}x^{1/2} - \pi \left(-\frac{9}{5}\right)x^{-14/5} + \frac{1}{3}x^{-2/3}. \\ = \frac{3}{2}\sqrt{x} + \frac{9\pi}{5}x^{-14/5} + \frac{1}{3}x^{-2/3}. \]

The power rule is incredibly versatile! We’ve seen it for positive and negative integers, and now we extend it to rational exponents. This means it works for square roots, cube roots, and any fractional power.

The proof uses implicit differentiation, which we just learned. By raising \(y = x^{m/n}\) to the power of \(n\), we get \(y^n = x^m\), which is an implicit relationship between \(y\) and \(x\) involving integer powers. Differentiating this implicitly with respect to \(x\) allows us to solve for \(\frac{dy}{dx}\) and, after some algebraic manipulation of exponents, confirm that the power rule form \(rx^{r-1}\) still holds.

This is a powerful generalization. Now, differentiating expressions with roots or fractional powers is just as straightforward as differentiating polynomials. The example shows how to apply this rule to a sum of terms with various rational exponents, including negative ones.

5.3.2 Logarithmic Differentiation

Logarithmic differentiation is a technique useful for functions with complex products, quotients, or variable exponents.

Theorem 5.3.4: Power Rule for General Real Exponents

Let \(r\) be a real constant. The function \(f(x) = x^r\) is defined for \(x > 0\), and \[ f ^ {\prime} (x) = r x ^ {r - 1}. \]

Verification using Logarithmic Differentiation (for \(x > 0\)):

Let \(y = x^r\).

Take the natural logarithm of both sides: \[ \ln y = \ln (x^r) \Rightarrow \ln y = r \ln x. \] Differentiate implicitly with respect to \(x\): \[ \frac {1}{y} \frac {d y}{d x} = r \frac {1}{x}. \] Solve for \(\frac{dy}{dx}\): \[ \frac {d y}{d x} = r \left(\frac {y}{x}\right) = r \left(\frac {x ^ {r}}{x}\right) = r x ^ {r - 1}. \]

Theorem 5.3.2: Power Rule for Rational Powers

Example 5.3.5: Find the derivative \(\frac{d}{dx}\left(x^{\pi} - \pi^{x}\right)\).

Solution:

Apply the sum/difference rule:

\(\frac{d}{dx}\left(x^{\pi} - \pi^{x}\right) = \frac{d}{dx}(x^{\pi}) - \frac{d}{dx}(\pi^{x})\). - For \(x^{\pi}\): This is \(x^r\) where \(r=\pi\) is a constant. Use power rule: \(\pi x^{\pi-1}\). - For \(\pi^{x}\): This is \(a^x\) where \(a=\pi\) is a constant. Use exponential rule: \(\pi^x \ln \pi\).

So, \(\frac{d}{dx}\left(x^{\pi} - \pi^{x}\right) = \pi x^{\pi-1} - \pi^x \ln \pi\).

Example 5.3.6: Find the derivative of \(y = x^x\) for \(x > 0\).

Solution: This has a variable in both the base and the exponent, so we use logarithmic differentiation.

Take the natural logarithm of both sides: \[ \ln y = \ln (x^x) \Rightarrow \ln y = x \ln x. \] Differentiate implicitly with respect to \(x\): \[ \frac {1}{y} \frac {d y}{d x} = (1) \ln x + x \left(\frac {1}{x}\right) \quad (\text{using Product Rule on } x \ln x). \] \[ \frac {1}{y} \frac {d y}{d x} = \ln x + 1. \] Solve for \(\frac{dy}{dx}\): \[ \frac {d y}{d x} = y (\ln x + 1) = x^x (\ln x + 1). \]

Logarithmic differentiation is a technique that simplifies the differentiation of functions that are products, quotients, or powers of other functions, especially when the exponent itself is a variable.

We start by verifying the power rule for any real constant \(r\) (not just integers or rationals) using logarithmic differentiation. By taking the natural logarithm of both sides of \(y=x^r\), we simplify the exponent using log properties, then differentiate implicitly. This re-derivation confirms the general power rule.

Example 5.3.5 is a tricky one because it combines two seemingly similar forms (\(x^{\text{constant}}\) and \(\text{constant}^x\)). It’s crucial to distinguish between them and apply the correct rule.

Example 5.3.6, \(y=x^x\), is where logarithmic differentiation truly shines. Neither the power rule (because the base is not constant) nor the exponential rule (because the exponent is not constant) directly applies. By taking logarithms, we transform the exponentiation into a product, which can then be differentiated using the product rule. This is a common and important application of this technique.

5.3.3 Derivative of Inverse Function

How do we find the derivative of an inverse function?

Theorem 5.3.7 (Derivative of Inverse Function):

If \(f\) is increasing (respectively decreasing) and continuous on an interval \((a, b)\) and \(f'(x_0) \neq 0\) for some \(x_0 \in (a, b)\), then \(f^{-1}\) is differentiable at the point \(y_0 = f(x_0)\), and \[ (f ^ {- 1}) ^ {\prime} (y _ {0}) = \frac {1}{f ^ {\prime} (f ^ {- 1} (y _ {0}))} = \frac {1}{f ^ {\prime} (x _ {0})}. \]

The condition \(f'(x_0) \neq 0\) ensures the tangent to \(f\) is not horizontal, so the tangent to \(f^{-1}\) is not vertical.

5.3.3 Derivative of Inverse Function

Example 5.3.8: Derivative of Inverse Cosine

Let \(f(x) = \cos x\), where \(x \in (0, \pi)\). Find \((f^{-1})'(0)\).

Solution:

Here, \(y_0 = 0\). We need to find \(x_0\) such that \(f(x_0) = y_0\), i.e., \(\cos x_0 = 0\).

For \(x_0 \in (0, \pi)\), \(x_0 = \frac{\pi}{2}\).

So, \(f^{-1}(0) = \frac{\pi}{2}\).

Now, find \(f'(x)\):

\(f'(x) = \frac{d}{dx}(\cos x) = -\sin x\).

Evaluate \(f'(x_0)\) at \(x_0 = \frac{\pi}{2}\):

\(f'(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1\).

Apply the inverse derivative formula: \[ (f ^ {- 1}) ^ {\prime} (0) = \frac {1}{f ^ {\prime} (f ^ {- 1} (0))} = \frac {1}{f ^ {\prime} (\pi / 2)} = \frac {1}{- \sin (\pi / 2)} = \frac {1}{-1} = -1. \]

The derivative of an inverse function gives us a way to find the rate of change of \(f^{-1}(y)\) with respect to \(y\), even if we don’t have an explicit formula for \(f^{-1}(y)\).

The formula is \((f^{-1})'(y_0) = 1 / f'(x_0)\), where \(y_0 = f(x_0)\). Essentially, the slope of the inverse function at a point is the reciprocal of the slope of the original function at the corresponding point. The condition that \(f'(x_0) \neq 0\) is important; if the original function has a horizontal tangent, its inverse would have a vertical tangent, and the derivative would be undefined.

In the example, we’re asked to find the derivative of the inverse of \(\cos x\) at \(y_0=0\). First, we find the \(x_0\) that corresponds to \(y_0=0\) under \(f(x)=\cos x\). Then, we find the derivative of the original function, \(f'(x)=-\sin x\). Finally, we plug these values into the formula. This gives us the derivative of the inverse cosine function at a specific point. This method is crucial for deriving the derivatives of all inverse trigonometric functions.

5.3.4 Inverse Trigonometric Functions

Using the inverse function theorem, we can derive derivatives for inverse trigonometric functions.

Theorem 5.3.9:

1. \(\frac{d}{dx}\left(\sin^{-1}x\right) = \frac{1}{\sqrt{1 - x^2}}\), for \(-1 < x < 1\).
2. \(\frac{d}{dx} (\cos^{-1}x) = \frac{-1}{\sqrt{1 - x^2}}\), for \(-1 < x < 1\).
3. \(\frac{d}{dx}\left(\tan^{-1}x\right) = \frac{1}{1 + x^2}\), for \(x\in \mathbb{R}\).
4. \(\frac{d}{dx} (\cot^{-1}x) = \frac{-1}{1 + x^2}\), for \(x\in \mathbb{R}\).
5. \(\frac{d}{dx} (\sec^{-1}x) = \frac{1}{|x|\sqrt{x^2 - 1}}\), for \(x < -1\) or \(x > 1\).
6. \(\frac{d}{dx} (\csc^{-1}x) = \frac{-1}{|x|\sqrt{x^2 - 1}}\), for \(x < -1\) or \(x > 1\).

Now we use the inverse function theorem (or implicit differentiation directly) to find the derivatives of the inverse trigonometric functions. These are also fundamental derivatives that should be memorized, especially arcsin, arccos, and arctan.

Let’s look at the proof for \(\arcsin x\). We set \(y = \arcsin x\), which implies \(\sin y = x\). Now we differentiate implicitly with respect to \(x\). The derivative of \(\sin y\) is \(\cos y \frac{dy}{dx}\), and the derivative of \(x\) is 1. Solving for \(\frac{dy}{dx}\) gives us \(1/\cos y\).

To express this in terms of \(x\), we use the Pythagorean identity \(\sin^2 y + \cos^2 y = 1\). Since \(\sin y = x\), we have \(\cos^2 y = 1 - x^2\), so \(\cos y = \pm \sqrt{1-x^2}\). Because the range of \(\arcsin x\) is \((-\pi/2, \pi/2)\), \(\cos y\) must be positive, so we choose the positive square root. Substituting this back gives the desired derivative.

The other inverse trigonometric derivatives can be derived similarly. It’s good practice to try to derive the derivative of \(\arctan x\) yourself using implicit differentiation.