MGU Mathematics Series — Calculus
Part II: Derivatives and Differentiation
Trigonometric · Exponential · Logarithmic · Inverse Trig
Chapter 5 of 14
MGU Mathematics Series  ·  Calculus  ·  Library Textbook Edition

Chapter 5:
Derivatives of Trigonometric,
Exponential, and Logarithmic
Functions

Extending differentiation to every major function family from precalculus — sine, cosine, \(e^x\), \(\ln x\), and inverse trigonometric functions — and connecting each derivative to meaning in models of growth, decay, oscillation, and scale

■ Math Level: Calculus I ■

■ 14 Sections  •  Part II of 5 ■

Follows Chapter 4: Differentiation Rules  ·  Prepares students for Chapter 6: Implicit Differentiation and Related Rates

Purpose and Overview

Chapter 4 gave you the core differentiation rules for algebraic functions: constants, powers, polynomials, products, quotients, and compositions. Those rules are powerful, but the world of calculus does not stop at polynomials and rational functions. The most important functions in science, engineering, economics, biology, computing, and physics include trigonometric functions that model oscillation and wave behavior, exponential functions that model growth and decay, and logarithmic functions that model scale, sensitivity, and information.

This chapter extends differentiation to those function families. You will learn to differentiate \(\sin x\), \(\cos x\), \(\tan x\), \(e^x\), \(a^x\), \(\ln x\), and their compositions with other functions. You will also meet logarithmic differentiation as a technique, study the derivative of an inverse function, and differentiate the inverse trigonometric functions.

Every formula in this chapter connects to meaning. An exponential derivative describes how fast a growing population or a radioactive material is changing at this moment. A trigonometric derivative describes the instantaneous rate of oscillation at a point on a wave. A logarithmic derivative describes sensitivity to scale. By the end of this chapter, you will be able to differentiate any function built from the families you know in precalculus, and you will understand what those derivatives mean.

Opening Question

A population of bacteria is growing according to the model \(P(t) = 500e^{0.3t}\), where \(t\) is measured in hours and \(P\) is the number of bacteria. At what rate is the population growing at \(t = 4\) hours? What does that rate mean in context, and how does it compare to the rate at \(t = 0\)?

This kind of question is not answerable with the power rule. The exponential function \(e^{0.3t}\) is not a polynomial. To answer it, you need the derivative of an exponential function — and that derivative turns out to be one of the most remarkable results in all of calculus.

14Sections
10Objectives
13Key Terms
20+Worked Examples

Why This Chapter Matters

Every function family you studied in precalculus reappears in calculus. Trigonometric functions describe oscillation, sound waves, light waves, alternating current, and circular motion. Exponential functions describe compound interest, radioactive decay, population growth, heat transfer, and charging and discharging capacitors. Logarithmic functions describe the pH scale, decibel levels, earthquake magnitude, and information entropy.

Without the derivatives of these functions, calculus would apply only to polynomials and rational expressions — a small and unrepresentative slice of the mathematical world. This chapter completes the core differentiation toolkit.

Learning Objectives

  1. 1Recall the graphs, domains, ranges, and key properties of trigonometric, exponential, and logarithmic functions.
  2. 2State and apply the derivatives of \(\sin x\), \(\cos x\), \(\tan x\), \(\sec x\), \(\csc x\), and \(\cot x\).
  3. 3Apply the chain rule to differentiate compositions involving trigonometric functions.
  4. 4State and apply the derivatives of \(e^x\), \(a^x\), \(\ln x\), and \(\log_a x\).
  5. 5Use logarithmic differentiation to differentiate functions that would be difficult to handle otherwise.
  6. 6Apply the inverse function derivative theorem to find derivatives of inverse functions.
  7. 7State and apply the derivatives of the inverse trigonometric functions \(\arcsin x\), \(\arccos x\), and \(\arctan x\).
  8. 8Combine all differentiation rules to differentiate complex mixed functions.
  9. 9Identify and avoid common mistakes in transcendental differentiation.
  10. 10Interpret derivatives of transcendental functions in modeling contexts.

Sections at a Glance

Sections 5.1 – 5.4
Trigonometric Functions and Their Derivatives
Function Families Review · Sine and Cosine · Tan, Sec, Csc, Cot · Chain Rule with Trig

Reviewing Special Function Families

Before differentiating transcendental functions, it is worth pausing to recall what these function families look like, how they behave, and what properties make them useful for modeling. Understanding the graph and behavior of a function helps you recognize whether a derivative you compute makes sense.

Trigonometric Functions

The sine and cosine functions are periodic with period \(2\pi\). Their graphs oscillate smoothly between \(-1\) and \(1\). The function \(\sin x\) equals zero at every integer multiple of \(\pi\), reaches its maximum of \(1\) at \(x = \pi/2 + 2k\pi\), and reaches its minimum of \(-1\) at \(x = -\pi/2 + 2k\pi\).

The function \(\tan x = \frac{\sin x}{\cos x}\) has vertical asymptotes wherever \(\cos x = 0\), which occurs at odd multiples of \(\pi/2\). Its period is \(\pi\) rather than \(2\pi\), and it is strictly increasing on each interval between its asymptotes.

Critical Convention — Radian Measure

All derivative formulas for trigonometric functions in calculus assume that angles are measured in radians. If you use degrees, the derivative formulas change and carry an extra factor of \(\pi/180\). In all calculus work, use radian measure unless explicitly told otherwise.

Exponential Functions

The most important base in calculus is Euler's number \(e \approx 2.71828\). The function \(f(x) = e^x\) is called the natural exponential function. It is the unique exponential function whose rate of growth at any point equals its value at that point — a remarkable property you will understand precisely in Section 5.5.

Logarithmic Functions

The natural logarithm \(\ln x\) is the inverse of \(e^x\). Its domain is \(x > 0\), and it passes through \((1, 0)\) because \(e^0 = 1\). Logarithms convert products to sums, quotients to differences, and powers to multiples — properties that become essential in Section 5.9.

The general logarithm \(\log_a x\) uses base \(a > 0\), \(a \neq 1\), and converts to the natural logarithm via: \(\log_a x = \dfrac{\ln x}{\ln a}\).

Graph Instruction

Draw graphs of \(y = \sin x\) and \(y = \cos x\) on \([-2\pi, 2\pi]\), marking zeros, maxima, and minima. On separate axes, draw \(y = e^x\) and \(y = \ln x\) as reflections of each other across \(y = x\). These graphs will help you verify derivative results throughout this chapter.

Derivatives of \(\sin x\) and \(\cos x\)

Using the definition of the derivative along with two fundamental limits from Chapter 2,

$$\lim_{h \to 0} \frac{\sin h}{h} = 1 \qquad \text{and} \qquad \lim_{h \to 0} \frac{\cos h - 1}{h} = 0$$

it can be shown that:

Derivatives of Sine and Cosine

\[\frac{d}{dx}[\sin x] = \cos x\]

\[\frac{d}{dx}[\cos x] = -\sin x\]

The derivative of sine has a beautiful geometric interpretation: the slope of the sine curve at any \(x\) equals the height of the cosine curve at the same \(x\). Where the sine curve is rising steeply (near \(x = 0\)), the cosine is near \(1\). Where the sine curve is at its peak (at \(x = \pi/2\)), the cosine is \(0\) — meaning the tangent line is horizontal, which is exactly what the graph shows.

Notice the negative sign in the cosine derivative. The cosine function starts at its maximum of \(1\) at \(x = 0\) and is immediately decreasing, so a negative derivative is correct.

Example 5.2.1
Differentiating a function involving sine
  • ProblemFind the derivative of \(f(x) = 3\sin x - x^2\).
  • ThinkThis is a difference of two terms. Apply the constant multiple rule to the first term and the power rule to the second.
  • Calculate\(f'(x) = 3\cos x - 2x\)
  • CheckAt \(x = 0\): \(f'(0) = 3\cos(0) - 2(0) = 3(1) - 0 = 3\). The original function starts rising steeply at \(x = 0\), so a slope of 3 is reasonable.
  • InterpretThe function is changing at a rate of \(3\cos x - 2x\) at any value \(x\).
Answer: \(f'(x) = 3\cos x - 2x\)
Example 5.2.2
Interpreting a cosine derivative in context
  • ProblemA wave height in centimeters is modeled by \(h(t) = 8\cos t\), where \(t\) is time in seconds. Find the rate of change at \(t = \pi/4\) seconds.
  • Represent\(h'(t) = 8 \cdot (-\sin t) = -8\sin t\)
  • Calculate\(h'(\pi/4) = -8\sin(\pi/4) = -8 \cdot \dfrac{\sqrt{2}}{2} = -4\sqrt{2} \approx -5.66\) cm/sec
  • CheckAt \(t = \pi/4\), the cosine just passed its maximum at \(t = 0\) and is heading downward. A negative rate of change is correct.
  • InterpretThe wave height is decreasing at approximately 5.66 centimeters per second at \(t = \pi/4\) seconds.
Answer: \(h'(\pi/4) = -4\sqrt{2} \approx -5.66\) cm/sec (decreasing)

Derivatives of \(\tan x\), \(\sec x\), \(\csc x\), and \(\cot x\)

The remaining trigonometric functions are all defined in terms of sine and cosine, so their derivatives follow from the quotient rule. The results are worth memorizing. Below is the key derivation for tangent, which illustrates the method for all four.

Derivation 5.3.1
Derivative of \(\tan x\) via the quotient rule
  • SetupWrite \(\tan x = \dfrac{\sin x}{\cos x}\) and apply the quotient rule.
  • Calculate \[\frac{d}{dx}[\tan x] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}\]
  • SimplifyUsing the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\): \(\dfrac{1}{\cos^2 x} = \sec^2 x\)
Result: \(\dfrac{d}{dx}[\tan x] = \sec^2 x\)
The Six Trigonometric Derivatives

\[\frac{d}{dx}[\sin x] = \cos x\]

\[\frac{d}{dx}[\cos x] = -\sin x\]

\[\frac{d}{dx}[\tan x] = \sec^2 x\]

\[\frac{d}{dx}[\cot x] = -\csc^2 x\]

\[\frac{d}{dx}[\sec x] = \sec x \tan x\]

\[\frac{d}{dx}[\csc x] = -\csc x \cot x\]

Memory Pattern — The Co-Function Sign Rule

The derivatives of the three "co-functions" — cosine, cotangent, cosecant — each carry a negative sign that the corresponding non-co function does not. This pattern: (cos/sin), (cot/tan), (csc/sec) all gain a negative sign when differentiated. No other pattern is needed to recall all six formulas.

Example 5.3.1
Differentiating a combination of trig functions
  • ProblemFind \(f'(x)\) if \(f(x) = 5\tan x + 2\sec x\).
  • Calculate\(f'(x) = 5\sec^2 x + 2\sec x \tan x\)
  • InterpretThe rate of change of \(f\) at any angle \(x\) is \(5\sec^2 x + 2\sec x \tan x\). Both terms are positive for small positive \(x\), so \(f\) is increasing there.
Answer: \(f'(x) = 5\sec^2 x + 2\sec x \tan x\)

Trigonometric Derivatives with the Chain Rule

The chain rule extends every trigonometric derivative to compositions. When a trigonometric function is applied to an expression other than \(x\) alone, you must multiply by the derivative of the inside expression. The general pattern is:

Chain Rule — Trigonometric Compositions

\[\frac{d}{dx}[\sin(g(x))] = \cos(g(x)) \cdot g'(x)\]

\[\frac{d}{dx}[\cos(g(x))] = -\sin(g(x)) \cdot g'(x)\]

\[\frac{d}{dx}[\tan(g(x))] = \sec^2(g(x)) \cdot g'(x)\]

And similarly for the remaining three functions.

Example 5.4.1
Chain rule with sine
  • ProblemFind \(\dfrac{d}{dx}[\sin(3x^2)]\).
  • ThinkThe outer function is sine; the inner function is \(g(x) = 3x^2\), so \(g'(x) = 6x\).
  • Calculate\(\dfrac{d}{dx}[\sin(3x^2)] = \cos(3x^2) \cdot 6x = 6x\cos(3x^2)\)
  • CheckThe chain rule contributes the factor \(6x\). Without it, the answer would be incomplete. The outer derivative is cosine — as expected.
Answer: \(6x\cos(3x^2)\)
Example 5.4.2
Vibrating string — physics application
  • ProblemA vibrating string has displacement \(y(t) = A\cos(\omega t + \phi)\), where \(A\) is amplitude, \(\omega\) is angular frequency in rad/sec, and \(\phi\) is phase shift. Find the velocity \(y'(t)\).
  • ThinkThe outer function is cosine. The inner function is \(\omega t + \phi\), with derivative \(\omega\).
  • Calculate\(y'(t) = A \cdot (-\sin(\omega t + \phi)) \cdot \omega = -A\omega\sin(\omega t + \phi)\)
  • InterpretThe maximum speed is \(A\omega\) (when \(|\sin| = 1\)), which increases with both amplitude and frequency. Higher-frequency vibrations are faster — a result from physics that calculus produces naturally.
Answer: \(y'(t) = -A\omega\sin(\omega t + \phi)\)
Example 5.4.3
Nested composition — \(\sin^3 x\)
  • ProblemDifferentiate \(f(x) = \sin^3 x\). (Note: \(\sin^3 x\) means \((\sin x)^3\).)
  • ThinkThe outer function is a cube: \(u^3\) where \(u = \sin x\).
  • Calculate\(f'(x) = 3(\sin x)^2 \cdot \cos x = 3\sin^2 x\cos x\)
  • CheckThree factors are present: the power rule contributes \(3\), the exponent drops by 1 to give \(\sin^2 x\), and the chain rule contributes \(\cos x\). All three are required.
Answer: \(3\sin^2 x\cos x\)
Sections 5.5 – 5.9
Exponential and Logarithmic Derivatives
\(e^x\) · \(a^x\) · \(\ln x\) · \(\log_a x\) · Logarithmic Differentiation

Derivatives of \(e^x\)

The Self-Referential Derivative

\[\frac{d}{dx}[e^x] = e^x\]

The natural exponential function is its own derivative. This means the rate at which \(e^x\) is changing at any point equals its value at that point.

This property can be understood from the limit definition using \(\lim_{h \to 0} \frac{e^h - 1}{h} = 1\). The number \(e\) is specifically chosen so that this limit equals exactly \(1\). For any other base \(a\), the analogous limit equals \(\ln a\), not \(1\) — which is why \(e\) is the natural base for calculus.

Graph Instruction

Draw \(y = e^x\) and mark the tangent line at \(x = 0\) (slope 1), at \(x = 1\) (slope \(e \approx 2.718\)), and at \(x = -1\) (slope \(1/e \approx 0.368\)). In each case, the slope equals the height of the function — a visually striking illustration of the self-referential property.

Chain Rule with \(e^x\)

When \(e\) is raised to an expression involving \(x\), multiply by the derivative of the exponent:

\[\frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x)\]

Example 5.5.1
Chain rule with a linear exponent
  • ProblemFind the derivative of \(f(x) = e^{3x}\).
  • ThinkInside: \(g(x) = 3x\), so \(g'(x) = 3\).
  • Calculate\(f'(x) = e^{3x} \cdot 3 = 3e^{3x}\)
  • InterpretThe function \(e^{3x}\) grows three times as fast as \(e^x\) at the same function value.
Answer: \(3e^{3x}\)
Example 5.5.2
Quadratic exponent — minimum check
  • ProblemDifferentiate \(g(x) = e^{x^2 - 4}\).
  • Calculate\(g'(x) = e^{x^2 - 4} \cdot 2x = 2xe^{x^2-4}\)
  • CheckAt \(x = 0\), \(g'(0) = 0\). The function has a minimum at \(x = 0\) (since \(x^2 - 4\) is minimized there), and a minimum implies a horizontal tangent — slope zero. ✓
Answer: \(2xe^{x^2-4}\)
Example 5.5.3
Bacterial growth rate — opening question answered
  • ProblemThe bacterial population \(P(t) = 500e^{0.3t}\) (bacteria, \(t\) in hours). Find the growth rate at \(t = 4\) hours.
  • Calculate \[P'(t) = 500 \cdot e^{0.3t} \cdot 0.3 = 150e^{0.3t}\] \[P'(4) = 150e^{1.2} \approx 150 \times 3.320 \approx 498 \text{ bacteria/hour}\]
  • CompareAt \(t = 0\): \(P'(0) = 150\) bacteria/hour. At \(t = 4\): \(\approx 498\) bacteria/hour. The rate has more than tripled as the population grew.
  • InterpretAt 4 hours, the population is growing at about 498 bacteria per hour — roughly equal to the entire initial population gained in a single hour. This illustrates the exponential acceleration of unconstrained growth.
Answer: \(P'(4) \approx 498\) bacteria per hour

Derivatives of \(a^x\)

For exponential functions with a base other than \(e\), a constant factor involving the natural logarithm appears. Since \(a = e^{\ln a}\), we can rewrite \(a^x = e^{x \ln a}\) and differentiate using the chain rule:

Derivative of \(a^x\) for Any Base \(a > 0\), \(a \neq 1\)

\[\frac{d}{dx}[a^x] = a^x \ln a\]

When \(a = e\), \(\ln e = 1\) and the formula reduces to \(\frac{d}{dx}[e^x] = e^x\). The factor \(\ln a\) is always present for non-\(e\) bases. For \(a = 2\): \(\ln 2 \approx 0.693\). For \(a = 10\): \(\ln 10 \approx 2.303\).

Example 5.6.1
Computational cost — doubling algorithm
  • ProblemA computational model costs \(C(n) = 3 \cdot 2^n\) operations, where \(n\) is the number of iterations. Find \(C'(n)\).
  • Calculate\(C'(n) = 3 \cdot 2^n \ln 2 \approx 3 \cdot 2^n \cdot 0.693\)
  • InterpretThe marginal cost per additional iteration is roughly 69.3% of the current cost. This exponential rate of growth is why certain algorithms become impractical as \(n\) increases.
Answer: \(C'(n) = 3 \cdot 2^n \ln 2\)
Example 5.6.2
Chain rule with base \(a\)
  • ProblemDifferentiate \(f(x) = 5^{x^2}\).
  • ThinkUse \(\frac{d}{dx}[a^u] = a^u \ln a \cdot u'\) with inside \(u = x^2\), \(u' = 2x\).
  • Calculate\(f'(x) = 5^{x^2} \cdot \ln 5 \cdot 2x = 2x\ln 5 \cdot 5^{x^2}\)
Answer: \(2x\ln 5 \cdot 5^{x^2}\)
Sections 5.7 – 5.9
Logarithmic Derivatives and Logarithmic Differentiation
\(\ln x\) · \(\log_a x\) · Logarithmic Differentiation Technique

Derivatives of \(\ln x\)

Derivative of the Natural Logarithm

\[\frac{d}{dx}[\ln x] = \frac{1}{x}, \quad x > 0\]

This can be derived from \(y = \ln x \Rightarrow e^y = x\). Differentiating both sides: \(e^y \cdot \frac{dy}{dx} = 1\), so \(\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}\).

Notice that the derivative \(\frac{1}{x}\) is a rational function even though \(\ln x\) is transcendental — one of the most useful derivatives in calculus. When the chain rule is applied:

Chain Rule Pattern for \(\ln(g(x))\)

\[\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}\]

The derivative of the inside goes to the numerator; the inside itself goes to the denominator. This pattern appears so frequently it is worth memorizing as a unit.

Example 5.7.1
Chain rule with natural log
  • ProblemDifferentiate \(f(x) = \ln(x^3 + 1)\).
  • ThinkInside: \(g(x) = x^3 + 1\), so \(g'(x) = 3x^2\).
  • Calculate\(f'(x) = \dfrac{3x^2}{x^3 + 1}\)
Answer: \(\dfrac{3x^2}{x^3+1}\)
Example 5.7.2
Logarithm and product rule combined
  • ProblemDifferentiate \(h(x) = x^2 \ln x\).
  • ThinkThis is a product of \(x^2\) and \(\ln x\). Apply the product rule.
  • Calculate \[h'(x) = 2x \ln x + x^2 \cdot \frac{1}{x} = 2x\ln x + x = x(2\ln x + 1)\]
  • CheckAt \(x = 1\): \(h'(1) = 1(0 + 1) = 1\). The function passes through zero at \(x = 1\) and is increasing, so a positive slope of 1 is reasonable. ✓
Answer: \(x(2\ln x + 1)\)

Derivatives of \(\log_a x\)

Derivative of \(\log_a x\)

Using the change-of-base formula \(\log_a x = \dfrac{\ln x}{\ln a}\):

\[\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}\]

With the chain rule: \(\dfrac{d}{dx}[\log_a(g(x))] = \dfrac{g'(x)}{g(x)\ln a}\)

Example 5.8.1
Differentiating a common logarithm
  • ProblemDifferentiate \(f(x) = \log_{10}(x^2 + 5)\).
  • Calculate\(f'(x) = \dfrac{2x}{(x^2+5)\ln 10}\)
  • InterpretThe factor \(\frac{1}{\ln 10} \approx 0.434\) makes this 0.434 times the natural log version. This illustrates why natural logs are preferred in calculus: their derivatives carry no extra constant.
Answer: \(\dfrac{2x}{(x^2+5)\ln 10}\)

Logarithmic Differentiation

Logarithmic differentiation is a technique that uses properties of logarithms to simplify differentiation. It is most useful when the function is a product or quotient of many factors, or when the function has a variable in both the base and the exponent.

The Method — Four Steps

Step 1. Set \(y = f(x)\) and take \(\ln\) of both sides: \(\ln y = \ln(f(x))\).

Step 2. Expand the right side using log properties (products → sums, quotients → differences, powers → multiples).

Step 3. Differentiate both sides. The left side gives \(\dfrac{1}{y} \cdot y'\) by the chain rule.

Step 4. Solve for \(y'\) and substitute back \(y = f(x)\).

Example 5.9.1
Differentiating \(x^x\) — the classic logarithmic differentiation case
  • ProblemDifferentiate \(f(x) = x^x\) for \(x > 0\).
  • ThinkThe power rule requires a constant exponent. The exponential rule requires a constant base. Neither applies. Use logarithmic differentiation.
  • SetupLet \(y = x^x\). Take \(\ln\): \(\ln y = x \ln x\).
  • Differentiate \[\frac{1}{y} \cdot y' = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\]
  • Solve \[y' = y(\ln x + 1) = x^x(\ln x + 1)\]
  • CheckAt \(x = 1\): \(y'(1) = 1^1(0 + 1) = 1\). The function passes through \((1,1)\) and the slope is 1 there — consistent with the graph increasing at that point.
Answer: \(\dfrac{d}{dx}[x^x] = x^x(\ln x + 1)\)
Example 5.9.2
Logarithmic differentiation for a complex quotient
  • ProblemDifferentiate \(f(x) = \dfrac{\sqrt{x+1}\,(x^2+3)}{(x-2)^4}\).
  • SetupTake \(\ln|f(x)| = \frac{1}{2}\ln(x+1) + \ln(x^2+3) - 4\ln|x-2|\)
  • Differentiate \[\frac{f'(x)}{f(x)} = \frac{1}{2(x+1)} + \frac{2x}{x^2+3} - \frac{4}{x-2}\]
  • Solve \[f'(x) = \frac{\sqrt{x+1}(x^2+3)}{(x-2)^4} \left(\frac{1}{2(x+1)} + \frac{2x}{x^2+3} - \frac{4}{x-2}\right)\]
  • InterpretRather than applying the quotient rule to a three-factor numerator, logarithmic differentiation converts the problem into a sum of simple fractions.
Answer: as shown above — logarithm converts multiplication into addition, making complex products manageable.
Sections 5.10 – 5.11
Inverse Functions and Inverse Trigonometric Derivatives
Inverse Function Theorem · \(\arcsin x\) · \(\arccos x\) · \(\arctan x\)

Derivatives of Inverse Functions

Inverse Function Derivative Theorem

If \(f\) is differentiable with an inverse \(f^{-1}\), and \(f'(f^{-1}(x)) \neq 0\), then:

\[\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}\]

Geometric meaning: When the graph of \(f\) is reflected across \(y = x\) to produce \(f^{-1}\), a slope of \(m\) on \(f\) becomes a slope of \(1/m\) on \(f^{-1}\). A steep function has a flat inverse, and vice versa.

This theorem is derived by differentiating the identity \(f(f^{-1}(x)) = x\) using the chain rule:

$$f'(f^{-1}(x)) \cdot \frac{d}{dx}[f^{-1}(x)] = 1 \implies \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}$$
Example 5.10.1
Applying the inverse function theorem
  • ProblemSuppose \(f(x) = x^3 + x\). Find the derivative of \(f^{-1}\) at \(x = 2\).
  • ThinkNote that \(f(1) = 1 + 1 = 2\), so \(f^{-1}(2) = 1\). Apply the theorem.
  • Calculate \[f'(x) = 3x^2 + 1, \quad f'(1) = 4\] \[\frac{d}{dx}[f^{-1}(x)]\bigg|_{x=2} = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(1)} = \frac{1}{4}\]
  • InterpretThe inverse function has slope \(1/4\) at \(x = 2\), while the original function has slope \(4\) at \(x = 1\). Slopes become reciprocals when passing between a function and its inverse.
Answer: \(\frac{1}{4}\)

Derivatives of Inverse Trigonometric Functions

The inverse trigonometric functions appear in geometry, physics, and engineering whenever an angle must be recovered from a ratio. Their derivatives follow from the inverse function theorem combined with the Pythagorean identity.

Derivation of \(\dfrac{d}{dx}[\arcsin x]\)

Let \(y = \arcsin x\), so \(\sin y = x\). Differentiate both sides:

$$\cos y \cdot y' = 1 \implies y' = \frac{1}{\cos y}$$

Since \(\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1-x^2}\) (using \(\cos y \geq 0\) on the principal range):

$$\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$$

Derivation of \(\dfrac{d}{dx}[\arctan x]\)

Let \(y = \arctan x\), so \(\tan y = x\). Differentiate:

$$\sec^2 y \cdot y' = 1 \implies y' = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1+x^2}$$
The Three Inverse Trigonometric Derivatives

\[\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}, \quad -1 < x < 1\]

\[\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}, \quad -1 < x < 1\]

\[\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}, \quad \text{all } x\]

Note: \(\arccos x\) is decreasing, which accounts for the negative sign. The arctangent derivative is valid for all real \(x\) and appears frequently in integral calculus.

Example 5.11.1
Differentiating arctangent with chain rule
  • ProblemDifferentiate \(f(x) = \arctan(2x)\).
  • ThinkOutside: arctangent. Inside: \(g(x) = 2x\), so \(g'(x) = 2\).
  • Calculate\(f'(x) = \dfrac{2}{1+(2x)^2} = \dfrac{2}{1+4x^2}\)
  • InterpretAs \(x \to \infty\), \(f'(x) \to 0\), consistent with \(\arctan(2x) \to \pi/2\) and leveling off.
Answer: \(\dfrac{2}{1+4x^2}\)
Example 5.11.2
Differentiating arcsine with chain rule
  • ProblemDifferentiate \(g(x) = \arcsin(x^2)\).
  • Calculate\(g'(x) = \dfrac{2x}{\sqrt{1-(x^2)^2}} = \dfrac{2x}{\sqrt{1-x^4}}\)
  • NoteDomain requires \(x^2 < 1\), so \(-1 < x < 1\).
Answer: \(\dfrac{2x}{\sqrt{1-x^4}}\)
Example 5.11.3
Projectile angle — physics application
  • ProblemA projectile has constant horizontal velocity \(v_x = 20\) m/s and vertical velocity \(v_y = 40 - 10t\) m/s. The angle satisfies \(\theta = \arctan\!\left(\frac{v_y}{v_x}\right) = \arctan(2 - 0.5t)\). Find \(\frac{d\theta}{dt}\) at \(t = 2\) seconds.
  • Calculate \[\frac{d\theta}{dt} = \frac{-0.5}{1+(2-0.5t)^2}\] At \(t = 2\): \(2 - 0.5(2) = 1\), so \(\dfrac{d\theta}{dt}\bigg|_{t=2} = \dfrac{-0.5}{1+1^2} = -0.25\) rad/sec
  • InterpretAt \(t = 2\) seconds, the angle of the velocity vector is decreasing at 0.25 radians per second — the projectile is tilting downward toward its peak trajectory.
Answer: \(-0.25\) radians per second (angle decreasing)
Sections 5.12 – 5.14
Mixed Derivatives, Common Mistakes, and the Road Ahead
Combined Functions · Eight Common Errors · Preparing for Implicit Differentiation

Mixed Transcendental Derivatives

Real-world functions often combine exponential, logarithmic, and trigonometric expressions. The strategy is always the same: identify the outermost structure first, apply the appropriate rule, and then work inward, applying the chain rule at each layer.

Example 5.12.1
Exponential times sine — damped oscillation
  • ProblemDifferentiate \(f(x) = e^x \sin x\).
  • ThinkThis is a product of \(e^x\) and \(\sin x\). Apply the product rule.
  • Calculate \[f'(x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)\]
  • InterpretThis combination appears in damped oscillation models. In physics, the analogous form \(e^{-\gamma t}\cos(\omega t)\) describes a decaying oscillator. The exponential factor amplifies (or here, grows) the oscillation over time.
Answer: \(e^x(\sin x + \cos x)\)
Example 5.12.2
Logarithm of a trigonometric function
  • ProblemDifferentiate \(g(x) = \ln(\sin x)\), valid where \(\sin x > 0\).
  • Calculate\(g'(x) = \dfrac{\cos x}{\sin x} = \cot x\)
  • InterpretThe derivative of \(\ln(\sin x)\) is \(\cot x\). This result appears in integral calculus when evaluating \(\int \cot x\, dx\).
Answer: \(\cot x\)
Example 5.12.3
Three layers of composition
  • ProblemDifferentiate \(h(x) = e^{\sin(x^2)}\).
  • ThinkOutermost: \(e^u\). Middle: \(\sin(v)\). Innermost: \(v = x^2\). Apply chain rule from outside in.
  • Calculate \[h'(x) = e^{\sin(x^2)} \cdot \cos(x^2) \cdot 2x = 2x\cos(x^2)\,e^{\sin(x^2)}\]
  • CheckEach layer contributes its factor: \(2x\) from innermost, \(\cos(x^2)\) from middle, \(e^{\sin(x^2)}\) from outer. All three are present.
Answer: \(2x\cos(x^2)\,e^{\sin(x^2)}\)
Example 5.12.4
Arctangent with exponential inside
  • ProblemDifferentiate \(f(x) = \arctan(e^{2x})\).
  • ThinkOuter: \(\arctan(u)\). Inner: \(u = e^{2x}\), so \(u' = 2e^{2x}\).
  • Calculate\(f'(x) = \dfrac{2e^{2x}}{1 + (e^{2x})^2} = \dfrac{2e^{2x}}{1+e^{4x}}\)
  • InterpretAs \(x \to \infty\), \(e^{4x}\) dominates the denominator and \(f'(x) \to 0\), consistent with \(\arctan(e^{2x}) \to \pi/2\) leveling off.
Answer: \(\dfrac{2e^{2x}}{1+e^{4x}}\)

Common Transcendental Derivative Mistakes

Mistake 1 — Applying the power rule to exponential functions
\(\frac{d}{dx}[e^x] = xe^{x-1}\)

The power rule requires a constant exponent. When the variable is the exponent, the function is exponential, not a power. Correct: \(\frac{d}{dx}[e^x] = e^x\).

Mistake 2 — Forgetting the chain rule with trigonometric functions
\(\frac{d}{dx}[\sin(3x^2)] = \cos(3x^2)\)

The outer derivative alone is never the complete answer when there is a nontrivial inside function. The chain rule requires multiplying by \(g'(x) = 6x\). Correct: \(6x\cos(3x^2)\).

Mistake 3 — Confusing \(\frac{d}{dx}[\ln x]\) with \(\frac{d}{dx}[e^x]\)
\(\frac{d}{dx}[\ln x] = e^x\)

These are inverse functions, not interchangeable formulas. Correct: \(\frac{d}{dx}[\ln x] = \frac{1}{x}\) and \(\frac{d}{dx}[e^x] = e^x\). They are different functions with different derivatives.

Mistake 4 — Forgetting the factor \(\ln a\) for base \(a\)
\(\frac{d}{dx}[3^x] = 3^x\)

The factor \(\ln a\) is always present for non-\(e\) bases. Correct: \(\frac{d}{dx}[3^x] = 3^x \ln 3\). When \(a = e\), \(\ln e = 1\) and the factor disappears invisibly — which is one reason \(e\) is the natural base.

Mistake 5 — Applying a standard rule to \(x^x\)
\(\frac{d}{dx}[x^x] = x \cdot x^{x-1} = x^x\)   or   \(\frac{d}{dx}[x^x] = x^x \ln x\)

Neither the power rule nor the exponential rule applies when both base and exponent are variable. Use logarithmic differentiation. Correct: \(\frac{d}{dx}[x^x] = x^x(\ln x + 1)\).

Mistake 6 — Error in the denominator for \(\ln(g(x))\)
\(\frac{d}{dx}[\ln(x^2+1)] = \frac{1}{2x}\)

The derivative of the inside goes to the numerator, not into the denominator as a replacement. Correct: \(\frac{d}{dx}[\ln(x^2+1)] = \frac{2x}{x^2+1}\).

Mistake 7 — Sign errors in co-function derivatives
\(\frac{d}{dx}[\cos x] = \sin x\)

The derivatives of cosine, cotangent, and cosecant each carry a negative sign. Correct: \(\frac{d}{dx}[\cos x] = -\sin x\). Forgetting this sign error causes incorrect answers throughout integration as well.

Mistake 8 — Using inverse trig formulas outside their domains
Applying \(\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}\) at \(x = 1\)

At \(x = \pm 1\) the denominator is zero and the formula is undefined. The domain restriction \(-1 < x < 1\) is not optional for arcsine and arccosine. The arctangent formula \(\frac{1}{1+x^2}\) is valid for all real \(x\).

Preparing for Implicit and Related-Rate Problems

Chapter 6 extends differentiation in two directions. First, it handles relationships where \(y\) cannot be solved explicitly as a function of \(x\). Second, it uses derivatives to connect the rates of change of quantities that are related through a geometric or physical equation.

Both techniques rely on the chain rule, including the versions for transcendental functions covered in this chapter.

Preview — Implicit Differentiation

The equation \(x^2 + y^2 = 25\) defines a circle but does not define \(y\) as a single function of \(x\). To find the slope at any point on this circle, you differentiate both sides with respect to \(x\) and treat \(y\) as a function of \(x\), applying the chain rule whenever \(y\) appears. This is implicit differentiation. If \(\ln(xy) = x^2\) appeared in such an equation, the logarithmic derivative rules from this chapter would be needed.

Preview — Related Rates

A ladder 10 meters long leans against a wall. As the base slides away at 0.5 m/sec, how fast is the top descending? The height and base satisfy \(x^2 + y^2 = 100\). Differentiating with respect to time \(t\) gives a relationship between \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). When trigonometric expressions appear — for example, if the angle the ladder makes with the ground is also tracked — the derivatives of \(\sin\), \(\cos\), and \(\arctan\) from this chapter enter directly.

Section 5.14Readiness Problems
  1. Differentiate \(f(x) = \ln(x^2 + y^2)\) with respect to \(x\), treating \(y\) as a constant. Now consider what would change if \(y\) were a function of \(x\).
  2. The equation \(e^y = x^2 + 1\) implicitly defines \(y\) as a function of \(x\). Differentiate both sides with respect to \(x\) (chain rule on the left) and solve for \(\frac{dy}{dx}\).
  3. An angle \(\theta = \arctan(x/y)\) where both \(x\) and \(y\) depend on time \(t\). Write an expression for \(\frac{d\theta}{dt}\) in terms of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). (You will complete this reasoning fully in Chapter 6.)

Chapter Summary

Key Terms

transcendental function
A function that cannot be expressed as a finite combination of polynomial operations. Includes exponential, logarithmic, and trigonometric functions.
Euler's number (\(e\))
The irrational number \(\approx 2.71828\) that is the unique base for which \(\frac{d}{dx}[a^x] = a^x\) with no extra constant factor.
natural exponential function
The function \(f(x) = e^x\), unique in being its own derivative: \(\frac{d}{dx}[e^x] = e^x\).
natural logarithm
The function \(\ln x = \log_e x\), inverse of \(e^x\), with derivative \(\frac{1}{x}\) for \(x > 0\).
radian measure
The unit of angle measurement required for all calculus trigonometry formulas. Degrees must be converted before applying derivative formulas.
logarithmic differentiation
A technique for differentiating products, quotients with many factors, or variable-exponent functions by taking the natural log of both sides before differentiating.
inverse function derivative theorem
The theorem that \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\). Slopes between a function and its inverse are reciprocals.
arcsine
The inverse function of \(\sin x\) on \([-\pi/2, \pi/2]\); also written \(\sin^{-1} x\). Derivative: \(\frac{1}{\sqrt{1-x^2}}\) for \(-1 < x < 1\).
arctangent
The inverse function of \(\tan x\) on \((-\pi/2, \pi/2)\); also written \(\tan^{-1} x\). Derivative: \(\frac{1}{1+x^2}\) for all \(x\).
chain rule (transcendental)
The rule for differentiating compositions involving transcendental functions: always multiply the outer derivative by the derivative of the inside expression.
co-function sign pattern
The derivatives of cosine, cotangent, and cosecant each carry a negative sign that their counterparts (sine, tangent, secant) do not.
base \(a\) exponential derivative
The formula \(\frac{d}{dx}[a^x] = a^x \ln a\), derived by rewriting \(a^x = e^{x \ln a}\) and using the chain rule.
log derivative pattern
The chain rule applied to \(\ln(g(x))\) yields \(g'(x)/g(x)\): the derivative of the inside divided by the inside itself.

Trigonometric Derivative Practice

Concept CheckUnderstanding Trig Derivatives
  1. State the derivatives of all six trigonometric functions from memory.
  2. Explain why all trigonometric derivative formulas in calculus require radian measure rather than degrees.
  3. What does the graph of \(y = \cos x\) tell you about the sign of \(\frac{d}{dx}[\sin x]\) at \(x = \pi\)? Explain without computing.
  4. Without computing, what is the slope of \(y = \sin x\) at its maximum? Why?
  5. Explain the co-function sign pattern that connects the derivatives of cosine, cotangent, and cosecant to their counterparts.
Skill PracticeTrig Differentiation

Find each derivative.

  1. \(f(x) = 4\sin x\)
  2. \(g(x) = \cos x - 3x\)
  3. \(h(x) = \tan x + \sec x\)
  4. \(f(x) = x^2 \sin x\)
  5. \(g(x) = \dfrac{\cos x}{x}\)
  6. \(h(x) = \csc x - \cot x\)
  7. \(f(x) = \sin(5x)\)
  8. \(g(x) = \cos(x^3)\)
  9. \(h(x) = \tan(2x-1)\)
  10. \(f(x) = \sec(x^2)\)
  11. \(g(x) = \sin^4 x\)
  12. \(h(x) = \cos(\sin x)\)
  13. \(f(x) = x\cos(3x)\)
  14. \(g(x) = \dfrac{\sin x}{1+\cos x}\)
ApplicationsTrigonometric Models
  1. A buoy bobs up and down with height \(h(t) = 3\sin(2t)\) meters, where \(t\) is in seconds. Find the velocity at \(t = \pi/4\) seconds and interpret the result.
  2. The voltage in an AC circuit is \(V(t) = 120\cos(120\pi t)\), where \(t\) is in seconds. Find the rate of change of voltage at \(t = 1/240\) seconds.
  3. A pendulum has angular displacement \(\theta(t) = 0.15\cos(2\pi t)\) radians. Find both \(\theta'(t)\) and \(\theta''(t)\) and interpret each in context.

Exponential Derivative Practice

Concept CheckUnderstanding \(e^x\) and \(a^x\)
  1. Explain in your own words why \(\frac{d}{dx}[e^x] = e^x\). What property of \(e\) makes this possible?
  2. What is the slope of \(y = e^x\) at the point \((0,1)\)? At the point \((1, e)\)?
  3. A student writes \(\frac{d}{dx}[e^{x^2}] = e^{x^2}\). What is wrong with this answer?
  4. Compare \(\frac{d}{dx}[e^x]\) to \(\frac{d}{dx}[3^x]\). What is similar? What is different?
Skill PracticeExponential Differentiation

Find each derivative.

  1. \(f(x) = 7e^x\)
  2. \(g(x) = e^x + x^3\)
  3. \(h(x) = e^{-x}\)
  4. \(f(x) = e^{4x}\)
  5. \(g(x) = e^{x^2+1}\)
  6. \(h(x) = xe^x\)
  7. \(f(x) = \dfrac{e^x}{x}\)
  8. \(g(x) = e^{\sin x}\)
  9. \(h(x) = e^{\cos(3x)}\)
  10. \(f(x) = 2^x\)
  11. \(g(x) = 5^{x^2}\)
  12. \(h(x) = 3^{2x-1}\)
  13. \(f(x) = x^2 e^{3x}\)
  14. \(g(x) = e^{2x}\sin x\)
ApplicationsExponential Models
  1. A radioactive substance decays as \(A(t) = 200e^{-0.05t}\) grams, where \(t\) is years. Find the rate of decay at \(t = 10\) and interpret the result.
  2. A savings account grows as \(B(t) = 1000e^{0.06t}\) dollars. Find the instantaneous rate of growth at \(t = 5\) years.
  3. A population follows \(P(t) = 300e^{0.2t}\). Compare the growth rates at \(t = 0\), \(t = 5\), and \(t = 10\). What pattern do you observe?

Logarithmic Derivative Practice

Skill PracticeLogarithmic Differentiation
  1. \(f(x) = \ln(x^4)\)
  2. \(g(x) = \ln(2x+5)\)
  3. \(h(x) = \ln(x^2+x)\)
  4. \(f(x) = x\ln x\)
  5. \(g(x) = (\ln x)^3\)
  6. \(h(x) = \ln(\sin x)\)
  7. \(f(x) = \ln(\sec x)\)
  8. \(g(x) = \ln(e^x+1)\)
  9. \(h(x) = \log_3(x^2-4)\)
  10. \(f(x) = \log_{10}(5x)\)
Logarithmic DifferentiationVariable Exponents and Products

Use logarithmic differentiation for each.

  1. \(f(x) = x^x\)
  2. \(g(x) = x^{\sin x}\)
  3. \(h(x) = (x+1)^x\)
  4. \(f(x) = \dfrac{x^2(x-1)^3}{\sqrt{x+2}}\)
  5. \(g(x) = \dfrac{(x^2+1)^4 e^{2x}}{(x-3)^2}\)
ApplicationsLogarithmic Models
  1. The pH of a solution is \(\text{pH} = -\log_{10}[\text{H}^+]\). Find the rate of change of pH with respect to \([\text{H}^+]\) and interpret its sign.
  2. The decibel level of a sound is \(D = 10\log_{10}(I/I_0)\). Find \(dD/dI\) and explain why the rate of change decreases as intensity increases.

Mixed Practice

Combined RulesFull Differentiation Toolkit

Identify the outermost structure first, then apply the appropriate rules in order.

  1. \(f(x) = e^x \cos x\)
  2. \(g(x) = \ln(\arctan x)\)
  3. \(h(x) = \sin(e^x)\)
  4. \(f(x) = \arctan(\ln x)\)
  5. \(g(x) = e^{\arcsin x}\)
  6. \(h(x) = \dfrac{\sin x}{e^x}\)
  7. \(f(x) = \ln(\sin^2 x + 1)\)
  8. \(g(x) = \sec(e^{2x})\)
  9. \(h(x) = \arcsin(1-x^2)\)
  10. \(f(x) = x^2 \arctan(3x)\)
  11. \(g(x) = e^{2x}\tan(x^2)\)
  12. \(h(x) = \cos^3(\ln x)\)

Applications

  1. Oscillating Spring. A spring-mass system has position \(x(t) = 0.2e^{-0.5t}\cos(4t)\) meters. Find the velocity \(x'(t)\). Describe the motion: is the amplitude increasing or decreasing? What does \(e^{-0.5t}\) contribute?
  2. Learning Curve. A student's skill score follows \(S(t) = 100(1-e^{-0.1t})\) percent. Find \(S'(t)\). At what rate is the student improving after 10 days? After 50 days? What does the limiting behavior suggest?
  3. Drug Concentration. The concentration \(C(t) = 5te^{-0.3t}\) mg/L where \(t\) is hours. Find \(C'(t)\) and set it to zero to find the time of maximum concentration.
  4. Investment Return. An investment follows \(V(t) = Pe^{rt}\). Show that \(V'(t) = rV(t)\) and interpret: what does this equation say about the relationship between the account's level and its growth rate?
  5. Projectile Angle. A ball is thrown horizontally from 20 meters. The velocity angle satisfies \(\theta = \arctan(v_y/v_x)\) where \(v_x = 15\) m/s and \(v_y = -9.8t\) m/s. Find \(d\theta/dt\) at \(t = 1\) second and interpret the result.
  6. Angle of a Ramp. An engineer designs a ramp where \(\theta = \arctan(h/12)\) and \(h\) is variable. Find \(d\theta/dh\) at \(h = 3\) meters. Interpret the result.

Error Analysis

Each problem shows a student's error. Identify it, write the correct answer, and state a principle that prevents the mistake.

Problem 1
\(\frac{d}{dx}[e^{x^3}] = e^{x^3}\)

What is missing? Write the correct derivative and identify which rule was forgotten.

Problem 2
\(\frac{d}{dx}[\cos(5x)] = \sin(5x)\)

Find two errors: one involving sign and one involving the chain rule. Write the correct answer.

Problem 3
\(\frac{d}{dx}[2^x] = x \cdot 2^{x-1}\)

What rule did the student misapply? Write the correct derivative.

Problem 4
\(\frac{d}{dx}[\ln(x^3+1)] = \frac{1}{3x^2}\)

Find and explain the error. Where does \(3x^2\) belong in the correct answer?

Problem 5
\(\frac{d}{dx}[x^x] = x \cdot x^{x-1} = x^x\)

Why do both the power rule and the exponential rule fail here? What technique should be used, and what is the correct answer?

Problem 6
\(\frac{d}{dx}[\sin^2 x] = 2\sin x\)

What factor is missing? State the chain rule requirement and write the correct answer.

Problem 7
\(\frac{d}{dx}[\arctan(x^2)] = \frac{1}{1+x^4}\)

What did the student forget? Write the correct derivative.

Problem 8
\(\frac{d}{dx}[\cos(\ln x)] = -\sin(\ln x)\)

What does the chain rule require that the student omitted? Write the complete correct answer.

Chapter 5 Checkpoint

Chapter 5 Checkpoint
Complete without assistance — assesses all key skills, concepts, and applications from Chapter 5
Part A — Formula Knowledge
  1. Write from memory the derivatives of: \(\sin x\), \(\cos x\), \(\tan x\), \(\sec x\), \(\csc x\), \(\cot x\), \(e^x\), \(a^x\), \(\ln x\), \(\log_a x\), \(\arcsin x\), \(\arccos x\), \(\arctan x\).
  2. State when each formula requires a domain restriction and identify those restrictions.
Part B — Skill Problems
  1. Differentiate: \(f(x) = 3\sin(x^2) - e^{2x}\)
  2. Differentiate: \(g(x) = x^2 e^{-x}\)
  3. Differentiate: \(h(x) = \ln(x^2+4)\)
  4. Differentiate: \(f(x) = \arctan(5x)\)
  5. Differentiate: \(g(x) = e^x \tan x\)
  6. Differentiate: \(h(x) = \dfrac{\ln x}{\sin x}\)
  7. Differentiate: \(f(x) = \sec(e^x)\)
  8. Differentiate: \(g(x) = 4^{x^3}\)
  9. Differentiate: \(h(x) = \cos^2(\ln x)\)
  10. Use logarithmic differentiation: \(f(x) = x^{\ln x}\)
Part C — Applications
  1. A population of 1000 bacteria grows as \(P(t) = 1000e^{0.4t}\). Find the growth rate at \(t = 3\) hours and express the answer with appropriate units.
  2. A pendulum's angular displacement is \(\theta(t) = 0.25\cos(3\pi t)\) radians. Find angular velocity and evaluate at \(t = 1/6\) seconds. Interpret the result.
  3. A chemical concentration follows \(C(t) = C_0 e^{-kt}\) for constants \(C_0\) and \(k > 0\). Show that \(C'(t) = -kC(t)\) and explain in words what this equation says about the relationship between concentration level and its rate of change.
Part D — Error Identification
  1. A student differentiates \(f(x) = e^{\cos x}\) and writes \(f'(x) = e^{\cos x}\). Identify and correct the error.
  2. A student differentiates \(g(x) = \ln(3x^2-1)\) and writes \(g'(x) = \frac{1}{6x}\). Find the correct derivative and explain what the student did wrong.
Part E — Concept Questions
  1. Explain why \(\frac{d}{dx}[e^x] = e^x\) makes the natural exponential function special among all exponential functions. What property of \(e\) is responsible?
  2. Describe the inverse function derivative theorem in your own words. Give one example showing how it applies.
  3. A function \(f(x)\) is increasing rapidly near a point \(x = a\). What does the inverse function theorem tell you about the rate of change of \(f^{-1}\) near the corresponding point?
Looking Ahead
Chapter 6: Implicit Differentiation and Related Rates

Chapter 5 completed the core differentiation toolkit for every major function family from precalculus. Chapter 6 applies that toolkit in two new directions. Implicit differentiation handles equations where \(y\) is not isolated — such as circles, ellipses, and more complex curves — by treating \(y\) as a function of \(x\) and using the chain rule on every occurrence of \(y\). Related rates problems use the same idea to connect the rates at which two or more quantities change together, modeling situations from geometry, physics, and engineering. Both topics deepen your understanding of the chain rule and prepare you for the applications of derivatives in Chapters 7 and 8.

MGU Library Connection
  • Appendix G: Common Derivatives Reference — complete table of all transcendental derivative formulas from this chapter
  • Appendix C: Trigonometry Reference — graphs, identities, and inverse trig function properties underlying every derivative in Sections 5.2–5.4 and 5.11
  • Trigonometry and Precalculus (Chapter on Function Families) — reviews graphs and domains of \(\sin x\), \(\cos x\), \(e^x\), \(\ln x\), and the six inverse trig functions
  • Physics (Oscillation and Wave Motion chapter) — connects \(\frac{d}{dx}[A\cos(\omega t + \phi)] = -A\omega\sin(\omega t+\phi)\) to velocity and acceleration models
  • Data Science Foundations (Information Theory chapter) — uses logarithmic derivatives in entropy and cross-entropy loss functions
  • Calculus II (Integration Techniques chapter) — antiderivatives of \(1/x\), \(e^x\), \(\sin x\), \(\cos x\), and the inverse trig functions flow directly from the derivatives developed here