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Chain RuleWorksheet
(80 Practice Problems with Step-by-Step Solutions)

Last Updated: February 22, 2026

Intermediate–Advanced ⌛ Est. 3–4 Hours 80 Problems

A complete, rigorous resource covering the Chain Rule for differentiation — from the intuition behind composite functions through advanced multi-layer applications, real-world physics problems, and exam-style multiple choice. Suitable for AP Calculus AB/BC, university Calculus I and II, and advanced secondary mathematics.

Introduction

Among all the rules of differentiation, the Chain Rule stands in a class of its own. It is simultaneously the most conceptually rich and the most practically indispensable rule in all of differential calculus. A student who fully understands the Chain Rule can differentiate an enormous range of functions — far beyond what the Power Rule, Product Rule, or Quotient Rule alone can handle. More importantly, the Chain Rule is the gateway to understanding implicit differentiation, related rates, and the entirety of multivariable calculus.

The central idea is beautifully simple: in real life, quantities frequently depend on other quantities that themselves depend on something else. The temperature inside a chemical reactor depends on the pressure, and the pressure depends on time. The position of a satellite depends on its orbital angle, and the orbital angle depends on time. The Chain Rule tells us precisely how to compute the rate of change of the final quantity with respect to the original variable, no matter how many intermediate layers of dependence exist.

Formally, suppose we have a composite function \(y = f(g(x))\). Here, \(g\) is the inner function and \(f\) is the outer function. The Chain Rule states that:

The Chain Rule — Leibniz Notation
\[\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}\]

where \(u = g(x)\) is the inner function and \(y = f(u)\) is the outer function expressed in terms of \(u\).

In prime notation, the same statement reads:

The Chain Rule — Prime Notation
\[\frac{d}{dx}\bigl[f(g(x))\bigr] = f'(g(x))\cdot g'(x)\]

Differentiate the outer function at the inner function, then multiply by the derivative of the inner function.

There is an intuitive way to remember this. Think of the Chain Rule as an instruction: differentiate from the outside in, and multiply as you go. You peel away one layer of the composite at a time, differentiating each layer and multiplying the results together. This layered structure is what makes the Chain Rule so powerful and so widely applicable.

This worksheet is structured to take you from first principles through genuine mastery. The worked examples in Section 1 establish the fundamental technique. Sections 2 through 4 provide carefully graded practice problems — basic, intermediate, and challenging — designed to build both speed and confidence. Section 5 situates the Chain Rule in real-world contexts drawn from physics, engineering, and mathematical optimisation. Section 6 offers a mixed review and multiple-choice section mirroring AP Calculus and university exam formats. The complete answer key and detailed solutions appear at the end.

A word of advice before you begin: many students rush into computation before they have clearly identified the outer and inner functions. Slow down at the identification step. Once you can reliably see the structure of a composite function — once you can immediately say "the outer function is \(\sin(\cdot)\) and the inner function is \(x^3 + 2x\)" — the mechanics of the Chain Rule follow naturally. Invest in that identification step, and the rest will come.

Learning Objectives

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Key Concepts & Formulae

1.1 The Chain Rule — Statement and Notations

Let \(g\) be differentiable at \(x\) and \(f\) be differentiable at \(g(x)\). Then the composite function \(h(x) = f(g(x))\) is differentiable at \(x\) and:

Chain Rule — All Forms
\[h'(x) = f'\!\bigl(g(x)\bigr)\cdot g'(x)\] \[\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} \quad \text{where } u = g(x)\] \[\frac{d}{dx}\bigl[f(g(x))\bigr] = f'(g(x))\cdot g'(x)\]

1.2 The Generalised Power Rule

When the outer function is a power function, the Chain Rule specialises to the Generalised Power Rule, which is arguably the single most used derivative shortcut in calculus:

Generalised Power Rule
\[\frac{d}{dx}\bigl[u^n\bigr] = n\,u^{n-1}\cdot \frac{du}{dx}\]

where \(u = g(x)\) is any differentiable function of \(x\) and \(n\) is any real number.

1.3 Chain Rule with Standard Functions

Combining the Chain Rule with the standard derivatives of common functions gives the following essential formulae:

Outer Function \(f(u)\) Chain Rule Result \(\dfrac{d}{dx}[f(u)]\)
\(u^n\)\(n\,u^{n-1}\cdot u'\)
\(e^u\)\(e^u \cdot u'\)
\(a^u\)\(a^u \ln a \cdot u'\)
\(\ln u\)\(\dfrac{u'}{u}\)
\(\log_a u\)\(\dfrac{u'}{u\ln a}\)
\(\sin u\)\(\cos u\cdot u'\)
\(\cos u\)\(-\sin u\cdot u'\)
\(\tan u\)\(\sec^2 u\cdot u'\)
\(\sec u\)\(\sec u\tan u\cdot u'\)
\(\csc u\)\(-\csc u\cot u\cdot u'\)
\(\cot u\)\(-\csc^2 u\cdot u'\)
\(\arcsin u\)\(\dfrac{u'}{\sqrt{1-u^2}}\)
\(\arctan u\)\(\dfrac{u'}{1+u^2}\)
\(\sqrt{u}\)\(\dfrac{u'}{2\sqrt{u}}\)

1.4 How to Identify Composite Functions

The key question is: what is the last operation applied to \(x\)? That operation defines the outer function. Everything fed into that operation is the inner function. For instance:

1.5 Worked Examples

Example 1 — Generalised Power Rule

Find \(\dfrac{d}{dx}\!\left[(3x^2 + 5)^4\right]\).

Identify Layers

Outer: \(u^4\). Inner: \(u = 3x^2+5\), so \(u' = 6x\).

Apply Chain Rule

\(\dfrac{d}{dx}\!\left[(3x^2+5)^4\right] = 4(3x^2+5)^3 \cdot 6x\)

\(= 24x(3x^2+5)^3\)
Example 2 — Exponential Outer Function

Find \(\dfrac{d}{dx}\!\left[e^{x^3 - 2x}\right]\).

Identify Layers

Outer: \(e^u\). Inner: \(u = x^3 - 2x\), so \(u' = 3x^2 - 2\).

Apply Chain Rule

\(\dfrac{d}{dx}\!\left[e^{x^3-2x}\right] = e^{x^3-2x}\cdot(3x^2-2)\)

\(= (3x^2-2)e^{x^3-2x}\)
Example 3 — Trigonometric Outer Function

Find \(\dfrac{d}{dx}\!\left[\sin(5x^2)\right]\).

Identify Layers

Outer: \(\sin(u)\). Inner: \(u = 5x^2\), so \(u' = 10x\).

Apply Chain Rule

\(\dfrac{d}{dx}\!\left[\sin(5x^2)\right] = \cos(5x^2)\cdot 10x\)

\(= 10x\cos(5x^2)\)
Example 4 — Natural Logarithm Outer Function

Find \(\dfrac{d}{dx}\!\left[\ln(x^2 + 1)\right]\).

Identify Layers

Outer: \(\ln(u)\). Inner: \(u = x^2+1\), so \(u' = 2x\).

Apply Chain Rule

\(\dfrac{d}{dx}\!\left[\ln(x^2+1)\right] = \dfrac{2x}{x^2+1}\)

\(= \dfrac{2x}{x^2+1}\)
Example 5 — Multiple Layer Composition

Find \(\dfrac{d}{dx}\!\left[\sin(e^{3x})\right]\).

Identify Layers

Three layers: \(\sin(\,\cdot\,)\) applied to \(e^{(\,\cdot\,)}\) applied to \(3x\).

First Chain Rule (outermost layer)

\(\cos(e^{3x})\cdot\dfrac{d}{dx}\!\left[e^{3x}\right]\)

Second Chain Rule (middle layer)

\(\dfrac{d}{dx}\!\left[e^{3x}\right] = e^{3x}\cdot 3\)

Combine

\(\cos(e^{3x})\cdot 3e^{3x}\)

\(= 3e^{3x}\cos(e^{3x})\)
Example 6 — Chain Rule Combined with Product Rule

Find \(\dfrac{d}{dx}\!\left[x^2 \sin(3x)\right]\).

Strategy

This is a product. Apply the Product Rule: \((fg)' = f'g + fg'\).

Let \(f = x^2\) and \(g = \sin(3x)\).

Differentiate Each Factor

\(f' = 2x\). For \(g' = \cos(3x)\cdot 3 = 3\cos(3x)\) (Chain Rule).

Combine via Product Rule

\(2x\sin(3x) + x^2\cdot 3\cos(3x)\)

\(= 2x\sin(3x) + 3x^2\cos(3x)\)
Example 7 — Chain Rule with Quotient Rule

Differentiate \(h(x) = \dfrac{e^{2x}}{x^3+1}\).

Apply Quotient Rule

\(h'(x) = \dfrac{(e^{2x})'(x^3+1) - e^{2x}(x^3+1)'}{(x^3+1)^2}\)

Differentiate Numerator Terms

\((e^{2x})' = 2e^{2x}\) (Chain Rule). \((x^3+1)' = 3x^2\).

Substitute and Simplify

\(h'(x) = \dfrac{2e^{2x}(x^3+1) - 3x^2 e^{2x}}{(x^3+1)^2} = \dfrac{e^{2x}(2x^3+2-3x^2)}{(x^3+1)^2}\)

\(= \dfrac{e^{2x}(2x^3-3x^2+2)}{(x^3+1)^2}\)
Example 8 — Finding a Tangent Line

Find the equation of the tangent line to \(y = (x^2 - 3)^3\) at \(x = 2\).

Differentiate

\(y' = 3(x^2-3)^2\cdot 2x = 6x(x^2-3)^2\)

Evaluate at \(x=2\)

\(y'(2) = 6(2)(4-3)^2 = 12\cdot1 = 12\). This is the slope.

\(y(2) = (4-3)^3 = 1\). Point: \((2,1)\).

Write Equation

\(y - 1 = 12(x-2)\implies y = 12x - 23\)

Tangent line: \(y = 12x - 23\)
Common Mistakes to Avoid
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Practice Problems

Part A — Basic Chain Rule  (Problems 1–20)

Apply the Chain Rule to each composite function. Identify the outer and inner function clearly before computing. Express all answers in simplified form.

  1. Differentiate \(f(x) = (2x + 3)^5\).
  2. Differentiate \(g(x) = (x^2 - 4)^3\).
  3. Differentiate \(h(x) = (1 - x)^7\).
  4. Differentiate \(f(x) = \sqrt{3x + 1}\).
  5. Differentiate \(g(x) = \sqrt{x^2 + 9}\).
  6. Differentiate \(h(x) = \dfrac{1}{(x^2 + 1)^2}\).
  7. Differentiate \(f(x) = e^{4x}\).
  8. Differentiate \(g(x) = e^{-x^2}\).
  9. Differentiate \(h(x) = e^{3x^2 - x}\).
  10. Differentiate \(f(x) = \sin(2x)\).
  11. Differentiate \(g(x) = \cos(x^3)\).
  12. Differentiate \(h(x) = \tan(5x)\).
  13. Differentiate \(f(x) = \ln(3x)\).
  14. Differentiate \(g(x) = \ln(x^2 + 4)\).
  15. Differentiate \(h(x) = \ln(\sin x)\).
  16. Differentiate \(f(x) = (4x^3 - 1)^{10}\).
  17. Differentiate \(g(x) = \sec(x^2)\).
  18. Differentiate \(h(x) = \sqrt[3]{x^2 - 5}\).
  19. Differentiate \(f(x) = 2^{3x}\).
  20. Differentiate \(g(x) = \arctan(2x)\).
Part B — Intermediate Problems  (Problems 21–45)

These problems involve multiple-term inner functions, combinations of rules, and compositions requiring careful multi-step work. Show all intermediate steps.

  1. Differentiate \(f(x) = \sin^3(x)\). (Hint: rewrite as \([\sin x]^3\).)
  2. Differentiate \(g(x) = \cos^4(2x)\).
  3. Differentiate \(h(x) = e^{\sin x}\).
  4. Differentiate \(f(x) = \sin(e^x)\).
  5. Differentiate \(g(x) = \ln(\cos x)\).
  6. Differentiate \(h(x) = \ln(\sec x + \tan x)\).
  7. Differentiate \(f(x) = \cos(\ln x)\).
  8. Differentiate \(g(x) = e^{\tan x}\).
  9. Differentiate \(h(x) = x^2 e^{3x}\). (Product Rule + Chain Rule.)
  10. Differentiate \(f(x) = x\sin(x^2)\). (Product Rule + Chain Rule.)
  11. Differentiate \(g(x) = \dfrac{\sin(2x)}{x^2 + 1}\). (Quotient Rule + Chain Rule.)
  12. Differentiate \(h(x) = \dfrac{e^x}{\cos(3x)}\).
  13. Differentiate \(f(x) = (x^2+1)^3(2x-1)^4\). (Product Rule + Generalised Power Rule.)
  14. Differentiate \(g(x) = \tan^2(x) + \sec^2(x)\). What do you notice about your answer?
  15. Differentiate \(h(x) = \sqrt{\dfrac{x+1}{x-1}}\). (Rewrite, then use Chain Rule and Quotient Rule.)
  16. Differentiate \(f(x) = \arcsin(3x)\).
  17. Differentiate \(g(x) = \arctan\!\left(\dfrac{x}{2}\right)\).
  18. Differentiate \(h(x) = \left(\dfrac{x^2+1}{x^2-1}\right)^3\).
  19. Differentiate \(f(x) = \ln\!\left(\dfrac{x^2}{x^2+1}\right)\). Use log laws first.
  20. Differentiate \(g(x) = e^{x^2}\ln(x)\). (Product Rule + Chain Rule.)
  21. Find \(f'(x)\) if \(f(x) = \sin(3x+1)\cos(3x+1)\). Hint: recognise the double angle formula.
  22. Differentiate \(h(x) = \sqrt{x^4 - 2x^2 + 1}\). Factor the radicand first.
  23. Differentiate \(f(x) = x^2\arctan(x)\). (Product Rule.)
  24. Differentiate \(g(x) = \dfrac{\ln(x^2+1)}{x}\).
  25. Differentiate \(h(x) = \tan\!\left(\arcsin(x)\right)\). Simplify using a right-triangle argument after differentiating.
Part C — Challenging Problems  (Problems 46–65)

Advanced compositions, multiple nested layers, second derivatives, tangent lines, critical points, and proof-style questions. These are characteristic of university midterm and final examination problems.

  1. Differentiate \(f(x) = e^{e^x}\). (Double exponential.)
  2. Differentiate \(g(x) = \ln(\ln(x))\) for \(x > 1\).
  3. Differentiate \(h(x) = \sin(\cos(\tan x))\). (Triple composition.)
  4. Differentiate \(f(x) = e^{\cos(x^2)}\).
  5. Differentiate \(g(x) = \left(\sin(\ln x)\right)^3\).
  6. Find the equation of the tangent line to \(y = \sqrt{4 - x^2}\) at \(x = 1\).
  7. Find the equation of the tangent line to \(y = e^{-x^2/2}\) at \(x = 0\). (This is the standard normal kernel.)
  8. At what values of \(x\) does \(f(x) = \sin(2x)\) have horizontal tangent lines on \([0, 2\pi]\)?
  9. Find the critical points of \(g(x) = x^2 e^{-x}\) by setting \(g'(x) = 0\).
  10. Find all \(x\) where \(h(x) = (x^2-1)^3\) has horizontal tangent lines.
  11. Find \(f''(x)\) for \(f(x) = e^{2x}\).
  12. Find \(g''(x)\) for \(g(x) = \sin(3x)\).
  13. Find \(h''(x)\) for \(h(x) = \ln(x^2+1)\).
  14. Prove that \(\dfrac{d}{dx}\left[\ln|f(x)|\right] = \dfrac{f'(x)}{f(x)}\), for \(f(x) \neq 0\). (This is the log-differentiation foundation.)
  15. Use the Chain Rule to find \(\dfrac{d}{dx}\left[a^x\right]\) by writing \(a^x = e^{x\ln a}\) and differentiating. State the result as a formula.
  16. If \(y = f(x^2)\) and \(f'(u) = \sqrt{u+1}\), find \(\dfrac{dy}{dx}\) at \(x = 2\).
  17. Given that \(g(2) = 3\), \(g'(2) = -1\), \(f(3) = 5\), and \(f'(3) = 4\), find \(\dfrac{d}{dx}\left[f(g(x))\right]\) at \(x = 2\).
  18. Differentiate \(f(x) = x^x\) using logarithmic differentiation. (Write \(y = x^x\), take \(\ln\) of both sides, differentiate implicitly.)
  19. Differentiate \(y = (x^2+1)^{\sin x}\) using logarithmic differentiation.
  20. Let \(F(x) = f(g(h(x)))\) where \(f\), \(g\), and \(h\) are all differentiable. Write an expression for \(F'(x)\) in terms of \(f'\), \(g'\), and \(h'\). Then evaluate \(F'(1)\) given \(h(1)=2\), \(h'(1)=3\), \(g(2)=4\), \(g'(2)=-1\), \(f'(4)=6\).
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Real-World Applications

3.1 Physics: Kinematics and Rates of Change

The Chain Rule is the mathematical foundation of the relationship between position, velocity, and acceleration in physics. Whenever a physical quantity depends on an intermediate variable that itself depends on time, the Chain Rule is the natural tool for computing rates of change.

Physics Applications  (Problems 66–70)
  1. Projectile with Air Resistance. The height of a projectile subject to mild air resistance is modelled by \(h(t) = 50e^{-0.1t}\sin(2t)\) metres, where \(t\) is in seconds.
    (a) Find \(h'(t)\), the vertical velocity at time \(t\).
    (b) Evaluate \(h'(0)\) and \(h'(\pi/4)\). Interpret the sign of each answer.
  2. Temperature in a Cooling Rod. The temperature along a metal rod changes with position \(x\) (cm) according to \(T(x) = 80e^{-0.05x^2}\) °C. Find \(T'(x)\) and determine the rate of temperature change at \(x = 2\) cm.
  3. Oscillating Spring. A mass on a spring has position \(s(t) = 3\cos(4\pi t)\) cm at time \(t\) seconds. Find the velocity \(v(t) = s'(t)\) and the acceleration \(a(t) = s''(t)\). Verify that \(a(t) = -16\pi^2 s(t)\), which is the equation of simple harmonic motion.
  4. Intensity of Sound. The intensity of a sound wave decays as \(I(r) = \dfrac{I_0}{r^2}\), where \(r\) is the distance from the source. If the source is moving so that \(r(t) = 2 + 3t\) metres, find \(\dfrac{dI}{dt}\) using the Chain Rule: \(\dfrac{dI}{dt} = \dfrac{dI}{dr}\cdot\dfrac{dr}{dt}\).
  5. Electric Circuit. The charge on a capacitor in an RC circuit is \(Q(t) = Q_0\left(1 - e^{-t/RC}\right)\), where \(R\) and \(C\) are positive constants. Find the current \(I(t) = \dfrac{dQ}{dt}\) and interpret its behaviour as \(t \to \infty\).

3.2 Engineering: Optimisation and Structural Analysis

Engineering Applications  (Problems 71–75)
  1. Beam Deflection. The deflection of a simply supported beam under a distributed load is approximated by \(y(x) = A\sin\!\left(\dfrac{\pi x}{L}\right)\), where \(L\) is the beam length and \(A\) is the maximum deflection. Find \(y'(x)\) (the slope of the beam) and \(y''(x)\) (proportional to the bending moment).
  2. Signal Attenuation. The amplitude of a signal transmitted through a medium decreases according to \(A(d) = A_0 e^{-\alpha d}\), where \(d\) is the depth in metres and \(\alpha > 0\) is the attenuation coefficient. Find \(\dfrac{dA}{dd}\) and express it as a multiple of \(A(d)\). What does this tell you about exponential decay?
  3. Fluid Flow: Poiseuille's Law. The velocity of a viscous fluid in a pipe of radius \(R\) is \(v(r) = v_{\max}\!\left(1 - \dfrac{r^2}{R^2}\right)\), where \(r\) is the radial distance from the centre. Find \(v'(r)\). At what radius is the speed decreasing most rapidly?
  4. Chemical Reaction Rate. The concentration of a reactant follows \(C(t) = \dfrac{C_0}{(1+kt)^2}\), where \(k\) is the rate constant. Find \(\dfrac{dC}{dt}\) using the Generalised Power Rule and interpret the sign of the result.
  5. Angle of Elevation. A surveillance camera is mounted 10 m above ground level. An object moves horizontally at ground level at 3 m/s. When the object is 10 m from the point directly below the camera, the angle of elevation \(\theta\) satisfies \(\tan\theta = \dfrac{10}{x}\). Use the Chain Rule to find \(\dfrac{d\theta}{dt}\) at that instant.
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Mixed Review & Multiple Choice

Part D — Multiple Choice  (Problems 76–80)

Circle the best answer. These questions mirror AP Calculus AB/BC and university midterm examination formats. No calculator assumed.

  1. If \(f(x) = (3x-1)^4\), then \(f'(x) =\)
    • (A) \(4(3x-1)^3\)
    • (B) \(12(3x-1)^3\)
    • (C) \(12(3x-1)^4\)
    • (D) \(4x(3x-1)^3\)
  2. The derivative of \(y = e^{\sin x}\) is:
    • (A) \(e^{\sin x}\)
    • (B) \(e^{\cos x}\)
    • (C) \(\cos x \cdot e^{\sin x}\)
    • (D) \(\cos x \cdot e^{\cos x}\)
  3. If \(g(x) = \ln(x^3 + 1)\), then \(g'(1) =\)
    • (A) \(\dfrac{1}{2}\)
    • (B) \(\dfrac{3}{2}\)
    • (C) \(3\ln 2\)
    • (D) \(\ln 2\)
  4. Given \(h(x) = \sqrt{f(x)}\) and \(f(4) = 9\), \(f'(4) = 6\), then \(h'(4) =\)
    • (A) \(1\)
    • (B) \(3\)
    • (C) \(\dfrac{1}{3}\)
    • (D) \(\dfrac{1}{6}\)
  5. Which of the following functions requires the Chain Rule but NOT the Product Rule to differentiate?
    • (A) \(y = x\sin x\)
    • (B) \(y = x^2 e^x\)
    • (C) \(y = \sin(x^2)\)
    • (D) \(y = x^2 \ln x\)
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Quick Answer Key

Answers to all 80 problems. Full step-by-step solutions follow below.

1. \(5(2x+3)^4\cdot 2 = 10(2x+3)^4\)
2. \(6x(x^2-4)^2\)
3. \(-7(1-x)^6\)
4. \(\dfrac{3}{2\sqrt{3x+1}}\)
5. \(\dfrac{x}{\sqrt{x^2+9}}\)
6. \(\dfrac{-4x}{(x^2+1)^3}\)
7. \(4e^{4x}\)
8. \(-2xe^{-x^2}\)
9. \((6x-1)e^{3x^2-x}\)
10. \(2\cos(2x)\)
11. \(-3x^2\sin(x^3)\)
12. \(5\sec^2(5x)\)
13. \(\dfrac{1}{x}\)
14. \(\dfrac{2x}{x^2+4}\)
15. \(\cot x\)
16. \(120x^2(4x^3-1)^9\)
17. \(2x\sec(x^2)\tan(x^2)\)
18. \(\dfrac{2x}{3(x^2-5)^{2/3}}\)
19. \(3\ln(2)\cdot 2^{3x}\)
20. \(\dfrac{2}{1+4x^2}\)
21. \(3\sin^2(x)\cos(x)\)
22. \(-8\cos^3(2x)\sin(2x)\)
23. \(\cos(x)\,e^{\sin x}\)
24. \(\cos(e^x)\cdot e^x\)
25. \(-\tan x\)
26. \(\sec x\)
27. \(-\dfrac{\sin(\ln x)}{x}\)
28. \(\sec^2(x)\,e^{\tan x}\)
29. \(e^{3x}(2x+3x^2\cdot 2)\! = \!e^{3x}(2+3x)\cdot x\)  → \(xe^{3x}(3x+2)\)
30. \(\sin(x^2) + 2x^2\cos(x^2)\)
31. \(\dfrac{2\cos(2x)(x^2+1)-2x\sin(2x)}{(x^2+1)^2}\)
32. \(\dfrac{e^x\cos(3x)+3e^x\sin(3x)}{\cos^2(3x)}\)
33. \(3(x^2+1)^2(2x-1)^4\cdot 2x + (x^2+1)^3\cdot 4(2x-1)^3\cdot 2\)
34. \(0\) (constant; see solution)
35. \(-\dfrac{1}{(x-1)^2\sqrt{\tfrac{x+1}{x-1}}}\)
36. \(\dfrac{3}{\sqrt{1-9x^2}}\)
37. \(\dfrac{2}{4+x^2}\)
38. \(\dfrac{-12x(x^2+1)^2(x^2-1)-3(x^2+1)^3\cdot 2x}{(x^2-1)^4}\) simplified
39. \(\dfrac{2}{x} - \dfrac{2x}{x^2+1}\)
40. \(2xe^{x^2}\ln x + \dfrac{e^{x^2}}{x}\)
41. \(\dfrac{1}{2}\cos(6x+2)\cdot 6 = 3\cos(6x+2)\)
42. \(\dfrac{2x(x^2-1)}{|x^2-1|}\) (sign depends on domain; see solution)
43. \(\dfrac{2x}{1+x^2}\arctan(x) + \dfrac{x^2}{1+x^2}\)
44. \(\dfrac{\tfrac{2x}{x^2+1}\cdot x - \ln(x^2+1)}{x^2}\)
45. \(\dfrac{1}{\sqrt{1-x^2}}\) (after simplification)
46. \(e^{e^x}\cdot e^x\)
47. \(\dfrac{1}{x\ln x}\)
48. \(-\sin(\cos(\tan x))\cdot(-\sin(\tan x))\cdot\sec^2 x\)
49. \(-2x\sin(x^2)\,e^{\cos(x^2)}\)
50. \(\dfrac{3\cos(\ln x)\sin^2(\ln x)}{x}\)
51. \(y = -\tfrac{1}{\sqrt{3}}(x-1) + \sqrt{3}\)
52. \(y = 1\) (horizontal tangent at \(x=0\))
53. \(x = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4}\)
54. \(x = 0\) (local max) and \(x = 2\) (local min)
55. \(x = 0, \pm 1\)
56. \(4e^{2x}\)
57. \(-9\sin(3x)\)
58. \(\dfrac{2(1-x^2)}{(x^2+1)^2}\)
59. Proof (see solution)
60. \(\dfrac{d}{dx}[a^x] = a^x\ln a\)
61. \(\dfrac{dy}{dx}\big|_{x=2} = 2\cdot 2\cdot\sqrt{4+1} = 4\sqrt{5}\)
62. \(-4\)
63. \(\dfrac{dy}{dx} = x^x(\ln x + 1)\)
64. See solution
65. \(F'(1) = f'(g(h(1)))\cdot g'(h(1))\cdot h'(1) = 6\cdot(-1)\cdot 3 = -18\)
66. See solution (product rule + chain rule)
67. \(T'(x) = -8xe^{-0.05x^2}\); \(T'(2) = -16e^{-0.2}\)
68. \(v(t) = -12\pi\sin(4\pi t)\); \(a(t) = -48\pi^2\cos(4\pi t)\)
69. \(\dfrac{dI}{dt} = \dfrac{-6I_0}{(2+3t)^3}\)
70. \(I(t) = \dfrac{Q_0}{RC}e^{-t/RC}\); \(I\to 0\) as \(t\to\infty\)
71. \(y'(x) = \dfrac{A\pi}{L}\cos\!\left(\dfrac{\pi x}{L}\right)\); \(y''(x) = -\dfrac{A\pi^2}{L^2}\sin\!\left(\dfrac{\pi x}{L}\right)\)
72. \(\dfrac{dA}{dd} = -\alpha A(d)\); exponential decay
73. \(v'(r) = -\dfrac{2v_{\max}r}{R^2}\); fastest decrease at \(r = R\)
74. \(\dfrac{dC}{dt} = \dfrac{-2C_0 k}{(1+kt)^3}\)
75. \(\dfrac{d\theta}{dt} = -\tfrac{3}{20}\) rad/s
76. (B)
77. (C)
78. (B)
79. (A)
80. (C)
Created by DerivativeCalculus.com — https://www.derivativecalculus.com

Step-by-Step Solutions

Part A — Basic Chain Rule (Problems 1–20)

Problem 1

Outer: \(u^5\), Inner: \(u = 2x+3\), \(u' = 2\).

\(\dfrac{d}{dx}\left[(2x+3)^5\right] = 5(2x+3)^4 \cdot 2\)

\(= 10(2x+3)^4\)
Problem 2

Outer: \(u^3\), Inner: \(u = x^2-4\), \(u' = 2x\).

\(= 3(x^2-4)^2 \cdot 2x = 6x(x^2-4)^2\)
Problem 3

Outer: \(u^7\), Inner: \(u = 1-x\), \(u' = -1\).

\(= 7(1-x)^6 \cdot (-1) = -7(1-x)^6\)
Problem 4

Rewrite: \(f(x) = (3x+1)^{1/2}\). Outer: \(u^{1/2}\), Inner: \(u = 3x+1\), \(u' = 3\).

\(f'(x) = \dfrac{1}{2}(3x+1)^{-1/2}\cdot 3\)

\(= \dfrac{3}{2\sqrt{3x+1}}\)
Problem 5

Outer: \(u^{1/2}\), Inner: \(u = x^2+9\), \(u' = 2x\).

\(\dfrac{1}{2}(x^2+9)^{-1/2}\cdot 2x\)

\(= \dfrac{x}{\sqrt{x^2+9}}\)
Problem 6

Rewrite: \(h(x) = (x^2+1)^{-2}\). Outer: \(u^{-2}\), Inner: \(u = x^2+1\), \(u' = 2x\).

\(-2(x^2+1)^{-3}\cdot 2x\)

\(= \dfrac{-4x}{(x^2+1)^3}\)
Problem 7

Outer: \(e^u\), Inner: \(u = 4x\), \(u' = 4\).

\(f'(x) = 4e^{4x}\)
Problem 8

Outer: \(e^u\), Inner: \(u = -x^2\), \(u' = -2x\).

\(g'(x) = e^{-x^2}\cdot(-2x) = -2xe^{-x^2}\)
Problem 9

Outer: \(e^u\), Inner: \(u = 3x^2 - x\), \(u' = 6x - 1\).

\(h'(x) = (6x-1)e^{3x^2-x}\)
Problem 10

Outer: \(\sin(u)\), Inner: \(u = 2x\), \(u' = 2\).

\(f'(x) = \cos(2x)\cdot 2 = 2\cos(2x)\)
Problem 11

Outer: \(\cos(u)\), Inner: \(u = x^3\), \(u' = 3x^2\).

\(g'(x) = -\sin(x^3)\cdot 3x^2 = -3x^2\sin(x^3)\)
Problem 12

Outer: \(\tan(u)\), Inner: \(u = 5x\), \(u' = 5\).

\(h'(x) = \sec^2(5x)\cdot 5 = 5\sec^2(5x)\)
Problem 13

Outer: \(\ln(u)\), Inner: \(u = 3x\), \(u' = 3\).

\(f'(x) = \dfrac{3}{3x} = \dfrac{1}{x}\). (Note: \(\ln(3x) = \ln 3 + \ln x\) gives the same result.)

\(f'(x) = \dfrac{1}{x}\)
Problem 14

Outer: \(\ln(u)\), Inner: \(u = x^2+4\), \(u' = 2x\).

\(g'(x) = \dfrac{2x}{x^2+4}\)
Problem 15

Outer: \(\ln(u)\), Inner: \(u = \sin x\), \(u' = \cos x\).

\(h'(x) = \dfrac{\cos x}{\sin x} = \cot x\)
Problem 16

Outer: \(u^{10}\), Inner: \(u = 4x^3-1\), \(u' = 12x^2\).

\(f'(x) = 10(4x^3-1)^9\cdot 12x^2 = 120x^2(4x^3-1)^9\)
Problem 17

Outer: \(\sec(u)\), Inner: \(u = x^2\), \(u' = 2x\).

\(g'(x) = \sec(x^2)\tan(x^2)\cdot 2x = 2x\sec(x^2)\tan(x^2)\)
Problem 18

Rewrite: \(h(x) = (x^2-5)^{1/3}\). Outer: \(u^{1/3}\), Inner: \(u = x^2-5\), \(u' = 2x\).

\(h'(x) = \dfrac{1}{3}(x^2-5)^{-2/3}\cdot 2x = \dfrac{2x}{3(x^2-5)^{2/3}}\)
Problem 19

Write \(2^{3x} = e^{3x\ln 2}\). Outer: \(e^u\), Inner: \(u = 3x\ln 2\), \(u' = 3\ln 2\).

\(f'(x) = 2^{3x}\cdot 3\ln 2 = 3\ln(2)\cdot 2^{3x}\)
Problem 20

Outer: \(\arctan(u)\), Inner: \(u = 2x\), \(u' = 2\).

\(g'(x) = \dfrac{2}{1+(2x)^2} = \dfrac{2}{1+4x^2}\)

Part B — Intermediate Problems (21–45)

Problem 21

Rewrite \(f(x) = [\sin x]^3\). Outer: \(u^3\), Inner: \(u = \sin x\), \(u' = \cos x\).

\(f'(x) = 3\sin^2(x)\cos(x)\)
Problem 22

Rewrite \(g(x) = [\cos(2x)]^4\). Two applications of the Chain Rule.

Outer: \(u^4\), Middle layer: \(\cos(v)\), Inner: \(v = 2x\).

\(4[\cos(2x)]^3\cdot(-\sin(2x))\cdot 2\)

\(g'(x) = -8\cos^3(2x)\sin(2x)\)
Problem 23

Outer: \(e^u\), Inner: \(u = \sin x\), \(u' = \cos x\).

\(h'(x) = e^{\sin x}\cdot\cos x\)
Problem 24

Outer: \(\sin(u)\), Inner: \(u = e^x\), \(u' = e^x\).

\(f'(x) = \cos(e^x)\cdot e^x\)
Problem 25

Outer: \(\ln(u)\), Inner: \(u = \cos x\), \(u' = -\sin x\).

\(g'(x) = \dfrac{-\sin x}{\cos x} = -\tan x\)
Problem 26

Inner: \(u = \sec x + \tan x\). Then \(u' = \sec x\tan x + \sec^2 x = \sec x(\tan x + \sec x)\).

\(h'(x) = \dfrac{\sec x(\sec x + \tan x)}{\sec x + \tan x} = \sec x\)

\(h'(x) = \sec x\)
Problem 27

Outer: \(\cos(u)\), Inner: \(u = \ln x\), \(u' = 1/x\).

\(f'(x) = -\sin(\ln x)\cdot\dfrac{1}{x} = -\dfrac{\sin(\ln x)}{x}\)
Problem 28

Outer: \(e^u\), Inner: \(u = \tan x\), \(u' = \sec^2 x\).

\(g'(x) = e^{\tan x}\cdot\sec^2 x\)
Problem 29

Product Rule: \((x^2)'e^{3x} + x^2(e^{3x})'\).

\(= 2xe^{3x} + x^2\cdot 3e^{3x}\)

\(h'(x) = xe^{3x}(3x+2)\)
Problem 30

Product Rule: \(f = x\), \(g = \sin(x^2)\). \(f' = 1\), \(g' = \cos(x^2)\cdot 2x\).

\(1\cdot\sin(x^2) + x\cdot 2x\cos(x^2)\)

\(= \sin(x^2) + 2x^2\cos(x^2)\)
Problem 31

Quotient Rule. Numerator: \(\cos(2x)\cdot 2\cdot(x^2+1) - \sin(2x)\cdot 2x\).

\(g'(x) = \dfrac{2(x^2+1)\cos(2x) - 2x\sin(2x)}{(x^2+1)^2}\)
Problem 32

Quotient Rule. Numerator: \(e^x\cos(3x) - e^x(-\sin(3x)\cdot 3) = e^x[\cos(3x)+3\sin(3x)]\).

\(h'(x) = \dfrac{e^x(\cos(3x)+3\sin(3x))}{\cos^2(3x)}\)
Problem 33

Product Rule: \(f = (x^2+1)^3\), \(g = (2x-1)^4\).

\(f' = 3(x^2+1)^2\cdot 2x = 6x(x^2+1)^2\); \(g' = 4(2x-1)^3\cdot 2 = 8(2x-1)^3\).

\(= 6x(x^2+1)^2(2x-1)^4 + 8(x^2+1)^3(2x-1)^3 = 2(x^2+1)^2(2x-1)^3\bigl[3x(2x-1)+4(x^2+1)\bigr]\)
Problem 34

We know from the Pythagorean identity that \(\tan^2 x + 1 = \sec^2 x\), so \(\tan^2 x + \sec^2 x\) is not a standard identity that simplifies to zero. Let us differentiate directly.

\(\dfrac{d}{dx}[\tan^2 x] = 2\tan x\sec^2 x\)

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

Sum: \(2\tan x\sec^2 x + 2\sec^2 x\tan x = 4\tan x\sec^2 x\).

\(g'(x) = 4\tan x\sec^2 x\)

Note: the answer is not zero. The identity \(\sec^2 x - \tan^2 x = 1\) gives a constant, whose derivative is zero; but the sum is not constant.

Problem 35

Let \(u = \dfrac{x+1}{x-1}\), so \(h(x) = u^{1/2}\).

\(u' = \dfrac{(x-1)-(x+1)}{(x-1)^2} = \dfrac{-2}{(x-1)^2}\).

\(h'(x) = \dfrac{1}{2}u^{-1/2}\cdot u' = \dfrac{1}{2}\cdot\sqrt{\dfrac{x-1}{x+1}}\cdot\dfrac{-2}{(x-1)^2}\)

\(= \dfrac{-1}{(x-1)^{3/2}(x+1)^{1/2}}\)
Problem 36

Outer: \(\arcsin(u)\), Inner: \(u = 3x\), \(u' = 3\).

\(f'(x) = \dfrac{3}{\sqrt{1-(3x)^2}} = \dfrac{3}{\sqrt{1-9x^2}}\)
Problem 37

Outer: \(\arctan(u)\), Inner: \(u = x/2\), \(u' = 1/2\).

\(g'(x) = \dfrac{1/2}{1+(x/2)^2} = \dfrac{1/2}{1+x^2/4} = \dfrac{2}{4+x^2}\)
Problem 38

Let \(u = \dfrac{x^2+1}{x^2-1}\). Apply Quotient Rule inside, then Generalised Power Rule outside.

\(u' = \dfrac{2x(x^2-1) - 2x(x^2+1)}{(x^2-1)^2} = \dfrac{-4x}{(x^2-1)^2}\)

\(h'(x) = 3\!\left(\dfrac{x^2+1}{x^2-1}\right)^{\!2}\cdot\dfrac{-4x}{(x^2-1)^2} = \dfrac{-12x(x^2+1)^2}{(x^2-1)^4}\)
Problem 39

Apply log laws first: \(f(x) = 2\ln x - \ln(x^2+1)\).

\(f'(x) = \dfrac{2}{x} - \dfrac{2x}{x^2+1}\)

\(f'(x) = \dfrac{2}{x} - \dfrac{2x}{x^2+1}\)
Problem 40

Product Rule: \((e^{x^2})'\ln x + e^{x^2}(\ln x)'\).

\((e^{x^2})' = 2xe^{x^2}\) (Chain Rule).

\(g'(x) = 2xe^{x^2}\ln x + \dfrac{e^{x^2}}{x} = e^{x^2}\!\left(2x\ln x + \dfrac{1}{x}\right)\)
Problem 41

Write \(f(x) = \sin(3x+1)\cos(3x+1) = \dfrac{1}{2}\sin(2(3x+1)) = \dfrac{1}{2}\sin(6x+2)\).

Now differentiate: \(f'(x) = \dfrac{1}{2}\cos(6x+2)\cdot 6\)

\(f'(x) = 3\cos(6x+2)\)
Problem 42

Factor: \(x^4 - 2x^2 + 1 = (x^2-1)^2\), so \(h(x) = \sqrt{(x^2-1)^2} = |x^2-1|\).

Differentiating: for \(x^2 \neq 1\), \(h'(x) = \dfrac{(x^2-1)\cdot 2x}{|x^2-1|} = \text{sgn}(x^2-1)\cdot 2x\).

\(h'(x) = \dfrac{2x(x^2-1)}{|x^2-1|}\) for \(x\neq\pm 1\).
Problem 43

Product Rule: \((x^2)'\arctan x + x^2(\arctan x)'\).

\(f'(x) = 2x\arctan x + \dfrac{x^2}{1+x^2}\)
Problem 44

Quotient Rule: numerator derivative is \(\dfrac{2x}{x^2+1}\cdot x - \ln(x^2+1)\cdot 1\).

\(g'(x) = \dfrac{\dfrac{2x^2}{x^2+1} - \ln(x^2+1)}{x^2}\)
Problem 45

Let \(y = \tan(\arcsin x)\). Set \(\theta = \arcsin x\), so \(\sin\theta = x\) and \(\cos\theta = \sqrt{1-x^2}\).

Thus \(\tan(\arcsin x) = \dfrac{x}{\sqrt{1-x^2}}\). Differentiating directly:

\(\dfrac{d}{dx}\!\left[\dfrac{x}{\sqrt{1-x^2}}\right] = \dfrac{\sqrt{1-x^2} + \dfrac{x^2}{\sqrt{1-x^2}}}{1-x^2} = \dfrac{1}{(1-x^2)^{3/2}}\)

\(h'(x) = \dfrac{1}{(1-x^2)^{3/2}}\)

Part C — Challenging Problems (46–65)

Problem 46

Outer: \(e^u\), Inner: \(u = e^x\), \(u' = e^x\).

\(f'(x) = e^{e^x}\cdot e^x\)
Problem 47

Outer: \(\ln(u)\), Inner: \(u = \ln x\), \(u' = 1/x\).

\(g'(x) = \dfrac{1/x}{\ln x} = \dfrac{1}{x\ln x}\)
Problem 48

Three nested layers: \(\sin[\cos(\tan x)]\).

Layer 1 (outermost): \(-\sin(\cos(\tan x))\cdot\dfrac{d}{dx}[\cos(\tan x)]\).

Layer 2: \(-\sin(\tan x)\cdot\dfrac{d}{dx}[\tan x]\).

Layer 3: \(\sec^2 x\).

\(h'(x) = -\sin(\cos(\tan x))\cdot(-\sin(\tan x))\cdot\sec^2 x = \sin(\cos(\tan x))\sin(\tan x)\sec^2 x\)
Problem 49

Outer: \(e^u\), Inner: \(u = \cos(x^2)\), \(u' = -\sin(x^2)\cdot 2x\).

\(f'(x) = e^{\cos(x^2)}\cdot(-2x\sin(x^2)) = -2x\sin(x^2)\,e^{\cos(x^2)}\)
Problem 50

Outer: \(u^3\), Middle: \(\sin(v)\), Inner: \(v = \ln x\), \(v' = 1/x\).

\(3[\sin(\ln x)]^2\cdot\cos(\ln x)\cdot\dfrac{1}{x}\)

\(g'(x) = \dfrac{3\sin^2(\ln x)\cos(\ln x)}{x}\)
Problem 51

\(y = (4-x^2)^{1/2}\). \(y' = \dfrac{-x}{\sqrt{4-x^2}}\). At \(x=1\): slope \(= \dfrac{-1}{\sqrt{3}}\).

Point: \(y(1) = \sqrt{3}\). Tangent line: \(y - \sqrt{3} = -\dfrac{1}{\sqrt{3}}(x-1)\).

\(y = -\dfrac{x}{\sqrt{3}} + \dfrac{1}{\sqrt{3}} + \sqrt{3} = -\dfrac{x}{\sqrt{3}} + \dfrac{4}{\sqrt{3}}\)
Problem 52

\(y' = e^{-x^2/2}\cdot(-x)\). At \(x=0\): \(y'(0) = 0\) (horizontal tangent).

Point: \(y(0) = e^0 = 1\).

Tangent line: \(y = 1\)
Problem 53

\(y' = 2\cos(2x) = 0\) requires \(2x = \pi/2 + k\pi\), i.e.\ \(x = \pi/4 + k\pi/2\).

On \([0, 2\pi]\): \(x = \pi/4,\; 3\pi/4,\; 5\pi/4,\; 7\pi/4\).

\(x = \dfrac{\pi}{4},\dfrac{3\pi}{4},\dfrac{5\pi}{4},\dfrac{7\pi}{4}\)
Problem 54

Product Rule: \(g'(x) = 2xe^{-x} + x^2(-e^{-x}) = xe^{-x}(2-x)\).

Critical points: \(x = 0\) and \(x = 2\).

\(g'(x) = xe^{-x}(2-x) = 0 \implies x = 0 \text{ or } x = 2\)
Problem 55

\(h'(x) = 3(x^2-1)^2\cdot 2x = 6x(x^2-1)^2 = 0\).

Solutions: \(x = 0\) and \(x^2 = 1\implies x = \pm 1\).

\(x = 0,\; x = 1,\; x = -1\)
Problem 56

\(f'(x) = 2e^{2x}\). \(f''(x) = 2\cdot 2e^{2x}\)

\(f''(x) = 4e^{2x}\)
Problem 57

\(g'(x) = 3\cos(3x)\). \(g''(x) = 3\cdot(-\sin(3x))\cdot 3\)

\(g''(x) = -9\sin(3x)\)
Problem 58

\(h'(x) = \dfrac{2x}{x^2+1}\). Apply Quotient Rule:

\(h''(x) = \dfrac{2(x^2+1) - 2x\cdot 2x}{(x^2+1)^2} = \dfrac{2-2x^2}{(x^2+1)^2}\)

\(h''(x) = \dfrac{2(1-x^2)}{(x^2+1)^2}\)
Problem 59 — Proof

For \(f(x) > 0\): \(\ln|f(x)| = \ln(f(x))\). By Chain Rule with outer \(\ln(u)\) and inner \(u = f(x)\):

\(\dfrac{d}{dx}[\ln(f(x))] = \dfrac{f'(x)}{f(x)}\). \(\checkmark\)

For \(f(x) < 0\): \(\ln|f(x)| = \ln(-f(x))\). By Chain Rule: \(\dfrac{-f'(x)}{-f(x)} = \dfrac{f'(x)}{f(x)}\). \(\checkmark\)

Therefore \(\dfrac{d}{dx}[\ln|f(x)|] = \dfrac{f'(x)}{f(x)}\) for all \(f(x)\neq 0\). \(\blacksquare\)
Problem 60

Write \(a^x = e^{x\ln a}\). Outer: \(e^u\), Inner: \(u = x\ln a\), \(u' = \ln a\).

\(\dfrac{d}{dx}[a^x] = e^{x\ln a}\cdot\ln a = a^x\ln a\)
Problem 61

\(y = f(x^2)\). By Chain Rule: \(\dfrac{dy}{dx} = f'(x^2)\cdot 2x\).

At \(x=2\): inner value \(= x^2 = 4\). \(f'(4) = \sqrt{4+1} = \sqrt{5}\).

\(\dfrac{dy}{dx}\bigg|_{x=2} = \sqrt{5}\cdot 2\cdot 2 = 4\sqrt{5}\)
Problem 62

By the Chain Rule: \(\dfrac{d}{dx}[f(g(x))]\big|_{x=2} = f'(g(2))\cdot g'(2)\).

\(g(2) = 3\), so \(f'(g(2)) = f'(3) = 4\). And \(g'(2) = -1\).

\(= 4\cdot(-1) = -4\)
Problem 63 — Logarithmic Differentiation

Let \(y = x^x\). Take \(\ln\) of both sides: \(\ln y = x\ln x\).

Differentiate both sides (implicit differentiation on left, Product Rule on right):

\(\dfrac{y'}{y} = \ln x + x\cdot\dfrac{1}{x} = \ln x + 1\).

Multiply both sides by \(y = x^x\):

\(\dfrac{d}{dx}[x^x] = x^x(\ln x + 1)\)
Problem 64 — Logarithmic Differentiation

Let \(y = (x^2+1)^{\sin x}\). Take \(\ln\): \(\ln y = \sin x\cdot\ln(x^2+1)\).

Differentiate: \(\dfrac{y'}{y} = \cos x\cdot\ln(x^2+1) + \sin x\cdot\dfrac{2x}{x^2+1}\).

\(y' = (x^2+1)^{\sin x}\!\left[\cos x\ln(x^2+1) + \dfrac{2x\sin x}{x^2+1}\right]\)
Problem 65 — Triple Composition

By repeated application of the Chain Rule: \(F'(x) = f'(g(h(x)))\cdot g'(h(x))\cdot h'(x)\).

Evaluate at \(x=1\): \(h(1) = 2\), \(g(h(1)) = g(2) = 4\), \(f'(g(h(1))) = f'(4) = 6\).

\(F'(1) = 6\cdot g'(2)\cdot h'(1) = 6\cdot(-1)\cdot 3\)

\(F'(1) = -18\)

Application Problems (66–75)

Problem 66

\(h(t) = 50e^{-0.1t}\sin(2t)\). Product Rule + Chain Rule.

\(h'(t) = 50\left[(-0.1)e^{-0.1t}\sin(2t) + e^{-0.1t}\cdot 2\cos(2t)\right]\)

\(= 50e^{-0.1t}[-0.1\sin(2t) + 2\cos(2t)]\)

(a) \(h'(0) = 50\cdot 1\cdot[0 + 2] = 100\) m/s (moving upward).

(b) \(h'(\pi/4) = 50e^{-\pi/40}[-0.1\sin(\pi/2) + 2\cos(\pi/2)] = 50e^{-\pi/40}(-0.1) \approx -4.6\) m/s (moving downward).
Problem 67

Outer: \(e^u\), Inner: \(u = -0.05x^2\), \(u' = -0.1x\).

\(T'(x) = 80e^{-0.05x^2}\cdot(-0.1x) = -8xe^{-0.05x^2}\).

\(T'(2) = -16e^{-0.2} \approx -13.1\) °C/cm. Temperature decreasing at 2 cm.
Problem 68

\(v(t) = s'(t) = 3\cdot(-\sin(4\pi t))\cdot 4\pi = -12\pi\sin(4\pi t)\) cm/s.

\(a(t) = v'(t) = -12\pi\cos(4\pi t)\cdot 4\pi = -48\pi^2\cos(4\pi t)\) cm/s².

Check: \(-16\pi^2\cdot s(t) = -16\pi^2\cdot 3\cos(4\pi t) = -48\pi^2\cos(4\pi t) = a(t)\). \(\checkmark\)

Verified: \(a(t) = -16\pi^2 s(t)\), confirming simple harmonic motion.
Problem 69

\(\dfrac{dI}{dr} = -\dfrac{2I_0}{r^3}\). \(\dfrac{dr}{dt} = 3\).

\(\dfrac{dI}{dt} = \dfrac{dI}{dr}\cdot\dfrac{dr}{dt} = -\dfrac{2I_0}{r^3}\cdot 3 = \dfrac{-6I_0}{r^3}\).

Substituting \(r = 2+3t\): \(\dfrac{dI}{dt} = \dfrac{-6I_0}{(2+3t)^3}\)
Problem 70

Outer: \((1-e^u)\cdot Q_0\), but differentiate \(1-e^{-t/RC}\) with respect to \(t\).

Inner: \(u = -t/RC\), \(u' = -1/RC\).

\(I(t) = Q_0\cdot e^{-t/RC}\cdot\dfrac{1}{RC} = \dfrac{Q_0}{RC}e^{-t/RC}\).

As \(t\to\infty\), \(I(t)\to 0\): the current decays to zero as the capacitor fully charges.
Problem 71

Chain Rule with inner \(u = \pi x/L\).

\(y'(x) = A\cos\!\left(\dfrac{\pi x}{L}\right)\cdot\dfrac{\pi}{L} = \dfrac{A\pi}{L}\cos\!\left(\dfrac{\pi x}{L}\right)\).

\(y''(x) = -\dfrac{A\pi^2}{L^2}\sin\!\left(\dfrac{\pi x}{L}\right)\)
Problem 72

Chain Rule: \(\dfrac{dA}{dd} = A_0e^{-\alpha d}\cdot(-\alpha) = -\alpha A_0 e^{-\alpha d}\).

Since \(A(d) = A_0e^{-\alpha d}\), we have \(\dfrac{dA}{dd} = -\alpha A(d)\).

The signal decays at a rate proportional to its current amplitude — the hallmark of exponential decay.
Problem 73

\(v'(r) = v_{\max}\cdot\left(-\dfrac{2r}{R^2}\right) = -\dfrac{2v_{\max}r}{R^2}\).

This is negative for all \(r > 0\): velocity strictly decreases from centre outward.

Rate of decrease \(|v'(r)|\) is greatest at \(r = R\) (the wall), where \(|v'(R)| = \dfrac{2v_{\max}}{R}\).
Problem 74

Rewrite: \(C(t) = C_0(1+kt)^{-2}\). Generalised Power Rule.

\(\dfrac{dC}{dt} = C_0\cdot(-2)(1+kt)^{-3}\cdot k = \dfrac{-2C_0 k}{(1+kt)^3}\)

Negative: concentration is decreasing. The rate slows as \(t\) increases.
Problem 75

Differentiate \(\tan\theta = 10/x\) with respect to \(t\) using the Chain Rule (related rates).

\(\sec^2\theta\cdot\dfrac{d\theta}{dt} = -\dfrac{10}{x^2}\cdot\dfrac{dx}{dt}\)

At \(x = 10\): \(\tan\theta = 1\implies\theta = \pi/4\implies\sec^2\theta = 2\). \(\dfrac{dx}{dt} = 3\).

\(2\cdot\dfrac{d\theta}{dt} = -\dfrac{10}{100}\cdot 3 = -\dfrac{3}{10}\)

\(\dfrac{d\theta}{dt} = -\dfrac{3}{20}\) rad/s (angle is decreasing as the object moves away).

Multiple Choice Solutions (76–80)

Problem 76

Outer: \(u^4\), Inner: \(u = 3x-1\), \(u' = 3\).

\(f'(x) = 4(3x-1)^3\cdot 3 = 12(3x-1)^3\)

Answer: (B) — \(12(3x-1)^3\)
Problem 77

Outer: \(e^u\), Inner: \(u = \sin x\), \(u' = \cos x\).

Answer: (C) — \(\cos x\cdot e^{\sin x}\)
Problem 78

\(g'(x) = \dfrac{3x^2}{x^3+1}\). At \(x=1\): \(g'(1) = \dfrac{3}{2}\).

Answer: (B) — \(\dfrac{3}{2}\)
Problem 79

\(h(x) = [f(x)]^{1/2}\). \(h'(x) = \dfrac{f'(x)}{2\sqrt{f(x)}}\).

\(h'(4) = \dfrac{6}{2\sqrt{9}} = \dfrac{6}{6} = 1\)

Answer: (A) — \(1\)
Problem 80

(A) \(x\sin x\): Product Rule needed. (B) \(x^2 e^x\): Product Rule needed. (C) \(\sin(x^2)\): only Chain Rule (one function nested in another — no product). (D) \(x^2\ln x\): Product Rule needed.

Answer: (C) — \(y = \sin(x^2)\)

Frequently Asked Questions

What is the Chain Rule in calculus, and what is it used for?
The Chain Rule is the differentiation rule for composite functions — functions of the form \(f(g(x))\), where one function is nested inside another. It states that \(\dfrac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)\): differentiate the outer function at the inner function and multiply by the derivative of the inner function. In practice, the Chain Rule is one of the most frequently used rules in all of calculus, appearing whenever quantities depend on intermediate variables. It underlies implicit differentiation, related rates, the Fundamental Theorem of Calculus for composite integrands, and the backpropagation algorithm used to train modern neural networks.
How do I know when to use the Chain Rule versus the Product Rule?
The key distinction is structural. If two functions are multiplied together as independent factors — like \(x^2\cdot\sin x\) — use the Product Rule. If one function is nested inside another — like \(\sin(x^2)\), where \(x^2\) is the argument of sine — use the Chain Rule. Some expressions require both: for instance, \(x^2\sin(3x)\) needs the Product Rule for the overall product, and the Chain Rule for the factor \(\sin(3x)\). The mental test is simple: ask "are these two functions combined by multiplication (Product Rule) or by substitution/composition (Chain Rule)?"
What is the Generalised Power Rule and how does it relate to the Chain Rule?
The Generalised Power Rule is simply the Chain Rule applied when the outer function is a power: \(\dfrac{d}{dx}[u^n] = n\,u^{n-1}\cdot u'\), where \(u\) is any differentiable function of \(x\). It is the Chain Rule specialised to the most common outer function students encounter. For example, \(\dfrac{d}{dx}[(x^2+1)^5] = 5(x^2+1)^4\cdot 2x = 10x(x^2+1)^4\). Recognising this as a special case of the Chain Rule — rather than memorising a separate formula — is the conceptually clean approach. Most basic Chain Rule problems in a Calculus I course are Generalised Power Rule problems.
How do I apply the Chain Rule when there are three or more nested functions?
Apply the Chain Rule repeatedly, working from the outermost layer inward, and multiply all the derivatives together. For \(F(x) = f(g(h(x)))\), the rule is \(F'(x) = f'(g(h(x)))\cdot g'(h(x))\cdot h'(x)\). The pattern always follows the same principle: each derivative is evaluated at the remaining layers of the function below it, and all derivatives are multiplied in sequence. A useful mnemonic is to think of the Chain Rule as peeling an onion: you work from the outer skin inward, one layer at a time, multiplying each derivative as you go.
What are the most common Chain Rule mistakes, and how can I avoid them?
By far the most common error is forgetting to multiply by the derivative of the inner function — computing \(\cos(x^2)\) instead of \(-2x\sin(x^2)\) when differentiating \(\cos(x^2)\), for example. A reliable habit is to always write the inner function explicitly as \(u\), compute \(u'\), and then check that you have included the factor \(u'\) in your final answer. Other common errors include applying the Chain Rule to a product (use the Product Rule instead), stopping after one application when there are multiple composition layers, and misidentifying which function is outer versus inner. Practising with the \(u\)-substitution notation before moving to mental calculations is strongly recommended for beginners.
How is the Chain Rule connected to the Leibniz notation \(dy/dx\)?
One of the most elegant aspects of the Leibniz notation is that the Chain Rule looks literally like fraction cancellation: if \(y = f(u)\) and \(u = g(x)\), then \(\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}\). The "\(du\)" terms appear to cancel, leaving \(\dfrac{dy}{dx}\). While this is not a rigorous justification (since \(dy\), \(du\), \(dx\) are not independent fractions), it is a powerful mnemonic that motivates the structure of the rule. In multivariable calculus, the Chain Rule in Leibniz form extends naturally to partial derivatives: \(\dfrac{\partial z}{\partial x} = \dfrac{\partial z}{\partial u}\cdot\dfrac{\partial u}{\partial x} + \dfrac{\partial z}{\partial v}\cdot\dfrac{\partial v}{\partial x}\), etc.
Does the Chain Rule work for implicit differentiation?
Yes. Implicit differentiation is an application of the Chain Rule. When we differentiate an expression like \(y^2\) with respect to \(x\), we treat \(y\) as a function of \(x\) and apply the Chain Rule: \(\dfrac{d}{dx}[y^2] = 2y\cdot\dfrac{dy}{dx}\). The factor \(\dfrac{dy}{dx}\) appears because differentiating the outer function \((\cdot)^2\) gives \(2y\), and then the Chain Rule requires multiplying by the derivative of the inner function \(y\) with respect to \(x\). Every step of implicit differentiation involving a function of \(y\) is an application of the Chain Rule. For further practice with implicit differentiation, visit the full worksheets library at DerivativeCalculus.com.
Where can I find more chain rule practice problems and a derivative calculator?
You can find additional worksheets on the Power Rule, Product Rule, Quotient Rule, implicit differentiation, related rates, and more at DerivativeCalculus.com/worksheets.html. The main site at DerivativeCalculus.com also features a step-by-step derivative calculator that can verify your Chain Rule answers and show intermediate working. For self-study, work through this worksheet in order, check each answer immediately, and revisit any problem type you miss more than once.
Created by DerivativeCalculus.com — https://www.derivativecalculus.com/worksheets.html

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