📝 Chain rule for derivatives (19 MCQs)

📖 From Calculus • 3. The Derivation • 19 questions available

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What is Chain rule for derivatives?

Definition:
The chain rule states that the derivative of a composite function $f(g(x))$ is the derivative of the outer function evaluated at the inner function times the derivative of the inner function, expressed as $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$ .

Example:
For $f(x) = (x^2 + 1)^3$ , let $u = x^2 + 1$ , then $f'(x) = 3u^2 \cdot 2x = 6x(x^2 + 1)^2$ .

Reason:
The chain rule is indispensable for differentiating nested functions, which are ubiquitous in real-world models, allowing breakdown of complex compositions into manageable differentiation steps.

Easy

Medium

Hard

📝 All Chain rule for derivatives MCQs

Q1. If a car travels at 30 miles per gallon and gasoline costs $5 per gallon, what is the rate of miles per dollar?

A.5 miles per dollar

B.6 miles per dollar ✅

C.150 miles per dollar

D.0.6 miles per dollar

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Using the chain rule, the mileage per dollar equals (30 miles per gallon) × (1 gallon / 5 dollars) = 30 × 0.2 = 6 miles per dollar. The calculation treats the two rates as multiplicative factors, giving the correct answer of 6 miles per dollar.

Q2. Let $f(u)=u^{2}$ and $g(x)=3x+1$ . What is $\frac{dy}{dx}$ at $x=2$ for $y=f(g(x))$ ?

A.12

B.21

C.36

D.42 ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: First compute $g(2)=7$ . Then $f'(u)=2u$ gives $f'(7)=14$ . The derivative of $g$ is $g'(x)=3$ . By the chain rule, $\frac{dy}{dx}=f'(g(x))g'(x)=14\cdot3=42$ . Thus the correct value is 42.

Q3. Given $\frac{dy}{du}=4u$ and $\frac{du}{dx}=3x^{2}$ , find $\frac{dy}{dx}$ when $u=2$ and $x=1$ .

A.24 ✅

B.12

C.8

D.6

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Apply the chain rule: $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=4u\cdot3x^{2}$ . Substituting $u=2$ and $x=1$ yields $4\cdot2\cdot3\cdot1^{2}=24$ . Hence the derivative at the specified point is 24.

Q4. Suppose temperature $T$ depends on pressure $P$ via $T=f(P)$ and pressure depends on altitude $h$ via $P=g(h)$ . If $f'(P)=0.5$ and $g'(h)=-0.02$ , what is the sign of $\frac{dT}{dh}$ ?

A.Positive

B.Negative ✅

C.Zero

D.Undefined

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The chain rule gives $\frac{dT}{dh}=f'(P)g'(h)=0.5\cdot(-0.02)=-0.01$ . The product of a positive and a negative number is negative, so the rate of change of temperature with respect to altitude is negative.

Q5. If $y=\sin(u)$ with $u=e^{x}$ , which expression correctly represents $\frac{dy}{dx}$ ?

\cos(e^{x})

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

✅

\sin(e^{x})\cdot e^{x}

-\cos(e^{x})

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: First differentiate the outer function: $\frac{dy}{du}=\cos u$ . Then differentiate the inner function: $\frac{du}{dx}=e^{x}$ . The chain rule multiplies these results, giving $\frac{dy}{dx}=\cos(e^{x})\cdot e^{x}$ .

Q6. For the composition $y=f(g(h(x)))$ where $f'(u)=2u$ , $g'(v)=\frac{1}{v}$ , and $h'(x)=\ln x$ , find $\frac{dy}{dx}$ at $x=e$ given $h(e)=1$ and $g(1)=2$ .

A.2

B.6

C.8

D.4 ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Compute each derivative: $h'(e)=\ln e=1$ . Then $g'(h(e))=1/1=1$ . Next $f'(g(h(e)))=2\cdot2=4$ . Multiplying via the chain rule gives $\frac{dy}{dx}=4\cdot1\cdot1=4$ . Thus the correct answer is 4.

Q7. Which statement best describes the chain rule?

A.It applies only to sums

B.It can be derived from the product rule

C.It relates derivatives of compositions ✅

D.It is the same as the quotient rule

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The chain rule specifically addresses how to differentiate a composite function $f(g(x))$ . It states that the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function, yields the derivative of the composition. This distinguishes it from rules for sums, products, or quotients.

Q8. Which function is NOT suitable for direct application of the chain rule because it is not a composition of differentiable functions?

\sqrt{x^{2}+1}

\ln(\sin x)

e^{x^{3}}

x^{2}+3

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The expression $x^{2}+3$ is a simple polynomial, not a composition of two functions; therefore the chain rule is unnecessary. The other options each involve an inner function (e.g., $x^{2}$ , $\sin x$ , $x^{3}$ ) wrapped by an outer function, making the chain rule applicable.

Q9. Given $f(x)=x^{2}$ and $g(x)=\sqrt{x}$ , which expression equals $\frac{d}{dx}[f(g(x))]$ ?

2\sqrt{x}

B.1 ✅

\frac{1}{\sqrt{x}}

2x

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: First compute the composition: $f(g(x))=(\sqrt{x})^{2}=x$ . The derivative of $x$ with respect to $x$ is 1. Thus $\frac{d}{dx}[f(g(x))]=1$ , matching option B.

Q10. For $y=\ln(\cos(5x))$ , which expression correctly represents $\frac{dy}{dx}$ ?

-5\frac{\sin(5x)}{\cos(5x)}

✅

-5\frac{\cos(5x)}{\sin(5x)}

5\frac{\sin(5x)}{\cos(5x)}

5\frac{\cos(5x)}{\sin(5x)}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiate the outer $\ln$ to get $\frac{1}{\cos(5x)}$ and multiply by the derivative of the inner $\cos(5x)$ , which is $-5\sin(5x)$ . The product yields $-5\sin(5x)/\cos(5x)$ , i.e., $-5\tan(5x)$ , matching option A.

Q11. Suppose $y=(3x^{2}+2)^{5}$ . Using the chain rule, what is $\frac{d^{2}y}{dx^{2}}$ at $x=1$ ?

A.108750 ✅

B.54375

C.216000

D.54000

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: First derivative: $y' =5(3x^{2}+2)^{4}\cdot6x =30x(3x^{2}+2)^{4}$ . Second derivative uses product and chain rules, giving $y'' =30(3x^{2}+2)^{4}+30x\cdot4(3x^{2}+2)^{3}\cdot6x$ . At $x=1$ , $3x^{2}+2=5$ . Substituting yields $30\cdot5^{4}+30\cdot1\cdot4\cdot5^{3}\cdot6 =108750$ .

Q12. Let $h(x)=\tan^{-1}(e^{\sin x})$ . Find $\frac{dh}{dx}$ at $x=0$ .

\frac12

✅

B.1

C.0

D.2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Derivative of $\tan^{-1}u$ is $\frac{1}{1+u^{2}}\,u'$ . Here $u=e^{\sin x}$ , so $u' = e^{\sin x}\cos x$ . At $x=0$ , $\sin0=0$ gives $u=1$ and $\cos0=1$ . Thus $\frac{dh}{dx}= \frac{1}{1+1^{2}}\cdot1\cdot1 =\frac12$ .

Q13. Why can the chain rule be viewed as “multiplying” rates of change?

A.Because derivatives are fractions ✅

B.Because composition adds rates

C.Because integration does that

D.None of the above

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Derivatives can be interpreted as infinitesimal ratios, such as $\frac{dy}{du}$ and $\frac{du}{dx}$ . When these ratios are multiplied, the intermediate variable $u$ cancels, leaving $\frac{dy}{dx}$ . This cancellation mirrors the algebraic manipulation of fractions, justifying the description of the chain rule as multiplying rates.

Q14. Which scenario best illustrates the chain rule?

A.Adding distances

B.Multiplying speeds

C.Changing temperature with altitude ✅

D.Counting objects

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: When temperature depends on pressure and pressure depends on altitude, the overall change of temperature with altitude involves the composition of two relationships. Applying the chain rule captures this nested dependence, making the temperature‑altitude example a clear illustration of the rule.

Q15. In the chain rule, what does the intermediate variable $u$ represent?

A.The final output

B.The independent variable

C.A placeholder for the inner function ✅

D.The derivative itself

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The symbol $u$ is introduced to denote the inner function $g(x)$ when expressing a composition $y=f(g(x))$ as $y=f(u)$ . It serves as a convenient placeholder that isolates the inner relationship, allowing the derivative $\frac{dy}{du}$ to be computed before re‑substituting $u=g(x)$ .

Q16. If $y=f(g(x))$ is differentiable, which must be true?

A.Both

f

and

g

are differentiable at relevant points ✅

B.Only

f

is differentiable

C.Only

g

is differentiable

D.Neither need be differentiable

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The chain rule requires that the inner function $g$ be differentiable at the point $x$ and that the outer function $f$ be differentiable at the value $g(x)$ . Without both derivatives existing, the product $\frac{dy}{du}\frac{du}{dx}$ cannot be formed, so differentiability of both is essential.

Q17. Consider $y=\sqrt{1+\sin^{2}(x)}$ . Using the chain rule, which expression is the derivative?

\frac{\sin x\cos x}{\sqrt{1+\sin^{2}x}}

✅

\frac{2\sin x\cos x}{\sqrt{1+\sin^{2}x}}

\frac{\sin x\cos x}{2\sqrt{1+\sin^{2}x}}

\frac{2\sin x\cos x}{1+\sin^{2}x}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Let $u=1+\sin^{2}x$ ; then $y=\sqrt{u}=u^{1/2}$ . Derivative $dy/du=\frac12 u^{-1/2}$ . Next $du/dx=2\sin x\cos x$ . Multiplying gives $\frac{dy}{dx}=\frac12 u^{-1/2}\cdot2\sin x\cos x =\frac{\sin x\cos x}{\sqrt{1+\sin^{2}x}}$ .

Q18. For $F(x)=\ln\big((x^{2}+1)^{3}\big)$ , which expression correctly applies the chain rule before simplifying?

\frac{3\cdot2x}{x^{2}+1}

✅

\frac{6x}{(x^{2}+1)^{3}}

\frac{2x}{x^{2}+1}

\frac{3}{2x(x^{2}+1)}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Write $F(x)=\ln(u)$ with $u=(x^{2}+1)^{3}$ . Then $F' = \frac{1}{u}\cdot u'$ . Differentiate $u$ : $u' =3(x^{2}+1)^{2}\cdot2x =6x(x^{2}+1)^{2}$ . Substituting gives $F' =\frac{6x(x^{2}+1)^{2}}{(x^{2}+1)^{3}} =\frac{3\cdot2x}{x^{2}+1}$ , which matches option A.

Q19. According to the chain rule theorem, if $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then the composition $f\circ g$ is differentiable at...

g(x)

x

✅

f(x)

D.None of the above

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The theorem states that the composition

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