Calculus 1 Quiz: Chain Rule
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Chain RuleQuestion 1 of 20

Let ff be a differentiable function such that f(2)=2f(2)=2 and f′(2)=3f'(2)=3. If h(x)=f(f(x))h(x) = f(f(x)), what is the value of h′(2)h'(2)?

99
66
44
1212
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Calculus 1 Quiz

Calculus 1 Quiz: Chain Rule

Practice Chain Rule in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Chain Rule, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Let ff be a differentiable function such that f(2)=2f(2)=2 and f′(2)=3f'(2)=3. If h(x)=f(f(x))h(x) = f(f(x)), what is the value of h′(2)h'(2)?

  1. 99 (correct answer)
  2. 66
  3. 44
  4. 1212
Explanation: When you encounter a composite function like h(x)=f(f(x))h(x) = f(f(x)) and need to find its derivative, you're dealing with the chain rule. This is a fundamental differentiation technique that applies when one function is nested inside another. To find h′(x)h'(x) where h(x)=f(f(x))h(x) = f(f(x)), you need to use the chain rule: the derivative of the outer function times the derivative of the inner function. Here, the outer function is ff evaluated at f(x)f(x), and the inner function is f(x)f(x). So h′(x)=f′(f(x))⋅f′(x)h'(x) = f'(f(x)) \cdot f'(x). At x=2x = 2: h′(2)=f′(f(2))⋅f′(2)h'(2) = f'(f(2)) \cdot f'(2). Since f(2)=2f(2) = 2, we have f′(f(2))=f′(2)=3f'(f(2)) = f'(2) = 3. Therefore: h′(2)=f′(2)⋅f′(2)=3⋅3=9h'(2) = f'(2) \cdot f'(2) = 3 \cdot 3 = 9. The answer is A) 99. Looking at the wrong choices: B) 66 likely comes from adding f′(2)+f′(2)=3+3=6f'(2) + f'(2) = 3 + 3 = 6 instead of multiplying—a common error when students confuse addition and multiplication rules. C) 44 might result from incorrectly using f(2)+f′(2)=2+2=4f(2) + f'(2) = 2 + 2 = 4, mixing up function values with derivatives. D) 1212 could come from incorrectly calculating f′(2)⋅f(2)⋅f′(2)=3⋅2⋅2=12f'(2) \cdot f(2) \cdot f'(2) = 3 \cdot 2 \cdot 2 = 12, unnecessarily including the function value. Remember: when differentiating composite functions, always identify the outer and inner functions clearly, then multiply their derivatives at the appropriate points. The chain rule requires multiplication, never addition.

Question 2

Let h(x)=f(x2)h(x) = f(x^2). If f′(4)=6f'(4) = 6 and f′′(4)=2f''(4) = 2, what is the value of h′′(2)h''(2)?

  1. 8
  2. 20
  3. 32
  4. 44 (correct answer)
Explanation: First, we find the first derivative of h(x)h(x) using the chain rule: h′(x)=f′(x2)⋅2xh'(x) = f'(x^2) \cdot 2x. To find the second derivative, we must use the product rule and the chain rule again: h′′(x)=ddx[f′(x2)]⋅(2x)+f′(x2)⋅ddx[2x]h''(x) = \frac{d}{dx}[f'(x^2)] \cdot (2x) + f'(x^2) \cdot \frac{d}{dx}[2x]. The first term requires another chain rule: h′′(x)=[f′′(x2)⋅2x]⋅(2x)+f′(x2)⋅2=4x2f′′(x2)+2f′(x2)h''(x) = [f''(x^2) \cdot 2x] \cdot (2x) + f'(x^2) \cdot 2 = 4x^2 f''(x^2) + 2f'(x^2). Now, we evaluate at x=2x=2: h′′(2)=4(22)f′′(22)+2f′(22)=16f′′(4)+2f′(4)h''(2) = 4(2^2) f''(2^2) + 2f'(2^2) = 16f''(4) + 2f'(4). Substituting the given values: h′′(2)=16(2)+2(6)=32+12=44h''(2) = 16(2) + 2(6) = 32 + 12 = 44.

Question 3

Let f(x)=g(x3)f(x) = g(x^3) and g(x)=h(2x)g(x) = h(2x). If h′(2)=4h'(2)=4, what is f′(1)f'(1)?

  1. 6
  2. 12
  3. 24 (correct answer)
  4. 36
Explanation: This requires two applications of the chain rule. First, let's find f′(x)f'(x). f′(x)=g′(x3)⋅ddx(x3)=g′(x3)⋅3x2f'(x) = g'(x^3) \cdot \frac{d}{dx}(x^3) = g'(x^3) \cdot 3x^2. Now we need to find an expression for g′(x)g'(x). From g(x)=h(2x)g(x) = h(2x), we differentiate to get g′(x)=h′(2x)⋅ddx(2x)=h′(2x)⋅2g'(x) = h'(2x) \cdot \frac{d}{dx}(2x) = h'(2x) \cdot 2. Now substitute this back into the expression for f′(x)f'(x): f′(x)=[h′(2x3)⋅2]⋅3x2=6x2h′(2x3)f'(x) = [h'(2x^3) \cdot 2] \cdot 3x^2 = 6x^2 h'(2x^3). Finally, evaluate at x=1x=1: f′(1)=6(1)2h′(2(1)3)=6h′(2)f'(1) = 6(1)^2 h'(2(1)^3) = 6h'(2). Given h′(2)=4h'(2)=4, we have f′(1)=6(4)=24f'(1) = 6(4) = 24.

Question 4

If g(x)=f−1(x)g(x) = f^{-1}(x) and f(x)=x3+2x+1f(x) = x^3 + 2x + 1, what is the value of g′(4)g'(4)?

  1. 1/31
  2. 1/7
  3. 1/14
  4. 1/5 (correct answer)
Explanation: When you encounter inverse function derivatives, remember that if g(x)=f−1(x)g(x) = f^{-1}(x), then g′(x)=1f′(g(x))g'(x) = \frac{1}{f'(g(x))}. This formula connects the derivative of an inverse function to the derivative of the original function. To find g′(4)g'(4), you need two things: the value of g(4)g(4) and f′(g(4))f'(g(4)). Since g(4)=f−1(4)g(4) = f^{-1}(4), you must first solve f(x)=4f(x) = 4. Setting x3+2x+1=4x^3 + 2x + 1 = 4 gives you x3+2x−3=0x^3 + 2x - 3 = 0. Testing x=1x = 1: 1+2−3=01 + 2 - 3 = 0 ✓. So g(4)=1g(4) = 1. Next, find f′(x)=3x2+2f'(x) = 3x^2 + 2. Therefore, f′(1)=3(1)2+2=5f'(1) = 3(1)^2 + 2 = 5. Using the inverse derivative formula: g′(4)=1f′(g(4))=1f′(1)=15g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(1)} = \frac{1}{5}. Answer (A) 131\frac{1}{31} likely comes from incorrectly calculating f′(1)f'(1) as 32+22=133^2 + 2^2 = 13 or making an arithmetic error. Answer (B) 17\frac{1}{7} might result from computing f′(1)=3(1)+2+2=7f'(1) = 3(1) + 2 + 2 = 7, forgetting the squared term. Answer (C) 114\frac{1}{14} could arise from doubling the correct denominator or other computational mistakes. The key strategy here is methodical execution: solve for the inverse value first, then apply the inverse derivative formula. Always double-check that you've correctly identified where to evaluate the original function's derivative—it's at the inverse value, not the original input.

Question 5

Suppose yy is a function of xx such that ey2=x2+1e^{y^2} = x^2+1. What is the value of dydx\frac{dy}{dx} at the point (e−1,1)( \sqrt{e-1}, 1 )?

  1. e−1e\frac{\sqrt{e-1}}{e} (correct answer)
  2. 2e−1e\frac{2\sqrt{e-1}}{e}
  3. 2ee−12e\sqrt{e-1}
  4. e−1\sqrt{e-1}
Explanation: We use implicit differentiation. Differentiating both sides of ey2=x2+1e^{y^2} = x^2+1 with respect to xx, we must use the chain rule on the left side: ddx(ey2)=ddx(x2+1)\frac{d}{dx}(e^{y^2}) = \frac{d}{dx}(x^2+1). This gives ey2⋅2ydydx=2xe^{y^2} \cdot 2y \frac{dy}{dx} = 2x. Solving for dydx\frac{dy}{dx} yields dydx=2x2yey2=xyey2\frac{dy}{dx} = \frac{2x}{2y e^{y^2}} = \frac{x}{y e^{y^2}}. Now we evaluate this derivative at the point (x,y)=(e−1,1)(x,y) = (\sqrt{e-1}, 1). dydx∣(e−1,1)=e−11⋅e12=e−1e\left.\frac{dy}{dx}\right|_{(\sqrt{e-1}, 1)} = \frac{\sqrt{e-1}}{1 \cdot e^{1^2}} = \frac{\sqrt{e-1}}{e}.

Question 6

If f(x)=x2+1f(x) = \sqrt{x^2+1}, find f′′(x)f''(x).

  1. 12x2+1\frac{1}{2\sqrt{x^2+1}}
  2. x(x2+1)3/2\frac{x}{(x^2+1)^{3/2}}
  3. −x2(x2+1)3/2\frac{-x^2}{(x^2+1)^{3/2}}
  4. 1(x2+1)3/2\frac{1}{(x^2+1)^{3/2}} (correct answer)
Explanation: This problem tests your ability to find second derivatives using the chain rule, particularly with composite functions involving radicals. When you see a function like f(x)=x2+1f(x) = \sqrt{x^2+1}, rewrite it as f(x)=(x2+1)1/2f(x) = (x^2+1)^{1/2} to make differentiation clearer. To find the first derivative, use the chain rule: f′(x)=12(x2+1)−1/2⋅2x=x(x2+1)1/2=xx2+1f'(x) = \frac{1}{2}(x^2+1)^{-1/2} \cdot 2x = \frac{x}{(x^2+1)^{1/2}} = \frac{x}{\sqrt{x^2+1}} For the second derivative, you need the quotient rule since f′(x)=x(x2+1)1/2f'(x) = \frac{x}{(x^2+1)^{1/2}}. Using ddx[uv]=u′v−uv′v2\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{u'v - uv'}{v^2}:
  • u=xu = x, so u′=1u' = 1
  • v=(x2+1)1/2v = (x^2+1)^{1/2}, so v′=x(x2+1)1/2v' = \frac{x}{(x^2+1)^{1/2}}
Therefore: f′′(x)=1⋅(x2+1)1/2−x⋅x(x2+1)1/2(x2+1)=(x2+1)−x2(x2+1)3/2=1(x2+1)3/2f''(x) = \frac{1 \cdot (x^2+1)^{1/2} - x \cdot \frac{x}{(x^2+1)^{1/2}}}{(x^2+1)} = \frac{(x^2+1) - x^2}{(x^2+1)^{3/2}} = \frac{1}{(x^2+1)^{3/2}} Answer A gives you the first derivative but missing the denominator's correct power. Answer B is actually the first derivative f′(x)f'(x), not the second. Answer C has the wrong sign and numerator—this might come from sign errors in the quotient rule. Study tip: When finding second derivatives of radical functions, always convert to exponential form first, and double-check your quotient rule algebra. The denominator's exponent typically increases with each derivative.

Question 7

The derivative of f(x)=ln⁡(e2x+1)f(x) = \ln(\sqrt{e^{2x} + 1}) is:

  1. e2xe2x+1\frac{e^{2x}}{e^{2x}+1} (correct answer)
  2. 12(e2x+1)\frac{1}{2(e^{2x}+1)}
  3. e2xe2x+1\frac{e^{2x}}{\sqrt{e^{2x}+1}}
  4. 2e2xe2x+1\frac{2e^{2x}}{e^{2x}+1}
Explanation: First, simplify the function using logarithm properties: f(x)=ln⁡((e2x+1)1/2)=12ln⁡(e2x+1)f(x) = \ln((e^{2x} + 1)^{1/2}) = \frac{1}{2} \ln(e^{2x} + 1). Now, differentiate using the chain rule. Let u=e2x+1u = e^{2x} + 1, so u′=2e2xu' = 2e^{2x}. The derivative is f′(x)=12⋅1e2x+1⋅(2e2x)=2e2x2(e2x+1)=e2xe2x+1f'(x) = \frac{1}{2} \cdot \frac{1}{e^{2x}+1} \cdot (2e^{2x}) = \frac{2e^{2x}}{2(e^{2x}+1)} = \frac{e^{2x}}{e^{2x}+1}.

Question 8

Let h(x)=1g(x)+1h(x) = \frac{1}{g(x)+1}. If g(3)=−2g(3)=-2 and g′(3)=5g'(3)=5, find h′(3)h'(3).

  1. -5 (correct answer)
  2. -5/4
  3. 5/9
  4. 5
Explanation: We can write h(x)=(g(x)+1)−1h(x) = (g(x)+1)^{-1}. Now we use the chain rule (general power rule). Let u=g(x)+1u = g(x)+1, so u′=g′(x)u' = g'(x). The derivative is h′(x)=−1(g(x)+1)−2⋅g′(x)=−g′(x)(g(x)+1)2h'(x) = -1(g(x)+1)^{-2} \cdot g'(x) = \frac{-g'(x)}{(g(x)+1)^2}. Now, substitute the given values for x=3x=3: h′(3)=−g′(3)(g(3)+1)2=−5(−2+1)2=−5(−1)2=−51=−5h'(3) = \frac{-g'(3)}{(g(3)+1)^2} = \frac{-5}{(-2+1)^2} = \frac{-5}{(-1)^2} = \frac{-5}{1} = -5.

Question 9

Let g(x)=(x−1x+1)3g(x) = \left(\frac{x-1}{x+1}\right)^3. What is g′(3)g'(3)?

  1. 18\frac{1}{8}
  2. 932\frac{9}{32}
  3. 332\frac{3}{32} (correct answer)
  4. 34\frac{3}{4}
Explanation: This requires the chain rule and the quotient rule. Let the outer function be f(u)=u3f(u) = u^3 and the inner function be u(x)=x−1x+1u(x) = \frac{x-1}{x+1}. By the chain rule, g′(x)=3(x−1x+1)2⋅u′(x)g'(x) = 3\left(\frac{x-1}{x+1}\right)^2 \cdot u'(x). To find u′(x)u'(x), we use the quotient rule: u′(x)=(1)(x+1)−(x−1)(1)(x+1)2=x+1−x+1(x+1)2=2(x+1)2u'(x) = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}. Substituting this back, we get g′(x)=3(x−1x+1)2⋅2(x+1)2=6(x−1)2(x+1)4g'(x) = 3\left(\frac{x-1}{x+1}\right)^2 \cdot \frac{2}{(x+1)^2} = \frac{6(x-1)^2}{(x+1)^4}. Now, evaluate at x=3x=3: g′(3)=6(3−1)2(3+1)4=6(22)44=6⋅4256=24256g'(3) = \frac{6(3-1)^2}{(3+1)^4} = \frac{6(2^2)}{4^4} = \frac{6 \cdot 4}{256} = \frac{24}{256}, which simplifies to 332\frac{3}{32}.

Question 10

What is the slope of the line tangent to the curve y=arctan⁡(e2x)y = \arctan(e^{2x}) at the point where x=0x=0?

  1. 1/2
  2. 1 (correct answer)
  3. 2
  4. π/4\pi/4
Explanation: The slope of the tangent line is given by the derivative dydx\frac{dy}{dx}. We use the chain rule. The derivative of arctan⁡(u)\arctan(u) is 11+u2\frac{1}{1+u^2}, and the derivative of u=e2xu=e^{2x} is 2e2x2e^{2x}. So, dydx=11+(e2x)2⋅2e2x=2e2x1+e4x\frac{dy}{dx} = \frac{1}{1+(e^{2x})^2} \cdot 2e^{2x} = \frac{2e^{2x}}{1+e^{4x}}. Now we evaluate this derivative at x=0x=0. dydx∣x=0=2e2(0)1+e4(0)=2e01+e0=2(1)1+1=22=1\left.\frac{dy}{dx}\right|_{x=0} = \frac{2e^{2(0)}}{1+e^{4(0)}} = \frac{2e^0}{1+e^0} = \frac{2(1)}{1+1} = \frac{2}{2} = 1.

Question 11

Given g(x)=f(sin⁡2(x))g(x) = f(\sin^2(x)), with f′(0)=3f'(0) = 3. Find g′(0)g'(0).

  1. 0 (correct answer)
  2. 1
  3. 3
  4. 6
Explanation: We use the chain rule twice. Let u(x)=sin⁡2(x)=(sin⁡x)2u(x) = \sin^2(x) = (\sin x)^2. First, find u′(x)u'(x) using the chain rule: u′(x)=2(sin⁡x)1⋅cos⁡x=2sin⁡xcos⁡xu'(x) = 2(\sin x)^1 \cdot \cos x = 2\sin x \cos x. Now, find g′(x)g'(x) using the chain rule: g′(x)=f′(u(x))⋅u′(x)=f′(sin⁡2(x))⋅(2sin⁡xcos⁡x)g'(x) = f'(u(x)) \cdot u'(x) = f'(\sin^2(x)) \cdot (2\sin x \cos x). We evaluate this at x=0x=0. g′(0)=f′(sin⁡2(0))⋅(2sin⁡0cos⁡0)g'(0) = f'(\sin^2(0)) \cdot (2\sin 0 \cos 0). Since sin⁡(0)=0\sin(0)=0 and cos⁡(0)=1\cos(0)=1, this becomes g′(0)=f′(02)⋅(2⋅0⋅1)=f′(0)⋅0=3⋅0=0g'(0) = f'(0^2) \cdot (2 \cdot 0 \cdot 1) = f'(0) \cdot 0 = 3 \cdot 0 = 0.

Question 12

Let h(x)=f(x)2+g(x)2h(x) = \sqrt{f(x)^2 + g(x)^2}. Given f(1)=3f(1)=3, f′(1)=4f'(1)=4, g(1)=4g(1)=4, and g′(1)=−3g'(1)=-3, find h′(1)h'(1).

  1. 0 (correct answer)
  2. 1
  3. 5
  4. 7/5
Explanation: We write h(x)=(f(x)2+g(x)2)1/2h(x) = (f(x)^2 + g(x)^2)^{1/2} and use the chain rule. Let u=f(x)2+g(x)2u = f(x)^2 + g(x)^2. Then u′=2f(x)f′(x)+2g(x)g′(x)u' = 2f(x)f'(x) + 2g(x)g'(x). The derivative of h(x)h(x) is h′(x)=12(f(x)2+g(x)2)−1/2⋅(2f(x)f′(x)+2g(x)g′(x))=f(x)f′(x)+g(x)g′(x)f(x)2+g(x)2h'(x) = \frac{1}{2} (f(x)^2 + g(x)^2)^{-1/2} \cdot (2f(x)f'(x) + 2g(x)g'(x)) = \frac{f(x)f'(x) + g(x)g'(x)}{\sqrt{f(x)^2 + g(x)^2}}. Now we substitute the given values at x=1x=1: h′(1)=f(1)f′(1)+g(1)g′(1)f(1)2+g(1)2=(3)(4)+(4)(−3)32+42=12−129+16=025=0h'(1) = \frac{f(1)f'(1) + g(1)g'(1)}{\sqrt{f(1)^2 + g(1)^2}} = \frac{(3)(4) + (4)(-3)}{\sqrt{3^2 + 4^2}} = \frac{12 - 12}{\sqrt{9+16}} = \frac{0}{\sqrt{25}} = 0.

Question 13

Let y=sin⁡3(2x)y = \sin^3(2x). What is dydx\frac{dy}{dx}?

  1. 3cos⁡2(2x)3\cos^2(2x)
  2. 3sin⁡2(2x)cos⁡(2x)3\sin^2(2x)\cos(2x)
  3. 6cos⁡2(2x)6\cos^2(2x)
  4. 6sin⁡2(2x)cos⁡(2x)6\sin^2(2x)\cos(2x) (correct answer)
Explanation: When you encounter a composite function like y=sin⁡3(2x)y = \sin^3(2x), you need to apply the chain rule systematically. This function has two layers of composition: an outer function (cubing) applied to sin⁡(2x)\sin(2x), and an inner function where sin⁡\sin is applied to 2x2x. To find dydx\frac{dy}{dx}, work from the outside in. First, treat sin⁡(2x)\sin(2x) as a single unit and differentiate the cubic: ddx[sin⁡(2x)]3=3[sin⁡(2x)]2⋅ddx[sin⁡(2x)]\frac{d}{dx}[\sin(2x)]^3 = 3[\sin(2x)]^2 \cdot \frac{d}{dx}[\sin(2x)]. This gives you 3sin⁡2(2x)3\sin^2(2x) times the derivative of sin⁡(2x)\sin(2x). Next, differentiate sin⁡(2x)\sin(2x): ddx[sin⁡(2x)]=cos⁡(2x)⋅ddx[2x]=cos⁡(2x)⋅2=2cos⁡(2x)\frac{d}{dx}[\sin(2x)] = \cos(2x) \cdot \frac{d}{dx}[2x] = \cos(2x) \cdot 2 = 2\cos(2x). Combining these steps: dydx=3sin⁡2(2x)⋅2cos⁡(2x)=6sin⁡2(2x)cos⁡(2x)\frac{dy}{dx} = 3\sin^2(2x) \cdot 2\cos(2x) = 6\sin^2(2x)\cos(2x). Choice A (3cos⁡2(2x)3\cos^2(2x)) incorrectly differentiates the trigonometric function and misses the chain rule entirely. Choice B (3sin⁡2(2x)cos⁡(2x)3\sin^2(2x)\cos(2x)) correctly applies the power rule and differentiates sin⁡(2x)\sin(2x) to cos⁡(2x)\cos(2x), but forgets to multiply by the derivative of the innermost function 2x2x, missing the factor of 2. Choice C (6cos⁡2(2x)6\cos^2(2x)) has the correct coefficient but wrong trigonometric functions. The correct answer is D: 6sin⁡2(2x)cos⁡(2x)6\sin^2(2x)\cos(2x). Strategy tip: For nested functions, always identify each layer and apply the chain rule step-by-step. Don't forget that the derivative of kxkx is kk, not 1.

Question 14

What is the y-intercept of the tangent line to the graph of f(x)=ln⁡(x2−3x+e)f(x) = \ln(x^2 - 3x + e) at the point where x=3x=3?

  1. 3e\frac{3}{e}
  2. 1−3e1 - \frac{3}{e}
  3. −9e-\frac{9}{e}
  4. 1−9e1 - \frac{9}{e} (correct answer)
Explanation: First, find the point of tangency. At x=3x=3, y=f(3)=ln⁡(32−3(3)+e)=ln⁡(9−9+e)=ln⁡(e)=1y = f(3) = \ln(3^2 - 3(3) + e) = \ln(9-9+e) = \ln(e) = 1. The point is (3,1)(3,1). Next, find the slope of the tangent line by finding the derivative, f′(x)f'(x). Using the chain rule, f′(x)=1x2−3x+e⋅(2x−3)f'(x) = \frac{1}{x^2 - 3x + e} \cdot (2x-3). The slope at x=3x=3 is m=f′(3)=132−3(3)+e⋅(2(3)−3)=1e⋅(3)=3em = f'(3) = \frac{1}{3^2 - 3(3) + e} \cdot (2(3)-3) = \frac{1}{e} \cdot (3) = \frac{3}{e}. The equation of the tangent line is y−y0=m(x−x0)y - y_0 = m(x-x_0), which is y−1=3e(x−3)y - 1 = \frac{3}{e}(x-3). To find the y-intercept, set x=0x=0: y−1=3e(0−3)=−9ey - 1 = \frac{3}{e}(0-3) = -\frac{9}{e}. Thus, the y-intercept is y=1−9ey = 1 - \frac{9}{e}.

Question 15

Let H(x)=(f∘g)(x)=f(g(x))H(x) = (f \circ g)(x) = f(g(x)). Suppose that g(1)=3g(1)=3, g′(1)=2g'(1)=2, and H′(1)=0H'(1)=0. Which of the following statements must be true?

  1. f′(3)=0f'(3) = 0 (correct answer)
  2. f′(2)=0f'(2) = 0
  3. f′(1)=0f'(1) = 0
  4. f′(x)f'(x) must be a constant function.
Explanation: When you encounter a composite function with given derivative information, the chain rule is your essential tool. The chain rule states that if H(x)=f(g(x))H(x) = f(g(x)), then H′(x)=f′(g(x))⋅g′(x)H'(x) = f'(g(x)) \cdot g'(x). Let's apply this to find what must be true. We know:
  • g(1)=3g(1) = 3
  • g′(1)=2g'(1) = 2
  • H′(1)=0H'(1) = 0
Using the chain rule at x=1x = 1: H′(1)=f′(g(1))⋅g′(1)=f′(3)⋅2H'(1) = f'(g(1)) \cdot g'(1) = f'(3) \cdot 2 Since H′(1)=0H'(1) = 0, we have: f′(3)⋅2=0f'(3) \cdot 2 = 0 This means f′(3)=0f'(3) = 0, confirming that choice A is correct. Now let's examine why the other options are wrong: Choice B suggests f′(2)=0f'(2) = 0, but this comes from incorrectly using g′(1)=2g'(1) = 2 as the input to f′f'. The chain rule uses f′(g(1))f'(g(1)), not f′(g′(1))f'(g'(1)). Choice C claims f′(1)=0f'(1) = 0, which would mean using the original input value x=1x = 1 directly in f′f'. This ignores the composition entirely and misapplies the chain rule. Choice D states that f′(x)f'(x) must be constant. While f′(3)=0f'(3) = 0, this tells us nothing about f′f' at other points. The derivative could equal zero at just this one point. Study tip: In chain rule problems, always identify what goes where: the derivative of the outer function is evaluated at the inner function's output value, then multiplied by the inner function's derivative.

Question 16

Let h(x)=(f(x))3h(x) = (f(x))^3 for a differentiable function ff. If f(1)=2f(1)=2 and the tangent line to f(x)f(x) at x=1x=1 is parallel to the line y=5x−3y=5x-3, what is the slope of the tangent line to h(x)h(x) at x=1x=1?

  1. 1515
  2. 6060 (correct answer)
  3. 3030
  4. 2424
Explanation: This problem tests your understanding of the chain rule and how to find derivatives of composite functions. When you see a function raised to a power where the base itself is a function, the chain rule is your primary tool. To find the slope of the tangent line to h(x)=(f(x))3h(x) = (f(x))^3 at x=1x = 1, you need h′(1)h'(1). Using the chain rule: h′(x)=3(f(x))2⋅f′(x)h'(x) = 3(f(x))^2 \cdot f'(x), so h′(1)=3(f(1))2⋅f′(1)h'(1) = 3(f(1))^2 \cdot f'(1). You're given that f(1)=2f(1) = 2, so you need to find f′(1)f'(1). Since the tangent line to f(x)f(x) at x=1x = 1 is parallel to y=5x−3y = 5x - 3, and parallel lines have the same slope, f′(1)=5f'(1) = 5. Now you can calculate: h′(1)=3(f(1))2⋅f′(1)=3(2)2⋅5=3⋅4⋅5=60h'(1) = 3(f(1))^2 \cdot f'(1) = 3(2)^2 \cdot 5 = 3 \cdot 4 \cdot 5 = 60. Choice A (15) likely comes from forgetting to square f(1)f(1): 3⋅2⋅5=303 \cdot 2 \cdot 5 = 30, then making an arithmetic error. Choice C (30) results from the same error of using f(1)f(1) instead of (f(1))2(f(1))^2. Choice D (24) might come from incorrectly using the power rule without the chain rule, getting 3⋅23=243 \cdot 2^3 = 24. When working with composite functions, always identify which rule applies first. For functions of the form (g(x))n(g(x))^n, remember that the chain rule gives you n(g(x))n−1⋅g′(x)n(g(x))^{n-1} \cdot g'(x) - don't forget that final g′(x)g'(x) factor.

Question 17

If h(x)=x32x+3h(x) = x^3 \sqrt{2x+3}, what is the value of h′(3)h'(3)?

  1. 99
  2. 8181
  3. 8484
  4. 9090 (correct answer)
Explanation: To find the derivative of h(x)h(x), we must use the product rule in conjunction with the chain rule. Let u(x)=x3u(x) = x^3 and v(x)=2x+3=(2x+3)1/2v(x) = \sqrt{2x+3} = (2x+3)^{1/2}. The product rule states h′(x)=u′(x)v(x)+u(x)v′(x)h'(x) = u'(x)v(x) + u(x)v'(x). We have u′(x)=3x2u'(x) = 3x^2. To find v′(x)v'(x), we use the chain rule: v′(x)=12(2x+3)−1/2⋅2=12x+3v'(x) = \frac{1}{2}(2x+3)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x+3}}. Combining these, we get h′(x)=3x22x+3+x3⋅12x+3h'(x) = 3x^2 \sqrt{2x+3} + x^3 \cdot \frac{1}{\sqrt{2x+3}}. Now, we evaluate at x=3x=3: h′(3)=3(32)2(3)+3+332(3)+3=3(9)9+279=27(3)+273=81+9=90h'(3) = 3(3^2)\sqrt{2(3)+3} + \frac{3^3}{\sqrt{2(3)+3}} = 3(9)\sqrt{9} + \frac{27}{\sqrt{9}} = 27(3) + \frac{27}{3} = 81 + 9 = 90.

Question 18

If g(x)=sin⁡(cos⁡(x2))g(x) = \sin(\cos(x^2)), what is the value of g′(π/2)g'(\sqrt{\pi/2})?

  1. −1-1
  2. −2π-\sqrt{2\pi} (correct answer)
  3. 2π\sqrt{2\pi}
  4. 11
Explanation: When you see a complex composition of functions like this, you need to apply the chain rule multiple times. Think of this as peeling an onion - you'll differentiate from the outside in, keeping track of each layer. To find g′(x)g'(x) where g(x)=sin⁡(cos⁡(x2))g(x) = \sin(\cos(x^2)), apply the chain rule systematically:
  • The outermost function is sin⁡(u)\sin(u) where u=cos⁡(x2)u = \cos(x^2)
  • The middle function is cos⁡(v)\cos(v) where v=x2v = x^2
  • The innermost function is x2x^2
Using the chain rule: g′(x)=cos⁡(cos⁡(x2))⋅(−sin⁡(x2))⋅2xg'(x) = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot 2x Now evaluate at x=π/2x = \sqrt{\pi/2}:
  • x2=π/2x^2 = \pi/2
  • cos⁡(π/2)=0\cos(\pi/2) = 0
  • sin⁡(π/2)=1\sin(\pi/2) = 1
So: g′(π/2)=cos⁡(0)⋅(−sin⁡(π/2))⋅2π/2=1⋅(−1)⋅2π=−2πg'(\sqrt{\pi/2}) = \cos(0) \cdot (-\sin(\pi/2)) \cdot 2\sqrt{\pi/2} = 1 \cdot (-1) \cdot \sqrt{2\pi} = -\sqrt{2\pi} Option A) gives −1-1, which you'd get if you forgot to include the factor of 2x=2π/22x = 2\sqrt{\pi/2}. Option C) gives 2π\sqrt{2\pi}, which misses the negative sign from the derivative of cosine. Option D) gives 11, which ignores both the chain rule factors and gets only cos⁡(0)⋅(−1)\cos(0) \cdot (-1). The correct answer is B) −2π-\sqrt{2\pi}. Study tip: For nested compositions, write out each layer explicitly and remember that the chain rule requires you to multiply the derivatives of all layers. Don't rush - systematic application prevents sign errors and missing factors.

Question 19

A point moves along the curve defined by the equation exy=x−ye^{xy} = x-y. What is the slope of the curve at the point (0,−1)(0, -1)?

  1. −2-2
  2. 12\frac{1}{2}
  3. 22 (correct answer)
  4. −1-1
Explanation: We use implicit differentiation to find dydx\frac{dy}{dx}. Differentiating both sides of the equation with respect to xx: ddx(exy)=ddx(x−y)\frac{d}{dx}(e^{xy}) = \frac{d}{dx}(x-y). The left side requires the chain rule: exy⋅ddx(xy)e^{xy} \cdot \frac{d}{dx}(xy). The derivative of the product xyxy is 1⋅y+x⋅dydx1 \cdot y + x \cdot \frac{dy}{dx}. So, the derivative of the left side is exy(y+xdydx)e^{xy}(y + x\frac{dy}{dx}). The right side is 1−dydx1 - \frac{dy}{dx}. Setting them equal: exy(y+xdydx)=1−dydxe^{xy}(y + x\frac{dy}{dx}) = 1 - \frac{dy}{dx}. We want to solve for dydx\frac{dy}{dx} at (0,−1)(0, -1). Substitute x=0x=0 and y=−1y=-1: e0(−1)(−1+0⋅dydx)=1−dydxe^{0(-1)}(-1 + 0 \cdot \frac{dy}{dx}) = 1 - \frac{dy}{dx}. This simplifies to e0(−1)=1−dydxe^0(-1) = 1 - \frac{dy}{dx}, which is −1=1−dydx-1 = 1 - \frac{dy}{dx}. Solving for dydx\frac{dy}{dx} gives dydx=2\frac{dy}{dx} = 2.

Question 20

If h(x)=(f(x))3h(x) = (f(x))^3, and given f(2)=−1f(2)= -1 and f′(2)=4f'(2) = 4, what is the value of h′(2)h'(2)?

  1. -12
  2. -4
  3. 12 (correct answer)
  4. 48
Explanation: To find h′(x)h'(x), we apply the chain rule (specifically, the general power rule). Let u=f(x)u = f(x), then h(x)=u3h(x) = u^3. The derivative is h′(x)=3u2⋅u′=3(f(x))2⋅f′(x)h'(x) = 3u^2 \cdot u' = 3(f(x))^2 \cdot f'(x). Now, we substitute x=2x=2 and use the given values: h′(2)=3(f(2))2⋅f′(2)=3(−1)2⋅4=3(1)⋅4=12h'(2) = 3(f(2))^2 \cdot f'(2) = 3(-1)^2 \cdot 4 = 3(1) \cdot 4 = 12.