Chain Rule - Calculus 1

What is the slope of the line tangent to the curve $y = \arctan(e^{2x})$ at the point where $x=0$ ?
Answer: 1 The slope of the tangent line is given by the derivative $\frac{dy}{dx}$ . We use the chain rule. The derivative of $\arctan(u)$ is $\frac{1}{1+u^2}$ , and the derivative of $u=e^{2x}$ is $2e^{2x}$ . So, $\frac{dy}{dx} = \frac{1}{1+(e^{2x})^2} \cdot 2e^{2x} = \frac{2e^{2x}}{1+e^{4x}}$ . Now we evaluate this derivative at $x=0$ . $\left.\frac{dy}{dx}\right|_{x=0} = \frac{2e^{2(0)}}{1+e^{4(0)}} = \frac{2e^0}{1+e^0} = \frac{2(1)}{1+1} = \frac{2}{2} = 1$ .
Let $h(x) = \sqrt{f(x)^2 + g(x)^2}$ . Given $f(1)=3$ , $f'(1)=4$ , $g(1)=4$ , and $g'(1)=-3$ , find $h'(1)$ .
Answer: 0 We write $h(x) = (f(x)^2 + g(x)^2)^{1/2}$ and use the chain rule. Let $u = f(x)^2 + g(x)^2$ . Then $u' = 2f(x)f'(x) + 2g(x)g'(x)$ . The derivative of $h(x)$ is $h'(x) = \frac{1}{2} (f(x)^2 + g(x)^2)^{-1/2} \cdot (2f(x)f'(x) + 2g(x)g'(x)) = \frac{f(x)f'(x) + g(x)g'(x)}{\sqrt{f(x)^2 + g(x)^2}}$ . Now we substitute the given values at $x=1$ : $h'(1) = \frac{f(1)f'(1) + g(1)g'(1)}{\sqrt{f(1)^2 + g(1)^2}} = \frac{(3)(4) + (4)(-3)}{\sqrt{3^2 + 4^2}} = \frac{12 - 12}{\sqrt{9+16}} = \frac{0}{\sqrt{25}} = 0$ .
Let $f$ be a differentiable, even function. Which of the following statements provides the most complete and necessary description of its derivative, $f'$ ?
Answer: $f'$ is an odd function. By definition, an even function satisfies $f(-x) = f(x)$ for all $x$ in its domain. Differentiating both sides of this equation with respect to $x$ requires the chain rule on the left side: $\frac{d}{dx}[f(-x)] = \frac{d}{dx}[f(x)]$ . This gives $f'(-x) \cdot \frac{d}{dx}(-x) = f'(x)$ , which simplifies to $f'(-x) \cdot (-1) = f'(x)$ , or $-f'(-x) = f'(x)$ . This is the definition of an odd function. Therefore, the derivative of any differentiable even function must be an odd function.
What is the y-intercept of the tangent line to the graph of $f(x) = \ln(x^2 - 3x + e)$ at the point where $x=3$ ?
Answer: $1 - \frac{9}{e}$ First, find the point of tangency. At $x=3$ , $y = f(3) = \ln(3^2 - 3(3) + e) = \ln(9-9+e) = \ln(e) = 1$ . The point is $(3,1)$ . Next, find the slope of the tangent line by finding the derivative, $f'(x)$ . Using the chain rule, $f'(x) = \frac{1}{x^2 - 3x + e} \cdot (2x-3)$ . The slope at $x=3$ is $m = f'(3) = \frac{1}{3^2 - 3(3) + e} \cdot (2(3)-3) = \frac{1}{e} \cdot (3) = \frac{3}{e}$ . The equation of the tangent line is $y - y_0 = m(x-x_0)$ , which is $y - 1 = \frac{3}{e}(x-3)$ . To find the y-intercept, set $x=0$ : $y - 1 = \frac{3}{e}(0-3) = -\frac{9}{e}$ . Thus, the y-intercept is $y = 1 - \frac{9}{e}$ .
Let $y = \ln(\sec(x))$ . Which of the following is an expression for the second derivative, $\frac{d^2y}{dx^2}$ ?
Answer: $\sec^2(x)$ First, find the first derivative, $\frac{dy}{dx}$ . Using the chain rule, $\frac{dy}{dx} = \frac{1}{\sec(x)} \cdot \frac{d}{dx}(\sec(x)) = \frac{1}{\sec(x)} \cdot (\sec(x)\tan(x)) = \tan(x)$ . Now, we find the second derivative by differentiating $\tan(x)$ : $\frac{d^2y}{dx^2} = \frac{d}{dx}(\tan(x)) = \sec^2(x)$ .
Let $f$ be a differentiable function such that $f(2)=2$ and $f'(2)=3$ . If $h(x) = f(f(x))$ , what is the value of $h'(2)$ ?
Answer: $9$ When you encounter a composite function like $h(x) = f(f(x))$ 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 $h'(x)$ where $h(x) = f(f(x))$ , 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 $f$ evaluated at $f(x)$ , and the inner function is $f(x)$ . So $h'(x) = f'(f(x)) \cdot f'(x)$ . At $x = 2$ : $h'(2) = f'(f(2)) \cdot f'(2)$ . Since $f(2) = 2$ , we have $f'(f(2)) = f'(2) = 3$ . Therefore: $h'(2) = f'(2) \cdot f'(2) = 3 \cdot 3 = 9$ . The answer is A) $9$ . Looking at the wrong choices: B) $6$ likely comes from adding $f'(2) + f'(2) = 3 + 3 = 6$ instead of multiplying—a common error when students confuse addition and multiplication rules. C) $4$ might result from incorrectly using $f(2) + f'(2) = 2 + 2 = 4$ , mixing up function values with derivatives. D) $12$ could come from incorrectly calculating $f'(2) \cdot f(2) \cdot f'(2) = 3 \cdot 2 \cdot 2 = 12$ , 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.
Suppose $y$ is a function of $x$ such that $e^{y^2} = x^2+1$ . What is the value of $\frac{dy}{dx}$ at the point $( \sqrt{e-1}, 1 )$ ?
Answer: $\frac{\sqrt{e-1}}{e}$ We use implicit differentiation. Differentiating both sides of $e^{y^2} = x^2+1$ with respect to $x$ , we must use the chain rule on the left side: $\frac{d}{dx}(e^{y^2}) = \frac{d}{dx}(x^2+1)$ . This gives $e^{y^2} \cdot 2y \frac{dy}{dx} = 2x$ . Solving for $\frac{dy}{dx}$ yields $\frac{dy}{dx} = \frac{2x}{2y e^{y^2}} = \frac{x}{y e^{y^2}}$ . Now we evaluate this derivative at the point $(x,y) = (\sqrt{e-1}, 1)$ . $\left.\frac{dy}{dx}\right|_{(\sqrt{e-1}, 1)} = \frac{\sqrt{e-1}}{1 \cdot e^{1^2}} = \frac{\sqrt{e-1}}{e}$ .
Given $g(x) = f(\sin^2(x))$ , with $f'(0) = 3$ . Find $g'(0)$ .
Answer: 0 We use the chain rule twice. Let $u(x) = \sin^2(x) = (\sin x)^2$ . First, find $u'(x)$ using the chain rule: $u'(x) = 2(\sin x)^1 \cdot \cos x = 2\sin x \cos x$ . Now, find $g'(x)$ using the chain rule: $g'(x) = f'(u(x)) \cdot u'(x) = f'(\sin^2(x)) \cdot (2\sin x \cos x)$ . We evaluate this at $x=0$ . $g'(0) = f'(\sin^2(0)) \cdot (2\sin 0 \cos 0)$ . Since $\sin(0)=0$ and $\cos(0)=1$ , this becomes $g'(0) = f'(0^2) \cdot (2 \cdot 0 \cdot 1) = f'(0) \cdot 0 = 3 \cdot 0 = 0$ .
Let $y = \frac{e^{kx}}{x}$ for some constant $k$ . If $\frac{dy}{dx} = 0$ when $x=2$ , what is the value of $k$ ?
Answer: 1/2 We first find the derivative $\frac{dy}{dx}$ using the quotient rule and the chain rule. $\frac{dy}{dx} = \frac{(\frac{d}{dx} e^{kx}) \cdot x - e^{kx} \cdot (\frac{d}{dx} x)}{x^2}$ . The derivative of $e^{kx}$ requires the chain rule: $\frac{d}{dx} e^{kx} = e^{kx} \cdot k$ . So, $\frac{dy}{dx} = \frac{(ke^{kx})x - e^{kx}(1)}{x^2} = \frac{e^{kx}(kx-1)}{x^2}$ . We are given that $\frac{dy}{dx}=0$ when $x=2$ . Setting the derivative to zero gives $\frac{e^{2k}(2k-1)}{2^2} = 0$ . Since $e^{2k}$ is always positive, the fraction is zero only if the numerator term $(2k-1)$ is zero. So, $2k-1=0$ , which gives $k=1/2$ .
Let $g(x) = \left(\frac{x-1}{x+1}\right)^3$ . What is $g'(3)$ ?
Answer: $\frac{3}{32}$ This requires the chain rule and the quotient rule. Let the outer function be $f(u) = u^3$ and the inner function be $u(x) = \frac{x-1}{x+1}$ . By the chain rule, $g'(x) = 3\left(\frac{x-1}{x+1}\right)^2 \cdot u'(x)$ . To find $u'(x)$ , we use the quotient rule: $u'(x) = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}$ . Substituting this back, we get $g'(x) = 3\left(\frac{x-1}{x+1}\right)^2 \cdot \frac{2}{(x+1)^2} = \frac{6(x-1)^2}{(x+1)^4}$ . Now, evaluate at $x=3$ : $g'(3) = \frac{6(3-1)^2}{(3+1)^4} = \frac{6(2^2)}{4^4} = \frac{6 \cdot 4}{256} = \frac{24}{256}$ , which simplifies to $\frac{3}{32}$ .
If $g(x) = f^{-1}(x)$ and $f(x) = x^3 + 2x + 1$ , what is the value of $g'(4)$ ?
Answer: 1/5 When you encounter inverse function derivatives, remember that if $g(x) = f^{-1}(x)$ , then $g'(x) = \frac{1}{f'(g(x))}$ . This formula connects the derivative of an inverse function to the derivative of the original function. To find $g'(4)$ , you need two things: the value of $g(4)$ and $f'(g(4))$ . Since $g(4) = f^{-1}(4)$ , you must first solve $f(x) = 4$ . Setting $x^3 + 2x + 1 = 4$ gives you $x^3 + 2x - 3 = 0$ . Testing $x = 1$ : $1 + 2 - 3 = 0$ ✓. So $g(4) = 1$ . Next, find $f'(x) = 3x^2 + 2$ . Therefore, $f'(1) = 3(1)^2 + 2 = 5$ . Using the inverse derivative formula: $g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(1)} = \frac{1}{5}$ . Answer (A) $\frac{1}{31}$ likely comes from incorrectly calculating $f'(1)$ as $3^2 + 2^2 = 13$ or making an arithmetic error. Answer (B) $\frac{1}{7}$ might result from computing $f'(1) = 3(1) + 2 + 2 = 7$ , forgetting the squared term. Answer (C) $\frac{1}{14}$ 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.
Let $h(x) = (f(x))^3$ for a differentiable function $f$ . If $f(1)=2$ and the tangent line to $f(x)$ at $x=1$ is parallel to the line $y=5x-3$ , what is the slope of the tangent line to $h(x)$ at $x=1$ ?
Answer: $60$ 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 $h(x) = (f(x))^3$ at $x = 1$ , you need $h'(1)$ . Using the chain rule: $h'(x) = 3(f(x))^2 \cdot f'(x)$ , so $h'(1) = 3(f(1))^2 \cdot f'(1)$ . You're given that $f(1) = 2$ , so you need to find $f'(1)$ . Since the tangent line to $f(x)$ at $x = 1$ is parallel to $y = 5x - 3$ , and parallel lines have the same slope, $f'(1) = 5$ . Now you can calculate: $h'(1) = 3(f(1))^2 \cdot f'(1) = 3(2)^2 \cdot 5 = 3 \cdot 4 \cdot 5 = 60$ . Choice A (15) likely comes from forgetting to square $f(1)$ : $3 \cdot 2 \cdot 5 = 30$ , then making an arithmetic error. Choice C (30) results from the same error of using $f(1)$ instead of $(f(1))^2$ . Choice D (24) might come from incorrectly using the power rule without the chain rule, getting $3 \cdot 2^3 = 24$ . When working with composite functions, always identify which rule applies first. For functions of the form $(g(x))^n$ , remember that the chain rule gives you $n(g(x))^{n-1} \cdot g'(x)$ - don't forget that final $g'(x)$ factor.
The position of a particle is given by $s(t) = \sec(\pi t^2)$ for $t>0$ . What is the velocity of the particle at $t=1/2$ ?
Answer: $\pi\sqrt{2}$ Velocity is the derivative of position, $v(t) = s'(t)$ . We use the chain rule to differentiate $s(t)$ . The derivative of $\sec(u)$ is $\sec(u)\tan(u)$ . Here, $u = \pi t^2$ , so $u' = 2\pi t$ . Therefore, $v(t) = s'(t) = \sec(\pi t^2) \tan(\pi t^2) \cdot (2\pi t)$ . Now, evaluate at $t=1/2$ : $v(1/2) = \sec(\pi (1/2)^2) \tan(\pi (1/2)^2) \cdot (2\pi (1/2)) = \sec(\pi/4) \tan(\pi/4) \cdot \pi$ . We know $\sec(\pi/4) = \sqrt{2}$ and $\tan(\pi/4)=1$ . So, $v(1/2) = \sqrt{2} \cdot 1 \cdot \pi = \pi\sqrt{2}$ .
Let $f(x) = g(x^3)$ and $g(x) = h(2x)$ . If $h'(2)=4$ , what is $f'(1)$ ?
Answer: 24 This requires two applications of the chain rule. First, let's find $f'(x)$ . $f'(x) = g'(x^3) \cdot \frac{d}{dx}(x^3) = g'(x^3) \cdot 3x^2$ . Now we need to find an expression for $g'(x)$ . From $g(x) = h(2x)$ , we differentiate to get $g'(x) = h'(2x) \cdot \frac{d}{dx}(2x) = h'(2x) \cdot 2$ . Now substitute this back into the expression for $f'(x)$ : $f'(x) = [h'(2x^3) \cdot 2] \cdot 3x^2 = 6x^2 h'(2x^3)$ . Finally, evaluate at $x=1$ : $f'(1) = 6(1)^2 h'(2(1)^3) = 6h'(2)$ . Given $h'(2)=4$ , we have $f'(1) = 6(4) = 24$ .
If $h(x) = x^3 \sqrt{2x+3}$ , what is the value of $h'(3)$ ?
Answer: $90$ To find the derivative of $h(x)$ , we must use the product rule in conjunction with the chain rule. Let $u(x) = x^3$ and $v(x) = \sqrt{2x+3} = (2x+3)^{1/2}$ . The product rule states $h'(x) = u'(x)v(x) + u(x)v'(x)$ . We have $u'(x) = 3x^2$ . To find $v'(x)$ , we use the chain rule: $v'(x) = \frac{1}{2}(2x+3)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x+3}}$ . Combining these, we get $h'(x) = 3x^2 \sqrt{2x+3} + x^3 \cdot \frac{1}{\sqrt{2x+3}}$ . Now, we evaluate at $x=3$ : $h'(3) = 3(3^2)\sqrt{2(3)+3} + \frac{3^3}{\sqrt{2(3)+3}} = 3(9)\sqrt{9} + \frac{27}{\sqrt{9}} = 27(3) + \frac{27}{3} = 81 + 9 = 90$ .
Let $y = \sin^3(2x)$ . What is $\frac{dy}{dx}$ ?
Answer: $6\sin^2(2x)\cos(2x)$ When you encounter a composite function like $y = \sin^3(2x)$ , you need to apply the chain rule systematically. This function has two layers of composition: an outer function (cubing) applied to $\sin(2x)$ , and an inner function where $\sin$ is applied to $2x$ . To find $\frac{dy}{dx}$ , work from the outside in. First, treat $\sin(2x)$ as a single unit and differentiate the cubic: $\frac{d}{dx}[\sin(2x)]^3 = 3[\sin(2x)]^2 \cdot \frac{d}{dx}[\sin(2x)]$ . This gives you $3\sin^2(2x)$ times the derivative of $\sin(2x)$ . Next, differentiate $\sin(2x)$ : $\frac{d}{dx}[\sin(2x)] = \cos(2x) \cdot \frac{d}{dx}[2x] = \cos(2x) \cdot 2 = 2\cos(2x)$ . Combining these steps: $\frac{dy}{dx} = 3\sin^2(2x) \cdot 2\cos(2x) = 6\sin^2(2x)\cos(2x)$ . Choice A ( $3\cos^2(2x)$ ) incorrectly differentiates the trigonometric function and misses the chain rule entirely. Choice B ( $3\sin^2(2x)\cos(2x)$ ) correctly applies the power rule and differentiates $\sin(2x)$ to $\cos(2x)$ , but forgets to multiply by the derivative of the innermost function $2x$ , missing the factor of 2. Choice C ( $6\cos^2(2x)$ ) has the correct coefficient but wrong trigonometric functions. The correct answer is D: $6\sin^2(2x)\cos(2x)$ . Strategy tip: For nested functions, always identify each layer and apply the chain rule step-by-step. Don't forget that the derivative of $kx$ is $k$ , not 1.
Let $F(x) = \int_{1}^{\cos(x)} \sqrt{1+t^3} \,dt$ . What is $F'(\pi/2)$ ?
Answer: -1 This problem requires the Fundamental Theorem of Calculus Part 1 combined with the chain rule. Let $g(x) = \cos(x)$ be the upper limit of integration. Then $F(x) = \int_{1}^{g(x)} \sqrt{1+t^3} \,dt$ . The derivative is $F'(x) = \sqrt{1+(g(x))^3} \cdot g'(x) = \sqrt{1+\cos^3(x)} \cdot (-\sin(x))$ . Evaluating at $x=\pi/2$ : $F'(\pi/2) = \sqrt{1+\cos^3(\pi/2)} \cdot (-\sin(\pi/2))$ . Since $\cos(\pi/2)=0$ and $\sin(\pi/2)=1$ , we have $F'(\pi/2) = \sqrt{1+0^3} \cdot (-1) = \sqrt{1} \cdot (-1) = -1$ .
If $h(x) = (f(x))^3$ , and given $f(2)= -1$ and $f'(2) = 4$ , what is the value of $h'(2)$ ?
Answer: 12 To find $h'(x)$ , we apply the chain rule (specifically, the general power rule). Let $u = f(x)$ , then $h(x) = u^3$ . The derivative is $h'(x) = 3u^2 \cdot u' = 3(f(x))^2 \cdot f'(x)$ . Now, we substitute $x=2$ and use the given values: $h'(2) = 3(f(2))^2 \cdot f'(2) = 3(-1)^2 \cdot 4 = 3(1) \cdot 4 = 12$ .
Let $h(x) = f(x^2 g(x))$ . Suppose $g(1) = 2$ , $g'(1) = -1$ , and $f'(2) = 3$ . Find the value of $h'(1)$ .
Answer: 9 To find $h'(x)$ , we must use the chain rule and the product rule. Let the inner function be $u(x) = x^2 g(x)$ . By the product rule, $u'(x) = 2x g(x) + x^2 g'(x)$ . By the chain rule, $h'(x) = f'(u(x)) \cdot u'(x) = f'(x^2 g(x)) \cdot (2x g(x) + x^2 g'(x))$ . To find $h'(1)$ , we substitute $x=1$ : $h'(1) = f'(1^2 g(1)) \cdot (2(1)g(1) + 1^2 g'(1))$ . Using the given values: $h'(1) = f'(2) \cdot (2(2) + (-1)) = 3 \cdot (4-1) = 3 \cdot 3 = 9$ .
Let $H(x) = (f \circ g)(x) = f(g(x))$ . Suppose that $g(1)=3$ , $g'(1)=2$ , and $H'(1)=0$ . Which of the following statements must be true?
Answer: $f'(3) = 0$ When you encounter a composite function with given derivative information, the chain rule is your essential tool. The chain rule states that if $H(x) = f(g(x))$ , then $H'(x) = f'(g(x)) \cdot g'(x)$ . Let's apply this to find what must be true. We know: - $g(1) = 3$ - $g'(1) = 2$ - $H'(1) = 0$ Using the chain rule at $x = 1$ : $H'(1) = f'(g(1)) \cdot g'(1) = f'(3) \cdot 2$ Since $H'(1) = 0$ , we have: $f'(3) \cdot 2 = 0$ This means $f'(3) = 0$ , confirming that choice A is correct. Now let's examine why the other options are wrong: Choice B suggests $f'(2) = 0$ , but this comes from incorrectly using $g'(1) = 2$ as the input to $f'$ . The chain rule uses $f'(g(1))$ , not $f'(g'(1))$ . Choice C claims $f'(1) = 0$ , which would mean using the original input value $x = 1$ directly in $f'$ . This ignores the composition entirely and misapplies the chain rule. Choice D states that $f'(x)$ must be constant. While $f'(3) = 0$ , this tells us nothing about $f'$ 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.

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