Calculus 1 Quiz: Differentiating Inverse Functions

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Differentiating Inverse FunctionsQuestion 1 of 20

A differentiable function $f$ with inverse $g=f^{-1}$ satisfies the equation $x(f(x))^3 + f(x) = 2x^3$ for all $x$ in its domain. If $f(1)=1$ , what is the value of $g'(1)$ ?

5/4

1/2

2/3

4/5

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Calculus 1 Quiz

Calculus 1 Quiz: Differentiating Inverse Functions

Practice Differentiating Inverse Functions 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 Differentiating Inverse Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

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

A differentiable function $f$ with inverse $g=f^{-1}$ satisfies the equation $x(f(x))^3 + f(x) = 2x^3$ for all $x$ in its domain. If $f(1)=1$ , what is the value of $g'(1)$ ?

$5/4$
$1/2$
$2/3$
$4/5$ (correct answer)

Explanation: When you encounter a problem involving inverse functions and their derivatives, remember that if

g = f^{-1}

, then

g'(a) = \frac{1}{f'(g(a))}

. Since we know

f(1) = 1

and need

g'(1)

, we have

g(1) = 1

, so

g'(1) = \frac{1}{f'(1)}

. To find

f'(1)

, differentiate the given equation

x(f(x))^3 + f(x) = 2x^3

using the product rule and chain rule:

\frac{d}{dx}[x(f(x))^3] + \frac{d}{dx}[f(x)] = \frac{d}{dx}[2x^3]

(f(x))^3 + x \cdot 3(f(x))^2 \cdot f'(x) + f'(x) = 6x^2

(f(x))^3 + 3x(f(x))^2f'(x) + f'(x) = 6x^2

x = 1

, using

f(1) = 1

(1)^3 + 3(1)(1)^2f'(1) + f'(1) = 6(1)^2

1 + 3f'(1) + f'(1) = 6

1 + 4f'(1) = 6

f'(1) = \frac{5}{4}

Therefore,

g'(1) = \frac{1}{f'(1)} = \frac{1}{\frac{5}{4}} = \frac{4}{5}

. Choice A (

\frac{5}{4}

) gives you

f'(1)

instead of

g'(1)

- a common error when forgetting the reciprocal relationship. Choice B (

\frac{1}{2}

) and Choice C (

\frac{2}{3}

) result from algebraic mistakes when solving for

f'(1)

or applying the inverse derivative formula incorrectly. Key Strategy: Always remember that

(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}

. First find the derivative of the original function, then take its reciprocal at the appropriate point.

Question 2

Let $f(x) = \sin(x) + x$ for $-\frac{\pi}{2} < x < \frac{\pi}{2}$ . If $g = f^{-1}$ , what is $g'(\frac{\pi}{2} + 1)$ ?

$1$ (correct answer)
$\frac{1}{1 - \sin(1)}$
$0$
$2$

Explanation: We need

g'(\frac{\pi}{2} + 1) = \frac{1}{f'(g(\frac{\pi}{2} + 1))}

. First, find

b = g(\frac{\pi}{2} + 1)

, which means

f(b) = \sin(b) + b = \frac{\pi}{2} + 1

. By inspection,

b = \pi/2

. Next, find the derivative

f'(x) = \cos(x) + 1

. Evaluate this at

b = \pi/2

f'(\pi/2) = \cos(\pi/2) + 1 = 0 + 1 = 1

. Finally,

g'(\frac{\pi}{2} + 1) = \frac{1}{f'(\pi/2)} = \frac{1}{1} = 1

Question 3

Let $f(x) = \tan(x)$ for $-\pi/2 < x < \pi/2$ , and let $g(x)=f^{-1}(x)$ . Consider the function $h(x) = x^2 g(x)$ . Find $h'(1)$ .

$\frac{\pi+4}{2}$
$\frac{1}{2}$
$\frac{5}{2}$
$\frac{\pi+1}{2}$ (correct answer)

Explanation: This problem tests your understanding of inverse function derivatives and the product rule. When you see a function involving an inverse function that you need to differentiate, remember that you'll likely need the inverse function derivative formula. First, identify what

g(x)

is: since

f(x) = \tan(x)

, we have

g(x) = f^{-1}(x) = \arctan(x)

. So

h(x) = x^2 \arctan(x)

. To find

h'(x)

, use the product rule:

h'(x) = 2x \arctan(x) + x^2 \cdot \frac{d}{dx}[\arctan(x)]

. Since

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

, we get:

h'(x) = 2x \arctan(x) + \frac{x^2}{1+x^2}

x = 1

h'(1) = 2(1)\arctan(1) + \frac{1^2}{1+1^2} = 2 \cdot \frac{\pi}{4} + \frac{1}{2} = \frac{\pi}{2} + \frac{1}{2} = \frac{\pi+1}{2}

Choice A,

\frac{\pi+4}{2}

, likely comes from incorrectly using

\arctan(1) = \pi

instead of

\frac{\pi}{4}

. Choice B,

\frac{1}{2}

, results from forgetting the

2x\arctan(x)

term entirely and only computing the second term. Choice C,

\frac{5}{2}

, might arise from computational errors or misremembering that

\arctan(1) = 1

instead of

\frac{\pi}{4}

. The correct answer is D. Study tip: Always memorize that

\arctan(1) = \frac{\pi}{4}

and

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

. These appear frequently in calculus problems involving inverse trig functions.

Question 4

For $x > 0$ , let $f(x) = x^x$ . This function is one-to-one for $x \ge 1/e$ . Let $g=f^{-1}$ . Find $g'(4)$ .

$4(\ln 2 + 1)$
$\frac{1}{4(\ln 2 + 1)}$ (correct answer)
$\frac{1}{256(\ln 4 + 1)}$
$\frac{1}{4}$

Explanation: When you encounter inverse function derivatives, remember the key formula: if

g = f^{-1}

, then

g'(y) = \frac{1}{f'(x)}

where

x = g(y)

. To find

g'(4)

, you first need to determine what value of

x

gives

f(x) = 4

. Since

f(x) = x^x

, you're solving

x^x = 4

. By inspection,

x = 2

works because

2^2 = 4

. So

g(4) = 2

. Next, find

f'(x)

using logarithmic differentiation. Taking the natural log of both sides:

\ln(f(x)) = x \ln x

. Differentiating implicitly:

\frac{f'(x)}{f(x)} = \ln x + 1

. Therefore,

f'(x) = x^x(\ln x + 1)

. At

x = 2

f'(2) = 2^2(\ln 2 + 1) = 4(\ln 2 + 1)

. Using the inverse function derivative formula:

g'(4) = \frac{1}{f'(2)} = \frac{1}{4(\ln 2 + 1)}

. Choice A gives

4(\ln 2 + 1)

, which is

f'(2)

, not

g'(4)

- this represents forgetting to take the reciprocal. Choice C uses

\ln 4

instead of

\ln 2

and includes an extra factor of 64, likely from confusing

f(2) = 4

with the point where you're evaluating. Choice D is

\frac{1}{4}

, which ignores the logarithmic term entirely. Study tip: For inverse function derivatives, always remember the reciprocal relationship and carefully track which point corresponds to which function. The derivative of the inverse at

y

equals the reciprocal of the original derivative at

x = g(y)

Question 5

Let $f(x)$ be a differentiable function with inverse $g(x)$ . If $f(3)=5$ and $f'(3)=2$ , what is the value of the derivative of $y = \frac{1}{g(x)}$ at $x=5$ ?

$\frac{1}{50}$
$-\frac{2}{9}$
$-\frac{1}{2}$
$-\frac{1}{18}$ (correct answer)

Explanation: This question tests your understanding of inverse function derivatives and the chain rule working together. When you see a problem involving the derivative of a function containing an inverse function, you'll need to combine the inverse function derivative formula with standard differentiation rules. To find the derivative of

y = \frac{1}{g(x)}

x = 5

, start by differentiating:

\frac{dy}{dx} = -\frac{1}{[g(x)]^2} \cdot g'(x)

. Now you need to find

g(5)

and

g'(5)

. Since

g(x)

is the inverse of

f(x)

, and

f(3) = 5

, this means

g(5) = 3

. For the derivative, use the inverse function formula:

g'(5) = \frac{1}{f'(g(5))} = \frac{1}{f'(3)} = \frac{1}{2}

. Substituting into your derivative formula:

\frac{dy}{dx}\bigg|_{x=5} = -\frac{1}{[g(5)]^2} \cdot g'(5) = -\frac{1}{3^2} \cdot \frac{1}{2} = -\frac{1}{18}

. This confirms answer D. Answer A (

\frac{1}{50}

) incorrectly uses

[f(3)]^2 \cdot f'(3) = 25 \cdot 2 = 50

in the denominator and misses the negative sign. Answer B (

-\frac{2}{9}

) incorrectly puts

f'(3) = 2

in the numerator instead of using the inverse derivative formula. Answer C (

-\frac{1}{2}

) forgets to square

g(5)

in the denominator. Remember: when dealing with inverse functions, always identify corresponding points first (if

f(a) = b

, then

g(b) = a

), then apply the inverse derivative formula

g'(b) = \frac{1}{f'(a)}

Question 6

The function $f(x) = \ln(x^2 - 2x + 2)$ is one-to-one on the domain $x \ge 1$ . Let $g = f^{-1}$ . Find the equation of the normal line to the graph of $y=g(x)$ at $x = \ln 5$ .

$y - 3 = \frac{5}{4}(x - \ln 5)$
$y - 3 = -\frac{4}{5}(x - \ln 5)$ (correct answer)
$y - 3 = \frac{4}{5}(x - \ln 5)$
$y - 3 = -\frac{5}{4}(x - \ln 5)$

Explanation: When you encounter inverse function problems involving normal lines, you need to work systematically through finding the point, the derivative, and then the perpendicular slope. First, find the point on

g(x)

when

x = \ln 5

. Since

g = f^{-1}

, you need

g(\ln 5) = a

where

f(a) = \ln 5

. This means

\ln(a^2 - 2a + 2) = \ln 5

, so

a^2 - 2a + 2 = 5

. Solving

a^2 - 2a - 3 = 0

gives

(a-3)(a+1) = 0

. Since the domain is

x \geq 1

, we take

a = 3

. So the point is

(\ln 5, 3)

. Next, find

g'(\ln 5)

using the inverse function derivative formula:

g'(x) = \frac{1}{f'(g(x))}

. We have

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

, so

f'(3) = \frac{4}{5}

. Therefore,

g'(\ln 5) = \frac{5}{4}

. The normal line is perpendicular to the tangent, so its slope is

-\frac{4}{5}

. The equation is

y - 3 = -\frac{4}{5}(x - \ln 5)

, which is choice B. Choice A uses the reciprocal

\frac{5}{4}

instead of the negative reciprocal. Choice C uses

\frac{4}{5}

, which is actually

f'(3)

rather than the negative reciprocal of

g'(\ln 5)

. Choice D uses

-\frac{5}{4}

, incorrectly taking the negative of

g'(\ln 5)

instead of its negative reciprocal. Study tip: Remember that normal lines require the negative reciprocal of the derivative, and inverse function derivatives use the reciprocal formula

g'(x) = \frac{1}{f'(g(x))}

Question 7

The functions $f$ and $g$ are inverses. The point $(2, 5)$ lies on the graph of $y=f(x)$ , and the slope of the tangent line to the graph of $f$ at this point is $-\frac{3}{4}$ . What is the equation of the tangent line to the graph of $y=g(x)$ at $x=5$ ?

$y - 2 = -\frac{4}{3}(x - 5)$ (correct answer)
$y - 2 = -\frac{3}{4}(x - 5)$
$y - 5 = -\frac{4}{3}(x - 2)$
$y - 5 = \frac{4}{3}(x - 2)$

Explanation: We are given

f(2)=5

and

f'(2) = -3/4

. Since

g=f^{-1}

, the point

(5, 2)

lies on the graph of

g

. The slope of the tangent line to

g

x=5

g'(5)

. Using the inverse function rule,

g'(5) = \frac{1}{f'(g(5))}

. Since

f(2)=5

, we have

g(5)=2

. Therefore,

g'(5) = \frac{1}{f'(2)} = \frac{1}{-3/4} = -\frac{4}{3}

. The equation of the tangent line to

g

at the point

(5, 2)

y - 2 = -\frac{4}{3}(x - 5)

Question 8

Let $g$ be the inverse of a differentiable function $f$ . The tangent line to the graph of $f$ at the point where $x=3$ is given by the equation $y = 5x - 11$ . Which of the following statements about the graph of $g$ must be true?

The graph of $g$ has a tangent line at $x=3$ with a slope of $1/5$ .
The equation of the tangent line to the graph of $g$ is $y = \frac{1}{5}x + \frac{11}{5}$ .
The graph of $g$ has a tangent line at $x=4$ with a slope of $5$ .
The graph of $g$ has a tangent line at $x=4$ with a slope of $1/5$ . (correct answer)

Explanation: When you encounter inverse functions and tangent lines, the key relationship to remember is that inverse functions "swap" x and y coordinates, and their derivatives are reciprocals at corresponding points. From the tangent line equation

y = 5x - 11

x = 3

, you can extract crucial information: the slope is

f'(3) = 5

, and the point on

f

(3, 4)

since

y = 5(3) - 11 = 4

. This means

f(3) = 4

. Since

g

is the inverse of

f

, the point

(3, 4)

f

corresponds to the point

(4, 3)

g

. The derivative formula for inverse functions tells us that

g'(x) = \frac{1}{f'(g(x))}

. At

x = 4

g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(3)} = \frac{1}{5}

. Therefore,

g

has a tangent line at

x = 4

with slope

\frac{1}{5}

, making D correct. A is wrong because it places the tangent line at

x = 3

instead of

x = 4

. B incorrectly assumes the tangent line equation can be found by simply manipulating the original equation algebraically. C uses the wrong slope—it should be

\frac{1}{5}

, not

5

. Study tip: Always identify corresponding points first (swap coordinates), then apply the reciprocal derivative relationship. The x-value where you evaluate

g'

is the y-value from the original function.

Question 9

Let $f$ be a twice-differentiable, one-to-one function. It is known that for all $x$ in the domain of $f$ , $f'(x) < 0$ and $f''(x) > 0$ . Let $g(x) = f^{-1}(x)$ . Which of the following describes the function $g$ ?

Increasing and concave up
Increasing and concave down
Decreasing and concave up (correct answer)
Decreasing and concave down

Explanation: Let's determine the signs of the first and second derivatives of

g(x)

First derivative of $g$ : $g'(x) = \frac{1}{f'(g(x))}$ . We are given that $f'(x) < 0$ for all $x$ . Since the output of $g(x)$ is in the domain of $f$ , $f'(g(x))$ is also negative. The reciprocal of a negative number is negative, so $g'(x) < 0$ . This means $g(x)$ is a decreasing function.
Second derivative of $g$ : We use the chain rule to differentiate $g'(x) = [f'(g(x))]^{-1}$ . $g''(x) = -[f'(g(x))]^{-2} \cdot \frac{d}{dx}[f'(g(x))]$ $g''(x) = -[f'(g(x))]^{-2} \cdot f''(g(x)) \cdot g'(x)$ Now substitute $g'(x) = \frac{1}{f'(g(x))}$ : $g''(x) = -[f'(g(x))]^{-2} \cdot f''(g(x)) \cdot \frac{1}{f'(g(x))} = -\frac{f''(g(x))}{[f'(g(x))]^3}$ . Now let's determine the sign of this expression. We are given $f''(x) > 0$ , so the numerator $f''(g(x))$ is positive. We are given $f'(x) < 0$ , so the denominator $[f'(g(x))]^3$ (the cube of a negative number) is negative. Therefore, $g''(x) = -\frac{\text{positive}}{\text{negative}} = -(\text{negative}) = \text{positive}$ . Since $g''(x) > 0$ , the function $g(x)$ is concave up. Combining our findings, $g(x)$ is decreasing and concave up.

Question 10

Let $f(x) = x^3 + 2x - 1$ . If $g$ is the inverse function of $f$ , what is the value of $g'(2)$ ?

$1/14$
$1/5$ (correct answer)
$5$
$14$

Explanation: The formula for the derivative of an inverse function is

g'(a) = \frac{1}{f'(g(a))}

. Here,

a=2

. First, we need to find

g(2)

. If

g(2)=x

, then

f(x)=2

. So we solve

x^3 + 2x - 1 = 2

, which simplifies to

x^3 + 2x - 3 = 0

. By inspection,

x=1

is a solution. Thus,

g(2)=1

. Next, we find the derivative of

f(x)

, which is

f'(x) = 3x^2 + 2

. Now we evaluate

f'(g(2)) = f'(1) = 3(1)^2 + 2 = 5

. Finally,

g'(2) = \frac{1}{f'(1)} = \frac{1}{5}

Question 11

The function $f(x) = x^5 + x^3 + 2x$ is one-to-one. If $L$ is the tangent line to the graph of $y=f^{-1}(x)$ at the point where $y=1$ , what is the slope of $L$ ?

$10$
$-10$
$1/1330$
$1/10$ (correct answer)

Explanation: The slope of the tangent line to

y=f^{-1}(x)

(f^{-1})'(x)

. We need to find the slope at the point on the curve where the y-coordinate is 1. Let this point be

(a,1)

. Then

f^{-1}(a) = 1

, which implies

f(1)=a

. Calculating

a

, we get

a = f(1) = 1^5 + 1^3 + 2(1) = 4

. So we need to find the slope at

x=4

, which is

(f^{-1})'(4)

. The formula is

(f^{-1})'(4) = \frac{1}{f'(f^{-1}(4))}

. Since

f^{-1}(4)=1

, this is

\frac{1}{f'(1)}

. First, we find the derivative:

f'(x) = 5x^4 + 3x^2 + 2

. Then,

f'(1) = 5(1)^4 + 3(1)^2 + 2 = 10

. The slope is

\frac{1}{10}

Question 12

Let $f$ be an invertible function and let $g=f^{-1}$ . The line $y=2x+5$ is tangent to the graph of $f$ at the point $(-1, 3)$ . Which of the following is an equation for the line tangent to the graph of $g$ at the point $(3, -1)$ ?

$y = 2x - 7$
$y = -\frac{1}{2}x + \frac{1}{2}$
$y = \frac{1}{2}x - \frac{5}{2}$ (correct answer)
$y = \frac{1}{2}x + \frac{7}{2}$

Explanation: When you encounter problems involving inverse functions and their tangent lines, remember that inverse functions "flip" both coordinates and slopes in specific ways. Since the line

y = 2x + 5

is tangent to

f

(-1, 3)

, we know that

f(-1) = 3

and

f'(-1) = 2

. Because

g = f^{-1}

, the point

(-1, 3)

f

becomes the point

(3, -1)

g

- the coordinates are swapped. The key insight is the relationship between derivatives of inverse functions:

(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}

. At the point

(3, -1)

g

, we have

g'(3) = \frac{1}{f'(g(3))} = \frac{1}{f'(-1)} = \frac{1}{2}

. So the tangent line to

g

(3, -1)

has slope

\frac{1}{2}

and passes through

(3, -1)

. Using point-slope form:

y - (-1) = \frac{1}{2}(x - 3)

, which simplifies to

y = \frac{1}{2}x - \frac{5}{2}

. Looking at the wrong answers: A) uses slope 2 instead of

\frac{1}{2}

, missing the reciprocal relationship. B) has the correct reciprocal slope but uses

-\frac{1}{2}

, incorrectly making it negative. D) has the right slope but wrong y-intercept, likely from an arithmetic error when solving for the line equation. Study tip: For inverse function tangent lines, always remember: swap the point coordinates and take the reciprocal of the slope. The derivative relationship

(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}

is your key formula.

Question 13

Let $f$ be a twice-differentiable, invertible function. If $f(x)>0$ , $f'(x)>0$ , and $f''(x)<0$ for all $x$ in the domain of $f$ , which of the following must be true for the inverse function $g=f^{-1}$ ?

$g''(x) < 0$ (concave down)
$g''(x) = 0$ for some $x$
$g''(x) > 0$ (concave up) (correct answer)
The concavity of $g$ cannot be determined.

Explanation: When you encounter questions about the concavity of inverse functions, you need to connect the properties of the original function to its inverse using the derivative formulas for inverse functions. Given that

f(x) > 0

f'(x) > 0

, and

f''(x) < 0

, let's find

g''(x)

where

g = f^{-1}

. Start with the key formula: if

g = f^{-1}

, then

g'(x) = \frac{1}{f'(g(x))}

. Since

f'(x) > 0

everywhere,

g'(x) > 0

as well. To find

g''(x)

, differentiate

g'(x) = \frac{1}{f'(g(x))}

using the chain rule:

g''(x) = -\frac{f''(g(x)) \cdot g'(x)}{[f'(g(x))]^2}

Now analyze the sign: The denominator

[f'(g(x))]^2

is always positive. The term

g'(x)

is positive. The key factor is

f''(g(x))

, which is negative since

f''(x) < 0

everywhere. Therefore:

g''(x) = -\frac{(\text{negative}) \cdot (\text{positive})}{(\text{positive})} = -\frac{(\text{negative})}{(\text{positive})} = \text{positive}

This means

g''(x) > 0

, so

g

is concave up everywhere. Choice A is wrong because

g''(x) > 0

, not negative. Choice B is incorrect since

g''(x)

is never zero—it's strictly positive. Choice D is wrong because we can definitively determine the concavity using the given information. Study tip: Remember that inverse functions "flip" concavity when the original function has consistent monotonicity. The formula

g''(x) = -\frac{f''(g(x)) \cdot g'(x)}{[f'(g(x))]^2}

is essential for these problems.

Question 14

Let $f(x) = x^5 + 3x^3 + 2x - 1$ . What is the value of $(f^{-1})'(5)$ ?

$\frac{1}{16}$ (correct answer)
$16$
$\frac{1}{3352}$
$\frac{1}{5}$

Explanation: To find

(f^{-1})'(5)

, we use the formula

(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}

with

a=5

. First, we need to find

b = f^{-1}(5)

, which means we need to solve

f(b) = 5

. So,

b^5 + 3b^3 + 2b - 1 = 5

, which simplifies to

b^5 + 3b^3 + 2b = 6

. By inspection,

b=1

is the solution. Next, we find the derivative of

f(x)

f'(x) = 5x^4 + 9x^2 + 2

. Now we evaluate

f'

b=1

f'(1) = 5(1)^4 + 9(1)^2 + 2 = 16

. Finally,

(f^{-1})'(5) = \frac{1}{f'(1)} = \frac{1}{16}

Question 15

Let $g(x)$ be the inverse of $f(x) = x e^{x^2}$ . Find $g'(e)$ .

$\frac{1}{e^{e}(1+2e^2)}$
$3e$
$\frac{1}{e}$
$\frac{1}{3e}$ (correct answer)

Explanation: When you encounter a problem asking for the derivative of an inverse function, you need to use the inverse function derivative formula: if

g(x)

is the inverse of

f(x)

, then

g'(a) = \frac{1}{f'(g(a))}

. First, find where

f(x) = e

. Since

f(x) = xe^{x^2}

, you need to solve

xe^{x^2} = e

. By inspection or careful analysis,

x = 1

works:

1 \cdot e^{1^2} = e

. Therefore,

g(e) = 1

. Next, find

f'(x)

using the product rule:

f'(x) = e^{x^2} + x \cdot e^{x^2} \cdot 2x = e^{x^2}(1 + 2x^2)

. At

x = 1

f'(1) = e^1(1 + 2 \cdot 1^2) = e(1 + 2) = 3e

. Therefore:

g'(e) = \frac{1}{f'(g(e))} = \frac{1}{f'(1)} = \frac{1}{3e}

Choice D is correct. Choice A gives

\frac{1}{e^{e}(1+2e^2)}

, which incorrectly evaluates

f'

x = e

instead of

x = 1

. Choice B gives

3e

, which is actually

f'(1)

rather than its reciprocal—a common error of forgetting the inverse relationship. Choice C gives

\frac{1}{e}

, which might result from incorrectly computing

f'(1) = e

instead of

3e

. Study tip: Always remember that

g'(a) = \frac{1}{f'(g(a))}

—you evaluate

f'

at the output of the inverse function, not at the input. Finding

g(a)

first is crucial.

Question 16

Let $f$ be a one-to-one differentiable function and let $g=f^{-1}$ . The graph of $y=f(x)$ has a tangent line with a slope of 2 at the point $(1,4)$ , and a tangent line with a slope of $1/2$ at the point $(4,6)$ . What is the value of $g'(4)$ ?

$2$
$\frac{1}{2}$ (correct answer)
$\frac{1}{4}$
$\frac{1}{6}$

Explanation: When you see inverse functions and derivatives together, you're dealing with the derivative of an inverse function formula: if

g = f^{-1}

, then

g'(a) = \frac{1}{f'(g(a))}

. This connects the slope of a function to the slope of its inverse at corresponding points. Since

f

has slope 2 at point

(1,4)

, we know

f(1) = 4

and

f'(1) = 2

. For the inverse function

g

, this means

g(4) = 1

(the coordinates flip). Using the inverse derivative formula:

g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(1)} = \frac{1}{2}

. Let's examine why the other answers are wrong: A)

2

would result from incorrectly thinking

g'(4) = f'(1) = 2

, which confuses the derivative relationship between inverse functions. C)

\frac{1}{4}

comes from mistakenly using

g'(4) = \frac{1}{f(1)} = \frac{1}{4}

, which incorrectly uses the function value instead of its derivative. D)

\frac{1}{6}

results from confusion about which point to use—perhaps incorrectly thinking

g'(4) = \frac{1}{f(4)} = \frac{1}{6}

using the given information that

f(4) = 6

. The correct answer is B)

\frac{1}{2}

. Key strategy: Remember that inverse functions have reciprocal derivatives at corresponding points. When given

f(a) = b

and

f'(a)

, you can find

g'(b) = \frac{1}{f'(a)}

where the input and output values switch between the function and its inverse.

Question 17

An implicit function is defined by the equation $y^3 + xy = 4$ . This equation defines $y$ as a differentiable function of $x$ , say $y=f(x)$ , and also defines $x$ as a differentiable function of $y$ , say $x=g(y)$ . Since $g=f^{-1}$ , what is the value of $g'(3)$ ?

$-\frac{58}{9}$ (correct answer)
$-\frac{9}{58}$
$-\frac{1}{9}$
$\frac{23}{9}$

Explanation: We want to find

g'(3) = (f^{-1})'(3)

, which is

\frac{1}{f'(f^{-1}(3))}

. First, find

x = f^{-1}(3)

by setting

y=3

in the original equation:

3^3 + x(3) = 4 \implies 27 + 3x = 4 \implies 3x = -23 \implies x = -23/3

. So, we need

\frac{1}{f'(-23/3)}

. Now, find

f'(x) = dy/dx

by implicit differentiation with respect to

x

3y^2 \frac{dy}{dx} + y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx}(3y^2+x) = -y \implies \frac{dy}{dx} = \frac{-y}{3y^2+x}

. Evaluate this at the point

(-23/3, 3)

f'(-23/3) = \frac{-3}{3(3^2) + (-23/3)} = \frac{-3}{27 - 23/3} = \frac{-3}{58/3} = -\frac{9}{58}

. The derivative of the inverse is the reciprocal:

g'(3) = \frac{1}{-9/58} = -\frac{58}{9}

Question 18

Let $f(x)$ be a differentiable function and let $g(x) = f^{-1}(x)$ . Suppose the graph of $f$ passes through $(4, 1)$ and the tangent line to $f$ at $x=4$ is parallel to the line $y = 3x+5$ . What is the value of $g'(1)$ ?

$-\frac{1}{3}$
$3$
$\frac{1}{3}$ (correct answer)
$1$

Explanation: This question tests your understanding of the derivative of inverse functions, which follows a specific relationship. When you see

g(x) = f^{-1}(x)

, remember that the derivative of an inverse function is the reciprocal of the original function's derivative at the corresponding point. The key formula is:

g'(a) = \frac{1}{f'(b)}

where

f(b) = a

. Since the graph of

f

passes through

(4, 1)

, we know

f(4) = 1

. This means the inverse function satisfies

g(1) = 4

, so we need to find

g'(1)

. The tangent line to

f

x = 4

is parallel to

y = 3x + 5

, which has slope 3. Therefore,

f'(4) = 3

. Using the inverse function derivative formula:

g'(1) = \frac{1}{f'(4)} = \frac{1}{3}

Looking at the wrong answers: Choice A gives

-\frac{1}{3}

, which incorrectly adds a negative sign—there's no reason for the derivative to be negative here. Choice B gives

3

, which is

f'(4)

itself; this represents forgetting to take the reciprocal. Choice D gives

1

, which might come from incorrectly using the point

(4,1)

without applying the proper inverse derivative relationship. Study tip: For inverse function derivatives, always remember the reciprocal relationship and identify the correct corresponding points. If

f(a) = b

, then

g'(b) = \frac{1}{f'(a)}

. Draw a quick diagram if needed to keep track of which point maps to which.

Question 19

Let $f(x) = x^5 + 3x - 2$ . What is the equation of the tangent line to the graph of $y=f^{-1}(x)$ at the point where $x=2$ ?

$y - 1 = \frac{1}{8}(x - 2)$ (correct answer)
$y - 2 = \frac{1}{8}(x - 1)$
$y - 1 = 8(x - 2)$
$y - 1 = \frac{1}{83}(x - 2)$

Explanation: To find the equation of the tangent line, we need a point and a slope. The point is on the graph of

y=f^{-1}(x)

and has an x-coordinate of 2. Let

g(x)=f^{-1}(x)

Find the point: We need $(2, g(2))$ . To find $g(2)$ , we must solve $f(y)=2$ . So, $y^5 + 3y - 2 = 2$ , which simplifies to $y^5 + 3y - 4 = 0$ . By inspection, $y=1$ is the solution. Thus, $g(2)=1$ . The point of tangency is $(2, 1)$ .
Find the slope: The slope is $g'(2)$ . Using the formula, $g'(2) = \frac{1}{f'(g(2))} = \frac{1}{f'(1)}$ .
Calculate $f'(x)$ . $f'(x) = 5x^4 + 3$ .
Evaluate $f'(1) = 5(1)^4 + 3 = 8$ .
The slope is $g'(2) = 1/8$ .
Using the point-slope form $y - y_1 = m(x - x_1)$ , we get $y - 1 = \frac{1}{8}(x - 2)$ .

Question 20

Let $f(x) = e^{2x} + x^3 + 1$ . If $g(x) = f^{-1}(x)$ , what is the value of $g'(2)$ ?

$1$
$1/2$ (correct answer)
$2$
$1/(2e^4 + 12)$

Explanation: We need to find

g'(2)

using the formula

g'(2) = \frac{1}{f'(g(2))}

. First, we must find the value of

g(2) = f^{-1}(2)

. This means we need to find an

x

such that

f(x)=2

. So we solve

e^{2x} + x^3 + 1 = 2

, which simplifies to

e^{2x} + x^3 = 1

. By inspection, we can see that

x=0

is the solution because

e^{2(0)} + 0^3 = e^0 + 0 = 1

. So,

g(2)=0

. Next, we find the derivative of

f(x)

f'(x) = \frac{d}{dx}(e^{2x} + x^3 + 1) = 2e^{2x} + 3x^2

. Now, we evaluate

f'

x=g(2)=0

f'(0) = 2e^{2(0)} + 3(0)^2 = 2e^0 + 0 = 2

. Finally, we apply the formula:

g'(2) = \frac{1}{f'(0)} = \frac{1}{2}