Calculus 1 Quiz: Selecting Derivative Procedures

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Selecting Derivative ProceduresQuestion 1 of 20

To find the derivative of $g(x) = \arctan(e^{3x})$ , which is the correct application of the chain rule?

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

g'(x) = \sec^2(e^{3x}) \cdot \frac{d}{dx}(e^{3x})

g'(x) = \frac{1}{1 + x^2} \cdot e^{3x} \cdot 3

g'(x) = (\frac{d}{dx}(\arctan(x))) \cdot (\frac{d}{dx}(e^{3x}))

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Calculus 1 Quiz

Calculus 1 Quiz: Selecting Derivative Procedures

Practice Selecting Derivative Procedures 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 Selecting Derivative Procedures, 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

To find the derivative of $g(x) = \arctan(e^{3x})$ , which is the correct application of the chain rule?

$g'(x) = \frac{1}{1 + (e^{3x})^2} \cdot \frac{d}{dx}(e^{3x})$ (correct answer)
$g'(x) = \sec^2(e^{3x}) \cdot \frac{d}{dx}(e^{3x})$
$g'(x) = \frac{1}{1 + x^2} \cdot e^{3x} \cdot 3$
$g'(x) = (\frac{d}{dx}(\arctan(x))) \cdot (\frac{d}{dx}(e^{3x}))$

Explanation: The function

g(x)

is a composition

f(u(x))

where the outer function is

f(u) = \arctan(u)

and the inner function is

u(x) = e^{3x}

. The chain rule states that the derivative is

f'(u(x)) \cdot u'(x)

. The derivative of

\arctan(u)

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

. So,

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

. This must be multiplied by the derivative of the inner function,

\frac{d}{dx}(e^{3x})

Question 2

For which of the following equations is implicit differentiation the only practical method to find $\frac{dy}{dx}$ among the standard procedures taught in a first-semester calculus course?

$x^2 + y^2 = 16$
$x^2y - x = 1$
$y = \frac{x^2+1}{\cos(x)}$
$y^5 + xy + x^3 = 2$ (correct answer)

Explanation: When deciding whether to use implicit differentiation, you need to ask: "Can I easily solve for y explicitly?" If solving for y would be extremely difficult or impossible with basic algebraic methods, then implicit differentiation becomes your only practical option. Let's examine each equation. For option A,

x^2 + y^2 = 16

, you can solve for y explicitly:

y = ±\sqrt{16 - x^2}

, then differentiate directly. For option B,

x^2y - x = 1

, you can rearrange to get

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

and use the quotient rule. Option C,

y = \frac{x^2+1}{\cos(x)}

, is already solved for y explicitly, so you'd use the quotient rule combined with trigonometric differentiation. Option D,

y^5 + xy + x^3 = 2

, presents a completely different situation. The

y^5

term makes it algebraically impossible to solve for y using methods available in first-semester calculus. You cannot isolate y without dealing with fifth roots and complex rearrangements that go beyond standard techniques. Implicit differentiation is your only practical path: differentiate both sides with respect to x, getting

5y^4\frac{dy}{dx} + y + x\frac{dy}{dx} + 3x^2 = 0

, then solve for

\frac{dy}{dx}

. The key insight is recognizing when an equation's structure prevents explicit solving. Watch for high-degree terms in y (like

y^5,y^3

, etc.) mixed with other terms containing both x and y—these are strong signals that implicit differentiation is your best (and often only) option.

Question 3

Consider the function $y = (x^2+1)^{\sin(x)}$ . Which differentiation procedure is most suitable for finding $\frac{dy}{dx}$ ?

Logarithmic differentiation, because both the base and the exponent are functions of $x$ . (correct answer)
The power rule, treating $\sin(x)$ as a constant exponent, followed by the chain rule for the base.
The exponential rule for $a^u$ , treating $x^2+1$ as a constant base, followed by the chain rule.
Implicit differentiation after rewriting the function as $y^{1/\sin(x)} = x^2+1$ and applying the quotient rule.

Explanation: Functions of the form

f(x)^{g(x)}

, where both the base and the exponent are variable functions, require logarithmic differentiation. This process involves taking the natural logarithm of both sides, using log properties to bring the exponent down, and then differentiating implicitly.

Question 4

After finding $\frac{dy}{dx}$ for an implicitly defined curve such as $x^3 + y^3 = 6xy$ , the resulting expression for the derivative typically involves both $x$ and $y$ . What is the geometric reason for this?

Because the original equation cannot be solved for $y$ , the derivative must contain $y$ as a way of representing the unknown relationship.
The chain rule, when applied to terms involving $y$ , always introduces an extra $y$ variable into the final expression for $\frac{dy}{dx}$ .
The slope of the tangent line at a point on the curve can depend on both the x- and y-coordinates of that point. (correct answer)
Implicit differentiation yields a special type of derivative that is a function of two variables, unlike an explicit derivative which is a function of one.

Explanation: When you encounter implicit differentiation problems, you're dealing with curves that can't easily be solved for y in terms of x alone. The key insight is understanding what the derivative

\frac{dy}{dx}

geometrically represents: the slope of the tangent line at any point on the curve. The slope of a tangent line at a specific point depends on where you are on the curve - that is, it depends on both the x-coordinate and y-coordinate of that point. Consider the folium of Descartes

x^3 + y^3 = 6xy

: different points

(x,y)

on this curve will generally have different slopes, and those slopes are determined by both coordinates of the point in question. Option C correctly captures this geometric reality - the slope at any point naturally depends on both coordinates of that point. Option A misses the point by focusing on solvability rather than geometry. Even when we can solve for y explicitly, implicit differentiation would still yield expressions involving both variables if we chose to use that method. Option B incorrectly suggests the chain rule automatically introduces extra y variables. The chain rule introduces

\frac{dy}{dx}

terms, not additional y variables in the final answer. Option D confuses the nature of the derivative itself. The derivative

\frac{dy}{dx}

is still fundamentally about how y changes with respect to x, but its value at any point happens to depend on both coordinates of that point. Remember: when you see both x and y in your implicit derivative, think "the slope depends on where I am on the curve."

Question 5

Let $h(x) = (\sin x)^{x^2}$ . Which of the following differentiation methods is the most direct and effective for finding $h'(x)$ ?

Apply the power rule, treating the exponent $x^2$ as a constant, and then use the chain rule.
Apply the rule for differentiating exponential functions, treating the base $\sin x$ as a constant.
Apply logarithmic differentiation by first taking the natural logarithm of both sides of the equation $y = h(x)$ . (correct answer)
Apply the product rule to the functions $f(x)=\sin x$ and $g(x)=x^2$ .

Explanation: The function

h(x)

is of the form

f(x)^{g(x)}

, where both the base and the exponent are functions of

x

. For this type of function, logarithmic differentiation is the standard and most effective method. Taking the natural logarithm gives

\ln(y) = x^2 \ln(\sin x)

, which can then be differentiated implicitly using the product rule. Options A and B are incorrect because they treat a variable part of the function as a constant. Option D is incorrect because the function is not a product.

Question 6

The first derivative of the curve defined by $x^2 + y^2 = 25$ is $\frac{dy}{dx} = -\frac{x}{y}$ . To find the second derivative $\frac{d^2y}{dx^2}$ , the expression for $\frac{dy}{dx}$ is differentiated with respect to $x$ . Which of the following statements correctly describes this procedure?

Apply the quotient rule to $-\frac{x}{y}$ , treating $y$ as a constant because the differentiation is with respect to $x$ .
Apply the quotient rule to $-\frac{x}{y}$ , treating $y$ as a function of $x$ and using the chain rule to find its derivative, $\frac{dy}{dx}$ . (correct answer)
Differentiate the original equation twice, which gives $2 + 2\frac{d^2y}{dx^2} = 0$ , and then solve for the second derivative.
Substitute $y = \sqrt{25-x^2}$ into the first derivative expression to eliminate $y$ before taking the second derivative.

Explanation: When finding the second derivative implicitly, we must differentiate the first derivative,

\frac{dy}{dx} = -\frac{x}{y}

, with respect to

x

. Since

y

is implicitly a function of

x

, we cannot treat it as a constant (ruling out A). We must use the quotient rule, and when differentiating the

y

in the denominator, we must use the chain rule, resulting in a

\frac{dy}{dx}

term. This is correctly described in B. Option C is incorrect; differentiating

2x+2y\frac{dy}{dx}=0

requires the product rule on the second term. Option D is a valid strategy for the top half of the circle, but it is not the required procedure for implicit differentiation and does not work for the entire curve without considering

y = -\sqrt{25-x^2}

separately.

Question 7

Let $f(x) = \sqrt{\tan(e^{3x})}$ . Finding $f'(x)$ requires repeated application of the chain rule. Which of the following correctly lists the sequence of functions whose derivatives are taken, from outermost to innermost?

square root, tangent, exponential function, linear function (correct answer)
tangent, exponential function, linear function, square root
exponential function, tangent, square root, linear function
square root, linear function, tangent, exponential function

Explanation: The function can be decomposed into layers of composition. The outermost operation is the square root, so

f(u) = \sqrt{u}

where

u=\tan(e^{3x})

. The next layer is the tangent function,

u(v) = \tan(v)

where

v=e^{3x}

. The next layer is the exponential function,

v(w) = e^w

where

w=3x

. The innermost function is the linear function

w(x)=3x

. The chain rule proceeds by differentiating from the outermost function to the innermost. Thus, the correct sequence is square root, tangent, exponential, then linear.

Question 8

Let $f$ be a differentiable, one-to-one function, and let $g(x) = f^{-1}(x)$ . The tangent line to the graph of $y=f(x)$ at $x=2$ is $y = 5x - 7$ . Which procedure leads to the value of $g'(3)$ ?

Find the slope of the given tangent line, which is 5, and conclude that $g'(3)$ must be its reciprocal, $1/5$ .
Identify $f(2)=3$ and $f'(2)=5$ from the tangent line, then use these values in the formula $g'(3) = 1/f'(g(3))$ . (correct answer)
Find the inverse of the tangent line function, $y = (x+7)/5$ , and then evaluate its derivative at $x=3$ .
Use the slope $f'(2)=5$ and calculate $g'(3)$ as $1/f'(3)$ , assuming the derivative value is the same.

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

g = f^{-1}

, then

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

. The key is identifying the right values to plug into this formula. From the tangent line

y = 5x - 7

x = 2

, you can extract two crucial pieces of information: the slope is

f'(2) = 5

, and the point of tangency gives you

f(2) = 5(2) - 7 = 3

. Since

f(2) = 3

, this means

g(3) = 2

(because

g

undoes what

f

does). Now you can apply the formula:

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

. This is exactly what option B describes. Option A makes the common error of thinking

g'(x)

always equals

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

at corresponding points, but this ignores that the domains are switched for inverse functions. Option C incorrectly treats the tangent line as if it were the actual function

f(x)

—the tangent line only approximates

f

near

x = 2

, so its inverse isn't

g

. Option D uses

f'(3)

instead of

f'(2)

, but we don't know

f'(3)

and we need

f'

evaluated at

g(3) = 2

, not at 3. Remember: for inverse function derivatives, you need both the derivative of the original function AND the correct input-output relationship between

f

and

g

. Always identify these paired values from the given information first.

Question 9

A student must find the derivatives of three functions: I. $y = x^{\pi}$ , II. $y = \pi^x$ , III. $y = x^x$ . Which choice correctly pairs the primary differentiation procedure for each function?

I: Power Rule, II: Logarithmic Differentiation, III: Exponential Rule
I: Exponential Rule, II: Power Rule, III: Power Rule
I: Power Rule, II: Exponential Rule, III: Logarithmic Differentiation (correct answer)
I: Chain Rule, II: Power Rule, III: Implicit Differentiation

Explanation: When you encounter functions with variables as bases and exponents, identifying the correct differentiation technique is crucial. The key is recognizing whether the base, exponent, or both contain variables. For function I,

y = x^{\pi}

, you have a variable base with a constant exponent. This is a perfect candidate for the Power Rule:

\frac{d}{dx}[x^n] = nx^{n-1}

, giving you

y' = \pi x^{\pi-1}

. Function II,

y = \pi^x

, has a constant base with a variable exponent. This is an exponential function requiring the Exponential Rule:

\frac{d}{dx}[a^x] = a^x \ln(a)

, yielding

y' = \pi^x \ln(\pi)

. Function III,

y = x^x

, has both variable base and variable exponent. Neither the Power Rule nor Exponential Rule applies directly. You need logarithmic differentiation: take the natural log of both sides, then differentiate implicitly to get

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

. This confirms answer C is correct. Answer A incorrectly suggests using logarithmic differentiation for

\pi^x

, which is unnecessary since it's a standard exponential function. Answer B misclassifies

\pi^x

as needing the Power Rule and

x^x

as needing the Power Rule, both wrong. Answer D incorrectly pairs the Chain Rule with

x^{\pi}

and suggests implicit differentiation for

x^x

, though logarithmic differentiation is more direct. Study tip: Memorize this pattern: constant exponent → Power Rule, constant base → Exponential Rule, both variable → logarithmic differentiation.

Question 10

Let $f(x)$ be a differentiable function. To find the derivative of $y = [f(x)]^2$ , one student uses the power rule combined with the chain rule. A second student uses the product rule on $y = f(x) \cdot f(x)$ . Which statement accurately compares these procedures?

Both procedures are valid and will yield the same result, $2f(x)f'(x)$ . (correct answer)
Only the chain rule procedure is valid; the product rule cannot be applied when the functions are identical.
Only the product rule procedure is valid; the power rule only applies to variables like $x^n$ , not functions.
The procedures are both valid but yield different results that are only equivalent under certain conditions for $f(x)$ .

Explanation: Both procedures are correct and demonstrate the consistency of differentiation rules. Using the chain rule (specifically, the generalized power rule):

\frac{d}{dx}[f(x)]^2 = 2[f(x)]^1 \cdot f'(x) = 2f(x)f'(x)

. Using the product rule on

f(x) \cdot f(x)

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

. The results are identical.

Question 11

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

$\frac{1}{14}$
$\frac{1}{5}$ (correct answer)
$5$
$\frac{1}{35}$

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

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

. Here,

a=2

. First, we must find

b = f^{-1}(2)

, which is the value of

x

such that

f(x)=2

. We solve

x^3 + 2x - 1 = 2

, which simplifies to

x^3 + 2x - 3 = 0

. By inspection,

x=1

is the solution, so

f^{-1}(2)=1

. Next, we find the derivative of

f(x)

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

. Now we evaluate

f'(f^{-1}(2)) = f'(1) = 3(1)^2 + 2 = 5

. Finally,

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

. Distractor D is

1/f'(2)

, a common error. Distractor C is

f'(1)

without taking the reciprocal. Distractor A is

1/f(2)

Question 12

For which of the following functions is it necessary to use logarithmic differentiation or a related method (like rewriting the base as $e^{\ln(\dots)}$ ) to find its derivative?

$f(x) = (\pi)^x$
$g(x) = x^{\pi}$
$h(x) = (\ln x)^2$
$k(x) = (x+1)^{\sin x}$ (correct answer)

Explanation: Logarithmic differentiation is required for functions of the form

y=u(x)^{v(x)}

, where both the base

u(x)

and the exponent

v(x)

are non-constant functions of

x

. Option D,

k(x) = (x+1)^{\sin x}

, fits this form. Option A,

f(x) = (\pi)^x

, is an exponential function with a constant base, whose derivative is

(\ln \pi)(\pi)^x

. Option B,

g(x) = x^{\pi}

, is a power function with a constant exponent, whose derivative is

\pi x^{\pi-1}

. Option C,

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

, can be differentiated using the power rule and the chain rule.

Question 13

Given the equation $x \ln(y) + y^3 = 2x$ , which equation is the result of correctly differentiating with respect to $x$ to find $\frac{dy}{dx}$ ?

$\ln(y) + \frac{x}{y} + 3y^2 = 2$
$\ln(y) + \frac{x}{y}\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 2$ (correct answer)
$\frac{1}{y}\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 2$
$1 \cdot \frac{1}{y} + 3y^2 \frac{dy}{dx} = 2$

Explanation: To find

\frac{dy}{dx}

implicitly, we differentiate each term with respect to

x

. The term

x \ln(y)

requires the product rule: the derivative is

(1) \cdot \ln(y) + x \cdot \frac{1}{y} \frac{dy}{dx}

. The term

y^3

requires the chain rule: its derivative is

3y^2 \frac{dy}{dx}

. The derivative of

2x

2

. Combining these results yields the equation in option B. Option A omits the

\frac{dy}{dx}

factors that arise from the chain rule. Option C applies the product rule incorrectly to the first term. Option D also applies the product rule incorrectly, differentiating

x

and

\ln(y)

separately.

Question 14

To find the derivative of the function $f(x) = e^{\sin(x^2)}$ , which differentiation rule must be applied as the primary, outermost step?

The product rule, because the function involves multiple expressions multiplied together.
The specific differentiation rule for $e^x$ , applied directly to the function's argument.
The power rule, because the function contains the expression $x^2$ .
The chain rule, because the function is a composition of an exponential, a sine, and a power function. (correct answer)

Explanation: When you encounter a function like

f(x) = e^{\sin(x^2)}

, you need to identify its structure to choose the right differentiation approach. This function is a composition - an exponential function containing a sine function, which itself contains a quadratic function. The chain rule is your primary tool here because you're differentiating a composite function. Think of it as peeling an onion: the outermost layer is

e^u

(where

u = \sin(x^2)

), the middle layer is

\sin(v)

(where

v = x^2

), and the innermost layer is

x^2

. To find

f'(x)

, you'll need:

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

This uses the chain rule three times, working from outside to inside. Looking at the wrong answers: Choice A suggests the product rule, but there's no multiplication of separate functions here - just one composite function. Choice B mentions applying the exponential rule directly to the argument, but you can't ignore that the argument itself is a complex function requiring its own differentiation. Choice C focuses on the power rule for

x^2

, but while the power rule will eventually be used (as the innermost step), it's not the primary approach needed. Study tip: When you see nested functions like

f(g(h(x)))

, always start with the chain rule as your primary strategy. Look for the outermost function first, then work your way inward. The chain rule often combines with other rules, but it governs the overall approach.

Question 15

Consider the implicitly defined curve $y^2 - 2\ln(xy) = 1$ . Which of the following describes the first step in the correct procedure to find the slope of the tangent line at a point $(x,y)$ ?

Solve for $y$ explicitly in terms of $x$ , then differentiate the resulting function using standard rules.
Differentiate each term of the equation with respect to $y$ , treating $x$ as a constant, to find $dx/dy$ .
Differentiate each term of the equation with respect to $x$ , applying the chain and product rules where necessary. (correct answer)
Substitute the coordinates of a specific point of tangency into the equation, then differentiate the equation.

Explanation: When you encounter an implicitly defined equation like

y^2 - 2\ln(xy) = 1

, you're dealing with a curve where y isn't explicitly solved for in terms of x. To find the slope (which is

\frac{dy}{dx}

), you need implicit differentiation. The correct first step is to differentiate both sides of the equation with respect to x, which is exactly what option C describes. As you differentiate each term, you'll need the chain rule for terms involving y (since y depends on x), and the product rule for terms like

\ln(xy)

. For example, differentiating

y^2

gives

2y\frac{dy}{dx}

, and differentiating

\ln(xy)

requires both the chain rule and product rule, yielding

\frac{1}{xy}(y + x\frac{dy}{dx})

. Option A is impossible here because this equation cannot be solved explicitly for y in terms of x due to the logarithmic term. Option B suggests differentiating with respect to y to find

\frac{dx}{dy}

, but the question asks for the slope

\frac{dy}{dx}

, not its reciprocal. While you could find

\frac{dx}{dy}

and then take its reciprocal, this creates unnecessary work. Option D puts the cart before the horse—you need the general derivative formula first before you can evaluate it at any specific point. Study tip: Whenever you see an equation that mixes x and y terms (especially with transcendental functions like ln), think "implicit differentiation." Always differentiate with respect to x first, then solve for

\frac{dy}{dx}

Question 16

For the curve defined by the equation $y \cos(x) = x^2 + y^2$ , which of the following expressions represents $\frac{dy}{dx}$ ?

$\frac{2x + y \sin(x)}{\cos(x) - 2y}$ (correct answer)
$\frac{2x - \sin(x)}{-\cos(x) - 2y}$
$\frac{2x + y \sin(x)}{2y - \cos(x)}$
$\frac{2x - y \sin(x)}{\cos(x) + 2y}$

Explanation: We use implicit differentiation. Differentiating both sides with respect to

x

: The left side requires the product rule:

\frac{d}{dx}(y \cos(x)) = \frac{dy}{dx}\cos(x) + y(-\sin(x))

. The right side is differentiated term-by-term:

\frac{d}{dx}(x^2 + y^2) = 2x + 2y\frac{dy}{dx}

. Setting them equal:

\frac{dy}{dx}\cos(x) - y\sin(x) = 2x + 2y\frac{dy}{dx}

. Now, we solve for

\frac{dy}{dx}

. Grouping terms with

\frac{dy}{dx}

gives

\frac{dy}{dx}(\cos(x) - 2y) = 2x + y\sin(x)

. Dividing yields

\frac{dy}{dx} = \frac{2x + y \sin(x)}{\cos(x) - 2y}

Question 17

A curve is defined by $\sin(x+y) = y^2 \cos(x)$ . Which of the following describes the necessary first step to find the slope of the tangent line at a point $(x, y)$ ?

Differentiate $\sin(x+y)$ using the chain rule and differentiate $y^2 \cos(x)$ using the product rule. (correct answer)
Isolate $y$ on one side of the equation and then use standard differentiation rules on the resulting explicit function.
Differentiate each term with respect to $x$ , treating $y$ as a constant because it is the dependent variable.
Use the quotient rule on $\frac{\sin(x+y)}{\cos(x)}$ after moving $\cos(x)$ to the left side of the equation.

Explanation: The equation defines

y

implicitly. To find the slope (

\frac{dy}{dx}

), we must differentiate both sides with respect to

x

. The left side,

\sin(x+y)

, is a composite function, requiring the chain rule. The right side,

y^2 \cos(x)

, is a product of two functions of

x

(since

y

depends on

x

), requiring the product rule.

Question 18

A curve is defined by the equation $x^2y + y^3 = 10$ . To find $\frac{dy}{dx}$ , which of the following describes the necessary application of differentiation rules?

Differentiate $x^2y$ using the product rule and $y^3$ using the chain rule, as part of implicit differentiation. (correct answer)
First, solve the equation for $y$ explicitly in terms of $x$ , and then differentiate the resulting function directly.
Differentiate each term separately: $x^2$ becomes $2x$ , $y$ becomes $\frac{dy}{dx}$ , and $y^3$ becomes $3y^2\frac{dy}{dx}$ .
Apply the chain rule to the term $x^2y$ and the power rule to $y^3$ after isolating the $y$ terms on one side.

Explanation: The equation defines

y

implicitly as a function of

x

. Therefore, implicit differentiation must be used. The term

x^2y

is a product of two functions of

x

(since

y

is a function of

x

), so the product rule is required. The term

y^3

is a composition of the cubing function and

y(x)

, so the chain rule is required, yielding

3y^2 \frac{dy}{dx}

Question 19

Let $f(x) = x^5 + 2x + 1$ . If $g(x)$ is the inverse of $f(x)$ , what procedure is required to find $g'(4)$ ?

Calculate $f'(x) = 5x^4 + 2$ , evaluate $f'(4)$ , and then compute the reciprocal $1/f'(4)$ .
Find a value $c$ such that $f(c) = 4$ , then calculate $f'(c)$ , and finally compute the reciprocal $1/f'(c)$ . (correct answer)
Find an explicit formula for $g(x)$ by solving the equation $y = x^5 + 2x + 1$ for $x$ , then differentiate $g(x)$ and evaluate at $x=4$ .
Calculate $f'(x)$ , find the inverse of the derivative function, $(f')^{-1}(x)$ , and then evaluate this new function at $x=4$ .

Explanation: The theorem for the derivative of an inverse function states that

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

. Here,

g = f^{-1}

and

a=4

. The first step is to find

c = f^{-1}(4)

, which means we must solve

f(c) = c^5 + 2c + 1 = 4

. By inspection,

c=1

. The next step is to find

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

. Finally, we evaluate

f'(c) = f'(1) = 5(1)^4 + 2 = 7

, and take the reciprocal. So,

g'(4) = 1/7

. Procedure B correctly outlines these steps.

Question 20

Let $h(x) = f(g(x))$ , where $g(2) = 3$ , $g'(2) = -1$ , $f'(2) = 4$ , and $f'(3) = 5$ . Which calculation correctly finds $h'(2)$ ?

$h'(2) = f'(g(2)) \cdot g'(2) = f'(3) \cdot (-1)$ (correct answer)
$h'(2) = f'(2) \cdot g'(2) = 4 \cdot (-1)$
$h'(2) = f(g'(2)) \cdot g(2) = f(-1) \cdot 3$
$h'(2) = f'(g(2)) + g'(2) = f'(3) + (-1)$

Explanation: This is a direct application of the chain rule. The formula is

h'(x) = f'(g(x)) \cdot g'(x)

. To find

h'(2)

, we substitute

x=2

h'(2) = f'(g(2)) \cdot g'(2)

. Using the given values,

g(2)=3

and

g'(2)=-1

, so

h'(2) = f'(3) \cdot (-1)

. Since

f'(3)=5

, the result is -5. Choice A shows the correct procedure.