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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.6.73
Chapter 3, Problem 3.6.73

Assume that f'(3) = −1, g'(2) = 5, g(2) = 3, and y = f(g(x)). What is y' at x = 2?

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To find y' at x = 2, we need to use the chain rule for differentiation. The chain rule states that if y = f(g(x)), then y' = f'(g(x)) * g'(x).
First, identify the values given in the problem: f'(3) = -1, g'(2) = 5, and g(2) = 3.
Since y = f(g(x)), we need to evaluate g(x) at x = 2. We are given that g(2) = 3, so g(x) at x = 2 is 3.
Next, substitute g(x) = 3 into f'(g(x)). We know f'(3) = -1, so f'(g(x)) at x = 2 is -1.
Finally, apply the chain rule: y' = f'(g(x)) * g'(x). Substitute the values: y' = -1 * 5. Calculate this product to find y' at x = 2.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Chain Rule

The chain rule is a fundamental theorem in calculus used to differentiate composite functions. If y = f(g(x)), then the derivative y' is found by multiplying the derivative of the outer function f at g(x) by the derivative of the inner function g at x. This allows us to compute the rate of change of y with respect to x when y is a function of another function.
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Intro to the Chain Rule

Derivative Evaluation

Derivative evaluation involves calculating the derivative of a function at a specific point. In this context, it requires substituting the given values into the derivative formula obtained from applying the chain rule. For y = f(g(x)), y' at x = 2 is calculated using f'(g(2)) and g'(2), which are provided in the problem statement.
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Composite Function

A composite function is formed when one function is applied to the result of another function, denoted as y = f(g(x)). Understanding composite functions is crucial for applying the chain rule, as it involves differentiating the outer function f with respect to the inner function g, and then differentiating g with respect to x. This concept is key to solving problems involving nested functions.
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Related Practice
Textbook Question

Using the Alternative Formula for Derivatives


Use the formula

f'(x) = lim (z → x) (f(z) − f(x)) / (z − x)

to find the derivative of the functions in Exercises 23–26.


g(x) = 1 + √x

Textbook Question

The general polynomial of degree n has the form

P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₂x² + a₁x + a₀,

where aₙ ≠ 0. Find P'(x).

Textbook Question

In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.


g(x) = 8 / x², (2, 2)

Textbook Question

Find the derivatives of the functions in Exercises 19–40.


f(x) = √(7 + x sec x)

Textbook Question

If the original 24 m edge length x of a cube decreases at the rate of 5 m/min, when x = 3 m at what rate does the cube’s

a. surface area change?

Textbook Question

Computer Explorations


Use a CAS to perform the following steps in Exercises 55–62.


a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point P satisfies the equation.


xy³ + tan(x + y) = 1, P(π/4, 0)