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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

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


f(x) = √(7 + x sec x)

Textbook Question

Finding Derivative Values


In Exercises 67–72, find the value of (f ∘ g)' at the given value of x.


f(u) = u + 1/cos²u, u = g(x) = πx, x = 1/4

Textbook Question

The edge x of a cube is measured with an error of at most 0.5%. What is the maximum corresponding percentage error in computing the cube’s


a. surface area?

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)