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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 3.6.72
Chapter 3, Problem 3.6.72

Finding Derivative Values


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


f(u) = ((u − 1) / (u + 1))², u = g(x) = (1 / x²) − 1, x = −1

Verified step by step guidance
1
First, understand that you need to find the derivative of the composite function (f ∘ g)(x), which means you need to apply the chain rule. The chain rule states that (f ∘ g)'(x) = f'(g(x)) * g'(x).
Calculate g(x) at x = -1. Substitute x = -1 into g(x) = (1 / x²) - 1 to find g(-1).
Next, find g'(x). Differentiate g(x) = (1 / x²) - 1 with respect to x. Use the power rule and the derivative of a constant.
Now, find f'(u). First, differentiate f(u) = ((u - 1) / (u + 1))² with respect to u. Use the chain rule and the quotient rule to differentiate this expression.
Finally, substitute g(-1) into f'(u) to find f'(g(-1)), and multiply this by g'(-1) to find (f ∘ g)'(-1).

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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 technique in calculus used to differentiate composite functions. If you have a function h(x) = f(g(x)), the derivative h'(x) 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 rule is essential for finding derivatives of nested functions like f ∘ g.
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Intro to the Chain Rule

Composite Functions

Composite functions involve applying one function to the results of another, denoted as (f ∘ g)(x) = f(g(x)). Understanding how to work with composite functions is crucial for applying the chain rule effectively, as it requires recognizing the inner and outer functions and their respective derivatives.
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Evaluate Composite Functions - Special Cases

Derivative of Rational Functions

Rational functions are quotients of polynomials, and their derivatives can be found using the quotient rule. For a function f(u) = (u - 1)/(u + 1), the derivative involves differentiating the numerator and denominator separately and applying the quotient rule: (v'u - uv')/v², where u and v are the numerator and denominator, respectively.
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Intro to Rational Functions
Related Practice
Textbook Question

Differential Estimates of Change


In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.


The change in the lateral surface area S = 2πrh of a right circular cylinder when the height changes from h₀ to h₀ + dh and the radius does not change

Textbook Question

In Exercises 41–58, find dy/dt.


y = sin²(πt − 2)

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) = x / (x − 1)

Textbook Question

Find the derivatives of the functions in Exercises 1–42.


s = (sec t + tan t)⁵

Textbook Question

Second Derivatives


In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.


3 + sin y = y – x³

Textbook Question

Free-Fall Applications

Free fall on Mars and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (s in meters, t in seconds) are s = 1.86t² on Mars and s = 11.44t² on Jupiter. How long does it take a rock falling from rest to reach a velocity of 27.8 m/sec (about 100 km/h) on each planet?