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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
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Tutti i libri di testoHass 15th EditionCh. 3 - DerivativesProblema 3.57d
Capitolo 3, Problema 3.57d

Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1.


x ƒ(x) g(x) ƒ'(x) g'(x)
0 1 1 -3 1/2
1 3 5 1/2 -4


Find the first derivatives of the following combinations at the given value of x.


d. ƒ(g(x)), x = 0

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To find the derivative of the composition of functions ƒ(g(x)) at x = 0, we use the chain rule. The chain rule states that the derivative of a composition of functions is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
The chain rule formula for the derivative of ƒ(g(x)) is: (ƒ(g(x)))' = ƒ'(g(x)) * g'(x).
First, evaluate g(x) at x = 0. From the given data, g(0) = 1.
Next, find ƒ'(g(x)) at x = 0. Since g(0) = 1, we need ƒ'(1). From the table, ƒ'(1) = 1/2.
Finally, find g'(x) at x = 0. From the table, g'(0) = 1/2. Now, apply the chain rule: (ƒ(g(x)))' at x = 0 is ƒ'(1) * g'(0) = (1/2) * (1/2).

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Chain Rule

The Chain Rule is a fundamental theorem in calculus used to differentiate composite functions. It states that if you have a function ƒ(g(x)), the derivative can be found by multiplying the derivative of the outer function ƒ with the derivative of the inner function g. This rule is essential for solving problems involving nested functions, as it allows for the systematic calculation of derivatives.
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Intro to the Chain Rule

Function Values and Derivatives

Understanding function values and their derivatives at specific points is crucial for applying calculus concepts. In this context, we need to evaluate ƒ(g(0)) and then find the derivative of that composition. The values of ƒ and g at x = 0, along with their derivatives, provide the necessary information to compute the derivative of the composite function using the Chain Rule.
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Percorso guidato
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Average Value of a Function

Evaluating Derivatives at Specific Points

Evaluating derivatives at specific points involves substituting the given x-value into the derivative expression. In this case, after applying the Chain Rule, we will need to evaluate the resulting expression at x = 0. This step is critical for obtaining the final numerical result, as it translates the abstract derivative into a concrete value that reflects the behavior of the function at that point.
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Critical Points
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e. When is it moving fastest (highest speed)? Slowest?


s = 4 - 7t + 6t² - t³, 0 ≤ t ≤ 4

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Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.


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Find the derivatives with respect to x of the following combinations at the given value of x.


f. (x¹¹ + f(x))⁻², x = 1

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Theory and Examples


In Exercises 51–54,


d. Over what intervals of x-values, if any, does the function y = f(x) increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more about this relationship in Section 4.3.)


y = −1/x

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Theory and Examples


In Exercises 51–54,


d. Over what intervals of x-values, if any, does the function y = f(x) increase as x increases? Decrease as x increases? How is this related to what you found in part (c)? (We will say more about this relationship in Section 4.3.)


y = x³/3

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d. During what year did home prices drop most rapidly and what is an estimate of this rate?

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Lunar projectile motion A rock thrown vertically upward from the surface of the moon at a velocity of 24 m/sec (about 86 km/h) reaches a height of s = 24t − 0.8t² m in t sec.

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