Problem 84 - University Calculus: Early Transcendentals

Question

In Exercises \(79-84,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\left(\frac{u-1}{u+1}\right)^{2}, u=g(x)=\frac{1}{x^{2}}-1, x=-1$$

Step-by-Step Solution

Answer

The derivative \((f \circ g)^{\prime}(-1)\) equals zero.

Step 1: Understand the Function Composition

We have two functions to compose: \( f(u) = \left( \frac{u-1}{u+1} \right)^{2} \) and \( u = g(x) = \frac{1}{x^2} - 1 \). We need to compute \((f \circ g)(x)\) which translates to \(f(g(x))\).

Step 2: Find \(f(g(x))\)

Substitute \(u = g(x)\) into \(f(u)\). This gives us \(f\left(\frac{1}{x^2} - 1\right) = \left( \frac{\frac{1}{x^2} - 1 - 1}{\frac{1}{x^2} - 1 + 1} \right)^{2}\). Simplify to get \(f(g(x)) = \left( \frac{x^2 - 2}{x^2} \right)^{2} \).

Step 3: Differentiate \((f \circ g)(x)\)

To find \((f \circ g)^{\prime}(x)\), use the chain rule: \( (f \circ g)^{\prime}(x) = f^{\prime}(g(x)) \cdot g^{\prime}(x) \). First, differentiate \(f(u)\) with respect to \(u\). For \(f(u) = \left(\frac{u-1}{u+1}\right)^2\), apply the chain rule and quotient rule: \(f^{\prime}(u) = 2 \cdot \frac{u-1}{u+1} \cdot \frac{1}{(u+1)^2}\).

Step 4: Find \(f^{\prime}(g(x))\)

Substitute \(u = g(x) = \frac{1}{x^2} - 1\) into \(f^{\prime}(u)\). We have \(f^{\prime}(g(x)) = 2 \cdot \frac{\frac{1}{x^2} - 1 - 1}{\left(\frac{1}{x^2}\right)^2}\). Simplify the expression.

Step 5: Differentiate \(g(x)\)

Calculate \(g^{\prime}(x)\) from \(g(x) = \frac{1}{x^2} - 1\). Differentiate to get \(g^{\prime}(x) = -\frac{2}{x^3}\).

Step 6: Evaluate \((f \circ g)^{\prime}(x)\) at \(x = -1\)

Substitute \(x = -1\) into the expressions for \(f^{\prime}(g(x))\) and \(g^{\prime}(x)\). Combing these using the chain rule gives: \((f \circ g)^{\prime}(x) = f^{\prime}(g(-1)) \cdot g^{\prime}(-1)\).

Key Concepts

Function Composition

Function composition involves combining two or more functions to form a new function. To understand it, think of function composition as plugging one function into another. In this exercise, we start with two functions:

• The first function is given by \( f(u) = \left(\frac{u-1}{u+1}\right)^{2} \) and is defined in terms of a variable \(u\).
• The second function is \( g(x) = \frac{1}{x^2} - 1 \).

To find the composition \((f \circ g)(x)\), we replace \(u\) in \( f(u) \) with \( g(x) \). This yields \( (f \circ g)(x) = f(g(x)) = \left(\frac{1/x^2 - 1 - 1}{1/x^2 - 1 + 1}\right)^2 \). This simplifies to \( \left(\frac{x^2 - 2}{x^2}\right)^2 \). This new function captures how \(f\) and \(g\) transform \(x\).

Chain Rule

The chain rule is a fundamental concept for finding the derivative of a composition of functions. If you need to differentiate a composite function, this rule is a life-saver! For functions \( f(u) \) and \( u = g(x) \), the chain rule tells us:

• \((f \circ g)'(x) = f'(g(x)) \cdot g'(x)\).

In simple terms, you first find the derivative of the outer function, \( f(u) \), but with respect to \(u\). Then, multiply that by the derivative of the inner function, \(g(x)\), with respect to \(x\). This method helps to break down complex derivatives into simpler parts. In the exercise, understanding how to apply this step simplifies the process of finding the derivative of composed functions.

Derivative

Derivatives represent the rate of change of a function. In calculus, they are used to determine how a function's value changes as its input changes. For this problem, we need the derivatives of both \(f(u)\) and \(g(x)\):

• For \(f(u) = \left(\frac{u-1}{u+1}\right)^2\), its derivative involves using both the chain rule and the quotient rule. The result is \( f'(u) = 2 \cdot \frac{u-1}{u+1} \cdot \frac{1}{(u+1)^2} \).
• For \(g(x) = \frac{1}{x^2} - 1\), the derivative is simpler: \( g'(x) = -\frac{2}{x^3} \).

Derivatives help to understand the nature and behavior of function compositions, showing how each part of the function influences the outcome when changes in \(x\) occur.

Quotient Rule

The quotient rule is specifically for taking derivatives of functions that can be expressed as a ratio of two other functions. For a function of the form \( h(x) = \frac{u(x)}{v(x)} \), the rule gives: \[ h'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \]In the exercise, \(f(u)\) required this rule. It is written as \(\left(\frac{u-1}{u+1}\right)\), thus both the numerator and denominator are functions of \(u\). The quotient rule was employed within the broader derivative calculation of \(f(u)\), alongside the chain rule, to achieve an accurate formulation of \(f'(u)\). Understanding the quotient rule allows solving more complex calculus problems involving fractions or division of functions.