Question

In Exercises \(33-38,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=1-\frac{1}{u}, \quad u=g(x)=\frac{1}{1-x}, \quad x=-1$$

Short answer

The value of \((f \circ g)'\) at \(x = -1\) is \(-1/4\).

Step-by-step solution

Identify the functions

This exercise involves the function \(f(u) = 1 - \frac{1}{u}\) and \(u = g(x) = \frac{1}{1-x}\). Here, \(f\) is a function of \(u\) and \(u\) is a function of \(x\). This sets up a situation of a composed function which is \(f(g(x))\). This is normally referred as \(f \circ g\). The task is to find the derivative of \(f \circ g\) at \(x = -1\).

Apply the chain rule

The chain rule in calculus states that if we have a composite function \(h(x) = f(g(x))\), then its derivative \(h'\) can be computed as \(h'(x) = f'(g(x)) \cdot g'(x)\). In this case, we get \((f \circ g)'(x) = f'(g(x)) \cdot g'(x)\).

Compute the derivative of each function

The derivative of the function \(f(u) = 1 - \frac{1}{u}\) is \(f'(u) = \frac{1}{u^2}\), and the derivative of \(g(x) = \frac{1}{1-x}\) is \(g'(x) = -\frac{1}{(1-x)^2}\). These derivatives are computed using the rule \(d/dx[1/u] = -1/u^2\) and \(d/dx[1/(1-x)] = -1/(1-x)^2\).

Substitute and Simplify

Substitute the derivatives into the chain rule equation given in Step 2, and also substitute \(g(x)\) in for \(u\). This gives: \((f \circ g)'(x) = f'(g(x)) \cdot g'(x) = \frac{1}{(1-x)^2} \cdot -\frac{1}{(1-x)^2}\).

Evaluate the derivative at \(x = -1\)

Substitute \(x = -1\) into the derivative equation obtained in Step 4 gives \((f \circ g)'(-1) = \frac{1}{(1-(-1))^2} \cdot -\frac{1}{(1-(-1))^2} = -1/4\).