Question

In Exercises \(33-38,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\left(\frac{u-1}{u+1}\right)^{2}, \quad u=g(x)=\frac{1}{x^{2}}-1, \quad x=-1$$

Short answer

\((f \circ g)'(-1) = 0.

Step-by-step solution

Find \(g(-1)\)

Substitute \(x = -1\) into \(g(x)=\frac{1}{x^{2}}-1\) to find the value of \(g(-1)\).\n So, \(g(-1) = \frac{1}{(-1)^2} - 1 = 1 - 1 = 0\)

Find \(f'(u)\)

The derivative of \(f(u)=(\frac{u-1}{u+1})^{2}\) can can be simplified as:\n\Let \(v = \frac{u-1}{u+1}\). Then the derivative of \(f = v^2\) is \(f'(u) = 2v\), by power rule. Also, the derivative of \(v\) with respect to \(u\) is \(\frac{-2}{(u+1)^2}\), by quotient rule. Hence, \(f'(u)=2v*\frac{-2}{(u+1)^2}=-\frac{4(u-1)}{(u+1)^3}\).

Find \(g'(x)\)

The derivative of \(g(x)=\frac{1}{x^{2}}-1\) can be found using the power rule and it is \(g'(x) = -2 * \frac{1}{x^{3}}\)

Substitute the values into the chain rule formula

Substitute \(f'(g(-1))\), \(g'(-1)\) into the chain rule formula \(f(g(x))'=f'(g(x))*g'(x)\).\n\So, \((f \circ g)'(-1)=f'(g(-1))*g'(-1)=-\frac{4(g(-1)-1)}{(g(-1)+1)^{3}}*-2*(-1)^{-3} = 0*\frac{8}{-1^3} = 0\).