Chapter 4.1, Problem 38E

To determine

To calculate: $(f.g)\text{'}$ for the given function at given value of x.

Answer to Problem 38E

The answer is $\frac{d(f(g(-1)))}{dx}=-8$ .

Explanation of Solution

Given information:

Given functions are $f(u)={\left(\frac{u-1}{u+1}\right)}^2,u=g(x)=\frac{1}{x^2}-1\text{and}\text{value}\text{of}x=-1$ .

Formula used:

  $\begin{array}{l}\frac{d(f(g(x)))}{dx}=f\text{'}(g(x))\times g\text{'}(x)\\ \frac{d(x^n)}{dx}=n{x}^{n-1}\end{array}$

Also quotient, chain and addition rule.

Calculation:

If $f(u)={\left(\frac{u-1}{u+1}\right)}^2,u=g(x)=\frac{1}{x^2}-1$

Then,

  $\begin{array}{c}f(g(x))={\left(\frac{\left(\frac{1}{x^2}-1\right)-1}{\left(\frac{1}{x^2}-1\right)+1}\right)}^2\\ ={\left(\frac{\left(\frac{1}{x^2}-2\right)}{\frac{1}{x^2}}\right)}^2={\left(\frac{\left(\frac{1-2x^2}{x^2}\right)}{\frac{1}{x^2}}\right)}^2\\ ={\left(1-2x^2\right)}^2\\ =1+4x^4-4x^2\end{array}$

Differentiating the function w.r.t x we get,

  $\begin{array}{c}\frac{d(f(g(x)))}{dx}=\frac{d\left(\left(1+4x^4-4x^2\right)\right)}{dx}\\ =4\times 4x^3-4\times 2x^1\\ =16x^3-8x\end{array}$

Now $x=-1$ in the above equation we get,

  $\begin{array}{c}\frac{d(f(g(-1)))}{dx}=16x^3-8x\\ =-16+8\\ =-8\end{array}$

The answer is $\frac{d(f(g(-1)))}{dx}=-8$ .