Question

In Exercises \(33-38,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\frac{2 u}{u^{2}+1}, \quad u=g(x)=10 x^{2}+x+1, \quad x=0$$

Short answer

\((f \circ g)^\prime(0) = 1.

Step-by-step solution

Find the function composition

First, find the function composition \(f(g(x))\) by replacing every instance of \(u\) in function \(f\) with \(g(x)=10x^{2}+x+1\), \[f(g(x))= \frac{2 *(10x^{2}+x+1)}{(10x^{2}+x+1)^{2}+1}.\]

Derive f and g

Next, use the Chain Rule for differentiation of functions. In general, the derivative of a composition of two functions is given by \((f \circ g)' = f'(g(x)) \cdot g'(x)\),where \(f'(u)\) is the derivative of f with respect to u and \(g'(x)\) is the derivative of g with respect to x. Here, Apply the formula of quotient rule in order to derive f(u), whereby \(x = u\), \[f'(u)=(u^{2} + 1) \cdot 2 - 2u \cdot (2u) / (u^{2} + 1)^{2}= \frac{2u^2 + 2 - 4u^2}{(u^2+1)^{2}}= \frac{2-2u^2}{(u^{2}+1)^{2}}.\] For \(g(x) = 10x^{2} + x + 1\), derive it as \[g'(x)=20x + 1.\]

Apply the Chain Rule

Substitute g, g' and f' into the Chain Rule \[(f \circ g)' = f'(g(x)) \cdot g'(x)= \frac{2-(2*(10x^{2}+x+1)^2)}{((10x^{2}+x+1)^{2}+1)^{2}} \cdot (20x+1).\]

Substituting x value into the function

The final step is to substitute the given x-value, which is \(x = 0\), into the function to find the value.\[(f \circ g)^\prime(0) = \frac{2-(2*(10*0^{2}+0+1)^2)}{((10*0^{2}+0+1)^{2}+1)^{2}} \cdot (20*0+1).\]