differentiate the function. f(x)=x/(x^(-1)+4) f(x)=

Explanation:

Step1: Rewrite the function

First, rewrite $f(x)=\frac{x}{x^{-1}+4}=\frac{x}{\frac{1}{x}+4}=\frac{x}{\frac{1 + 4x}{x}}=\frac{x^{2}}{1 + 4x}$.

Step2: Apply the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{2}$, so $u^\prime=2x$, and $v = 1 + 4x$, so $v^\prime = 4$. [ \begin{align*} f^\prime(x)&=\frac{(2x)(1 + 4x)-x^{2}(4)}{(1 + 4x)^{2}}\ &=\frac{2x+8x^{2}-4x^{2}}{(1 + 4x)^{2}}\ &=\frac{2x + 4x^{2}}{(1 + 4x)^{2}}\ &=\frac{2x(1 + 2x)}{(1 + 4x)^{2}} \end{align*} ]

Answer:

$\frac{2x(1 + 2x)}{(1 + 4x)^{2}}$