Question

the derivative of f(x) = $\frac{x}{x^{-1}+4}$ is □.

Explanation:

Step1: Rewrite the function

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

Step2: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^2}$. Here, $u = x^2$, $u'=2x$, $v = 1 + 4x$, $v'=4$.
$y'=\frac{2x(1 + 4x)-x^2\times4}{(1 + 4x)^2}$

Step3: Simplify the expression

$y'=\frac{2x+8x^2-4x^2}{(1 + 4x)^2}=\frac{4x^2+2x}{(1 + 4x)^2}$

Answer:

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