Question

find the derivative of the function.
f(x) = \frac{1}{(x^{2}-7)^{4}}
f(x) =

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{1}{(x^{2}-7)^{4}}=(x^{2}-7)^{- 4}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = u^n$ and $u = g(x)$, then $\frac{dy}{dx}=n\cdot u^{n - 1}\cdot u'$. Here $n=-4$ and $u = x^{2}-7$, and $u' = 2x$.
$f'(x)=-4(x^{2}-7)^{-4 - 1}\cdot(2x)$.

Step3: Simplify the expression

$f'(x)=-8x(x^{2}-7)^{-5}=-\frac{8x}{(x^{2}-7)^{5}}$.

Answer:

$-\frac{8x}{(x^{2}-7)^{5}}$