Question

if $f(x)=(4x + 6)^{-2}$, find $f(x)$. find $f(2)$.

Explanation:

Step1: Apply chain - rule

Let $u = 4x+6$, then $f(x)=u^{-2}$. The chain - rule states that $f^\prime(x)=\frac{df}{du}\cdot\frac{du}{dx}$. First, find $\frac{df}{du}$: $\frac{d}{du}(u^{-2})=-2u^{-3}$. Second, find $\frac{du}{dx}$: $\frac{d}{dx}(4x + 6)=4$.

Step2: Calculate $f^\prime(x)$

By the chain - rule $f^\prime(x)=\frac{df}{du}\cdot\frac{du}{dx}=-2u^{-3}\cdot4=-8(4x + 6)^{-3}$.

Step3: Find $f^\prime(2)$

Substitute $x = 2$ into $f^\prime(x)$. When $x = 2$, $f^\prime(2)=-8(4\times2 + 6)^{-3}=-8(8 + 6)^{-3}=-8\times14^{-3}=-\frac{8}{14^{3}}=-\frac{8}{2744}=-\frac{1}{343}$.

Answer:

$f^\prime(x)=-8(4x + 6)^{-3}$
$f^\prime(2)=-\frac{1}{343}$