Question

if (f(x)=4x^{-2}+\frac{1}{4}x^{2}+4), then (f(2)=)

Explanation:

Step1: Find the derivative of each term

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $y = 4x^{-2}$, $y^\prime=4\times(-2)x^{-2 - 1}=-8x^{-3}$; for $y=\frac{1}{4}x^{2}$, $y^\prime=\frac{1}{4}\times2x^{2 - 1}=\frac{1}{2}x$; for $y = 4$, $y^\prime = 0$. So $f^\prime(x)=-8x^{-3}+\frac{1}{2}x$.

Step2: Substitute $x = 2$ into $f^\prime(x)$

$f^\prime(2)=-8\times2^{-3}+\frac{1}{2}\times2$. First, calculate $2^{-3}=\frac{1}{8}$, then $-8\times2^{-3}=-8\times\frac{1}{8}=-1$, and $\frac{1}{2}\times2 = 1$. So $f^\prime(2)=-1 + 1=0$.

Answer:

$0$