Question

what is $lim_{h \to 0} \frac{8left(\frac{1}{2}+h
ight)^8 - 8left(\frac{1}{2}
ight)^8}{h}$?

Explanation:

Step1: Recognize derivative definition

This limit matches the definition of the derivative of $f(x)=8x^{\frac{1}{8}}$ at $x=2$, where $f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$.

Step2: Compute the general derivative

Use power rule: $\frac{d}{dx}[x^n]=nx^{n-1}$.
$$f'(x)=8\cdot\frac{1}{8}x^{\frac{1}{8}-1}=x^{-\frac{7}{8}}=\frac{1}{x^{\frac{7}{8}}}$$

Step3: Evaluate at $x=2$

Substitute $x=2$ into the derivative.
$$f'(2)=\frac{1}{2^{\frac{7}{8}}}=\frac{1}{\sqrt[8]{2^7}}=\frac{1}{\sqrt[8]{128}}$$

Answer:

$\frac{1}{2^{\frac{7}{8}}}$ (or $\frac{1}{\sqrt[8]{128}}$)