what is $lim_{h\rightarrow0}\frac{8(\frac{1}{2}+h)^{8}-8(\frac{1}{2})^{8}}{h}$?

Answer:

$4$

Explanation:

Step1: Recall the definition of the derivative

The given limit $\lim_{h\rightarrow0}\frac{8(\frac{1}{2}+h)^{8}-8(\frac{1}{2})^{8}}{h}$ is in the form of the definition of the derivative $f^\prime(a)=\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}$, where $f(x)=8x^{8}$ and $a=\frac{1}{2}$.

Step2: Find the derivative of $f(x)$

Using the power - rule for differentiation, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. For $f(x)=8x^{8}$, $f^\prime(x)=8\times8x^{7}=64x^{7}$.

Step3: Evaluate the derivative at $x = a$

Substitute $x=\frac{1}{2}$ into $f^\prime(x)$. So $f^\prime(\frac{1}{2})=64\times(\frac{1}{2})^{7}$. Since $(\frac{1}{2})^{7}=\frac{1}{128}$, then $f^\prime(\frac{1}{2})=64\times\frac{1}{128}= \frac{64}{128}=4$.