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

Answer:

$4\left(\frac{1}{2}\right)^{7}$

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 $y = x^{n}$, then $y^\prime=nx^{n - 1}$. For $f(x)=8x^{8}$, by the constant multiple rule $(cf(x))^\prime=cf^\prime(x)$ where $c = 8$ and $n = 8$, we have $f^\prime(x)=8\times8x^{7}=64x^{7}$.

Step3: Evaluate the derivative at $x=\frac{1}{2}$

Substitute $x=\frac{1}{2}$ into $f^\prime(x)$. So $f^\prime(\frac{1}{2})=64\times(\frac{1}{2})^{7}=4\times2^{4}\times\frac{1}{2^{7}}=4\times\frac{1}{2^{3}} = 4\times(\frac{1}{2})^{7}$.