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

Answer:

$4$

Explanation:

Step1: Recall the definition of the derivative

The limit $\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$ is the derivative of the function $y = f(x)$ at $x$. Here, $f(x)=8x^{8}$, and $x=\frac{1}{2}$.

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

Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $f(x)=8x^{8}$, 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 $f^{\prime}(x)$. Then $f^{\prime}(\frac{1}{2})=64\times(\frac{1}{2})^{7}$. Since $64 = 2^{6}$, we have $f^{\prime}(\frac{1}{2})=2^{6}\times\frac{1}{2^{7}}$. Using the rule $a^{m}\times a^{n}=a^{m + n}$ (here $a = 2$, $m = 6$, $n=-7$), we get $f^{\prime}(\frac{1}{2})=\frac{2^{6}}{2^{7}}=\frac{1}{2}=4$.