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 definition of the derivative of a function (y = f(x)) is (f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}). In this case, if we let (f(x)=8x^{8}), then (x=\frac{1}{2}), and the given limit (\lim_{h\rightarrow0}\frac{8(\frac{1}{2}+h)^{8}-8(\frac{1}{2})^{8}}{h}) is (f^{\prime}(\frac{1}{2})).

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

Using the power rule ((x^{n})^\prime=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)): (f^{\prime}(\frac{1}{2})=64\times(\frac{1}{2})^{7}). Since (64 = 2^{6}), then (64\times(\frac{1}{2})^{7}=\frac{2^{6}}{2^{7}}). Using the rule (a^{m}\div a^{n}=a^{m - n}), we get (\frac{2^{6}}{2^{7}}=\frac{1}{2}=4).