hw9 the derivative as a function (targets l6, d1, d2; $3.2)\nscore: 7/9 answered: 7/9\nquestion 8\ngiven…

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative is $f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\frac{8}{x}$, then $f(x + h)=\frac{8}{x + h}$.

Step2: Substitute into the formula

$\frac{f(x + h)-f(x)}{h}=\frac{\frac{8}{x + h}-\frac{8}{x}}{h}=\frac{\frac{8x-8(x + h)}{x(x + h)}}{h}=\frac{8x-8x-8h}{hx(x + h)}=\frac{-8h}{hx(x + h)}$.

Step3: Simplify the expression

Cancel out the $h$ terms: $\frac{-8h}{hx(x + h)}=\frac{-8}{x(x + h)}$ for $h\neq0$.

Step4: Take the limit as $h\rightarrow0$

$f^{\prime}(x)=\lim_{h\rightarrow0}\frac{-8}{x(x + h)}$. As $h\rightarrow0$, we get $f^{\prime}(x)=-\frac{8}{x^{2}}$.

Answer:

$-\frac{8}{x^{2}}$