let $f(x)=\frac{1}{2x + 7}$. according to the definition, $f(x)=lim_{h\rightarrow0}square$ (enter the limit…

Explanation:

Step1: Recall the definition of the derivative

The definition of the derivative of a function $y = f(x)$ is $f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$.

Step2: Find $f(x + h)$

Given $f(x)=\frac{1}{2x + 7}$, then $f(x + h)=\frac{1}{2(x + h)+7}=\frac{1}{2x+2h + 7}$.

Step3: Substitute into the derivative - formula

[ \begin{align*} f^{\prime}(x)&=\lim_{h\rightarrow0}\frac{\frac{1}{2x + 2h+7}-\frac{1}{2x + 7}}{h}\ &=\lim_{h\rightarrow0}\frac{\frac{(2x + 7)-(2x + 2h + 7)}{(2x + 2h+7)(2x + 7)}}{h}\ &=\lim_{h\rightarrow0}\frac{2x + 7-2x - 2h - 7}{h(2x + 2h+7)(2x + 7)}\ &=\lim_{h\rightarrow0}\frac{- 2h}{h(2x + 2h+7)(2x + 7)} \end{align*} ]

Step4: Simplify the expression

Cancel out the $h$ terms: $\lim_{h\rightarrow0}\frac{-2}{(2x + 2h+7)(2x + 7)}$.

Answer:

$\lim_{h\rightarrow0}\frac{-2}{(2x + 2h+7)(2x + 7)}$