Question

question
find the derivative of the function $f(x)=\frac{4}{x - 1}$ using the limit definition of the derivative.
provide your answer below.
f(x)=

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ is $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\frac{4}{x - 1}$, then $f(x + h)=\frac{4}{(x + h)-1}$.

Step2: Substitute into the formula

\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{\frac{4}{(x + h)-1}-\frac{4}{x - 1}}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{4(x - 1)-4((x + h)-1)}{((x + h)-1)(x - 1)}}{h}\\ &=\lim_{h ightarrow0}\frac{4x-4-(4x + 4h-4)}{h((x + h)-1)(x - 1)}\\ &=\lim_{h ightarrow0}\frac{4x-4 - 4x-4h + 4}{h((x + h)-1)(x - 1)}\\ &=\lim_{h ightarrow0}\frac{-4h}{h((x + h)-1)(x - 1)} \end{align*}$$

\]

Step3: Simplify the expression

Cancel out the $h$ terms: $\lim_{h
ightarrow0}\frac{-4}{((x + h)-1)(x - 1)}$.

Step4: Evaluate the limit

As $h
ightarrow0$, we get $f^{\prime}(x)=\frac{-4}{(x - 1)^2}$.

Answer:

$\frac{-4}{(x - 1)^2}$