Answer
$$\frac{-2}{(x-1)^{3}}$$
Work Step by Step
Given $$y=\frac{1}{(x-1)^{2}}$$
Now we find the derivative with limits:
\begin{align*}
\frac{d y}{d x}&=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{1}{((x+h)-1)^{2}}-\frac{1}{(x-1)^{2}}}{h}\\
&=\lim _{h \rightarrow 0} \frac{\frac{(x-1)^{2}-((x+h)-1)^{2}}{(x-1)^{2}((x+h)-1)^{2}}}{h}\\
&=\lim _{h \rightarrow 0} \frac{\left(x^{2}-2 x+1\right)-\left((x+h)^{2}-2(x+h)+1\right)}{h(x-1)^{2}((x+h)-1)^{2}}\\
&=\lim _{h \rightarrow 0} \frac{\left(x^{2}-2 x+1\right)-\left(x^{2}+2 x h+h^{2}-2 x-2 h+1\right)}{h(x-1)^{2}((x+h)-1)^{2}}\\
&=\lim _{h \rightarrow 0} \frac{-2 x-h+2}{(x-1)^{2}((x+h)-1)^{2}}\\
&=\frac{-2(x-1)}{(x-1)^{4}}\\
&=\frac{-2}{(x-1)^{3}}
\end{align*}
