# Differentiate the functions given in w.r.t. $x :$ $\sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

$\begin{array}{1 1} \large\frac{1}{2}\sqrt {\large\frac{(x-1)(x-2)}{(x-8)(x-4)(x-7)}}\begin{bmatrix}\large\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix} \\\large\frac{1}{2}\sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\begin{bmatrix}\large\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix} \\ \large\frac{1}{2}\sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\begin{bmatrix}\large\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix} \\ \large\frac{1}{4}\sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\begin{bmatrix}\large\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix} \end{array}$

• $\log m^n=n\log m$
• $\log mn=\log m+\log n$
• $\log \large\frac{m}{n}$$=\log m-\log n Step 1: Let y= \sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} Taking \log on both sides we get \log y=\log \sqrt {\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}} \log y=\log \begin{bmatrix}\large\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)} \end{bmatrix}^{\large\frac{1}{2}} \;\;\;=\large\frac{1}{2}$$\big(\log[(x-1)(x-2)]-\log[(x-3)(x-4)(x-5)]\big)$
$\;\;\;=\large\frac{1}{2}\big($$[\log(x-1)+\log(x-2)]-[\log(x-3)+\log(x-4)+\log(x-5)]\big) We know that \log mn=\log m+\log n \Rightarrow \large\frac{1}{2}$$[\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5)]$
$\large\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\begin{bmatrix}\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix}$
$\large\frac{dy}{dx}=y.\frac{1}{2}\begin{bmatrix}\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\end{bmatrix}$