Question

Differentiate the following w.r.t. $x$:
$\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$

Answer 1

The correct answer is $\therefore \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\bigg[(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\bigg]$
Let $y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$
Taking logarithm on both the sides,we obtain
$log\,y=log\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$
$\implies log\,y=\frac{1}{2}log\bigg[\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\bigg]$
$⇒log\,y=\frac{1}{2}[log(x-1)(x-2)-log(x-3)(x-4)(x-5)]$
$⇒log\,y=\frac{1}{2}[log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5)]$
Differentiating both sides with respect to x,we obtain
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}\bigg[\frac{1}{x-1}\frac{d}{dx}(x-1)+\frac{1}{x-2}.\frac{d}{dx}(x-2)-\frac{1}{x-3}.\frac{d}{dx}(x-3)-\frac{1}{x-4}.\frac{d}{dx}(x-4)-\frac{1}{x-5}.\frac{d}{dx}(x-5)\bigg]$
$⇒\frac{dy}{dx}=\frac{y}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})$
$\therefore \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\bigg[(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\bigg]$