Question

Q.No.9*
dy
If y = log( Vx+ Vx-a), then
is
dx​

Answer 1

Step-by-step explanation:

We have,

$y = log( \sqrt{x} + \sqrt{x - a} )$

$= > \frac{dy}{dx} = \frac{ \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{x - a} } }{ \sqrt{x} + \sqrt{x - a} }$

$= > \frac{dy}{dx} = \frac{ \sqrt{x} + \sqrt{x - a} }{2 \sqrt{x(x - a)} ( \sqrt{x} + \sqrt{x - a} )}$

$= > \frac{dy}{dx} = \frac{1}{2 \sqrt{x(x - a)} }$