Question

answer this question please it's very urgent

Answer 1

hope this will help you..
By the way Sorry for the Handwriting.. :))

Answer 2

Solution-

Given,

$y=\sin (\log x)$

From the properties of trigonometry,

$\Rightarrow \sin^2 x+\cos^2 x=1$

$\Rightarrow \cos^2 x=1-\sin^2 x$

Putting x = log x,

$\cos^2 (\log x)=1-\sin^2 (\log x)$

Putting the given value of y,

$\cos^2 (\log x)=1-y^2$

$\cos (\log x)=\sqrt{1-y^2}$

Now,

$\Rightarrow y=\sin (\log x)$

$\Rightarrow \frac{dy}{dx} =\frac{d(\sin (\log x))}{dx}$

$\Rightarrow \frac{dy}{dx} =\frac{d(\sin (\log x))}{d(\log x)} .\frac{d(\log x)}{dx}$

$\Rightarrow \frac{dy}{dx} =\cos (\log x).\frac{1}{x}$

$\Rightarrow \frac{dy}{dx} =\sqrt{1-y^2}.\frac{1}{x}$

$\Rightarrow \frac{dy}{dx} =\frac{\sqrt{1-y^2}}{x}$