Question

Differentiate log $\sin x$ by first principles ?

Answer 1

Let, $y=\log \sin x$

We have to find $\frac{dy}{dx}$ i.e. $\frac{d}{dx}\left(\log \sin x\right)$

Let us assume, $u=\sin x\Rightarrow \frac{du}{dx}=\cos x\therefore y=\log u\Rightarrow \frac{dy}{du}=\frac{1}{u}$

Now, $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}=\frac{1}{u}\times \cos x=\frac{1}{\sin x}\times \cos x=\frac{\cos x}{\sin x}=\cot x\therefore \frac{d}{dx}\left(\log \sin x\right)=\cot x.$