Question

d/dx ( log base e (sin x))​

Answer 1

Explanation:

$\frac{d}{dx}( log_{e}( sinx ))$

Supposed that:-

$\scriptsize \: sinx = u \: \: then \: \: log_{e}(sinx) = log_{e}(u)$

$\frac{dy}{dx} = \frac{d}{dx}( log_{e}(u)) \times \frac{du}{dx}$

$= \frac{1}{u} \times \frac{du}{dx} \: (\because \frac{d}{dx}( log_{e}(x)) = \frac{1}{x} )$

$= \frac{1}{sinx} \times \frac{d}{dx} (sinx) \: ( \because \: u = sinx)$

$\scriptsize \: = cosecx \times cosx \: ( \because \: \frac{1}{sinx} = cosecx \: and \: \frac{d}{dx} sinx = cosx)$

I hope it is helpful