Question

tan (log (sin x ))find dy/dx ​

Answer 1

Answer:

dy/dx=sec²x.logsinx.cotx

Step-by-step explanation:

y=tanx{log(sinx)}

On differentiation

dy/dx=sec²{log(sinx)}1/sinx.cosx

dy/dx=sec2.logsinx.cotx

Answer 2

$y = tan( log(sinx) ) \\ \\ \frac{dy}{dx} = {sec}^{2} log(sinx) \frac{d}{dx}log(sinx) \\ \\ \frac{dy}{dx} = {sec}^{2} log(sinx) \frac{1}{sinx} cosx \\ \\ \frac{dy}{dx} = {sec}^{2} log(sinx)cotx$