Question

y = log(sinx) find dy/ dx

Answer 1

y = log (sin x)
Let sinx = u

Therefore by applying chain rule,

dy/dx = d log u / du * d sin x / dx
dy / dx = 1/u * cos x = 1/sin x * cos x = cot x.

Therefore,
dy/dx = cot x when y = log ( sin x ).

Answer 2

Answer: The correct answer is cotx.

Explanation:

The given equation in the problem is as follows;

y = log(sinx)

Differentiate equation on both sides with respect to x.

The differentiation of sinx is cosx and the differentiation of logx is 1\x.

$\frac{dy}{dx}=\frac{cosx}{sinx}$

$\frac{d(log(sinx))}{dx}=\frac{cosx}{sinx}$

Put cosx\sinx= cotx

$\frac{d(log(sinx))}{dx}=cotx$

Therefore, the correct answer is cotx