Solution :
We have, y = log sin x
By using chain rule in differentiation,
Let u = sin x ⟹ dudx = cos x
And, y = log u ⟹ dydu = 1u
Now, dydx = dydu × dudx
⟹ dydx = 1u × cos x
⟹ dydx = 1sinx × cos x = cot x
Hence, the differentiation of log sin x with respect to x is cot x.
Similar Questions
What is the differentiation of 1/log x ?
What is the differentiation of logx2 ?