Question

differentiation of the following equation..

Answer 1

$\textbf{Answer :}$

Given, y = log sinx

Now, differentiating both sides with respect to x, we get

dy/dx = d/dx (log sinx)

= 1/sinx × d/dx (sinx)

= 1/sinx × cosx

= cotx

$\textbf{Rule :}$

d/dx (logx) = 1/x

d/dx (sinx) = cosx

#$\textbf{MarkAsBrainliest}$