Question

How do you find the derivatives of `y=(sintheta)^tantheta` by logarithmic differentiation?

Answer 1

`(dy)/(dx)=(sintheta)^(tantheta)(sec^2thetalnsintheta+1)`

Explanation:

usinf logarithmic differentiation

`y=(sintheta)^(tantheta)`

`lny=ln(sintheta)^(tantheta)`

by laws of logs

`lny=tanthetaln(sintheta)`

now differentiate `wrt" " x`

`d/dx(lny=tanthetaln(sintheta))`

`RHS" "` will need the product rule

`1/y(dy)/(dx)=sec^2thetalnsintheta+tantheta1/sinthetaxxcostheta`

`1/y(dy)/(dx)=sec^2thetalnsintheta+cancel((sintheta/costhetaxxcostheta/sintheta))^(=1)`

`1/y(dy)/(dx)=sec^2thetalnsintheta+1`

`(dy)/(dx)=y(sec^2thetalnsintheta+1)`

`(dy)/(dx)=(sintheta)^(tantheta)(sec^2thetalnsintheta+1)`