Find the derivative of the function?

y=ln tan^2 3xy=lntan23x

1 Answer
Mar 8, 2018

dy/dx=12csc(6x)dydx=12csc(6x)

Explanation:

We to find the derivative of

y=ln(tan^2(3x))=2ln(tan(3x))y=ln(tan2(3x))=2ln(tan(3x))

Use the chain rule, if y=f(u)y=f(u) and u=g(x)u=g(x) then

dy/dx=dy/(du)(du)/dxdydx=dydududx

Let y=2ln(u)=>(dy)/(du)=2/uy=2ln(u)dydu=2u

and u=tan(3x)=>(du)/(dx)=3sec^2(3x)u=tan(3x)dudx=3sec2(3x)

Thus

dy/dx=6/usec^2(3x)=6/tan(3x)sec^2(3x)=6csc(3x)sec(3x)dydx=6usec2(3x)=6tan(3x)sec2(3x)=6csc(3x)sec(3x)

Or by applying the identity csc(2x)=1/2csc(x)sec(x)csc(2x)=12csc(x)sec(x)

dy/dx=12csc(6x)dydx=12csc(6x)