Question

What is the derivative of `sin^2x/cosx`?

Answer 1

`f'(x)=((sinx)(2cos^2x+sin^2x))/cos^2x`

Explanation:

`f(x)=(sin^2x)/cosx`

`f'(x)=(cosxtimes2sinxcosx-sin^2xtimes-sinx)/cos^2x`

`f'(x)= (2sinxcos^2x+sin^3x)/cos^2x`

`f'(x)=((sinx)(2cos^2x+sin^2x))/cos^2x`

The quotient rule is given by:

`f(x)=u/v`

`f'(x)=(vu'-uv')/v^2`