Question

How do you find the derivative of `f(x) = sinxtanx`?

Answer 1

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

Apply the quotient and product rules to differentiate. However, let's first examine the technique used to differentiate `sin^2x`.

`y =sin^2x = (sinx)(sinx)`

`y' = cosx xx sinx + cosx xx sinx = 2cosxsinx = sin2x`

We now have the information we need to use the quotient rule.

`f'(x) = (sin2x xx cosx - (-sinx xx sin^2x))/(cosx)^2`

`f'(x) = (2sinxcosx xx cosx + sin^3x)/(cos^2x)`

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

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

`f'(x) = tanxsecx(2(1 - sin^2x) + sin^2x)`

`f'(x) = tanxsecx(2 - 2sin^2x + sin^2x)`

`f'(x) = tanxsecx(2 - sin^2x)`

Hopefully this helps!