Question

How do you find the derivative of `f(x) = 4x + piTan(x)`?

Answer 1

`4+π(1+tan^2x)`

Explanation:

`f'(x)=(4x+πtan(x))' = 4(x)'+π(tanx)' = 4+π(1+tan^2x)`

• because

`(tanx)'=(sinx/cosx)'` `=((sinx)'cosx-sinx(cosx)')/cos^2x =`

`(cos^2x+sin^2x)/cos^2x = 1/cos^2x = 1+tan^2x`

& `sin^2x+cos^2x=1 <=> sin^2x/cos^2x+1=1/cos^2x <=> 1 + tan^2x = 1/cos^2x`