Question

How do you differentiate `f(x)=cos^2x`?

Answer 1

`f'(x)=-sin2x`

Explanation:

differentiate using the `color(blue)"chain rule"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(2/2)|)))`

`"Express " f(x)=cos^2x=(cosx)^2`

`"let "u=cosxrArr(du)/(dx)=-sinx`

`"then "y=u^2rArr(dy)/(du)=2u`

`rArrdy/dx=2u(-sinx)`

change u back into terms of x

`rArrdy/dx=-2sinxcosx`

`color(orange)"Reminder" sin2x=2sinxcosx`

`rArrdy/dx=-sin2x`

Answer 2

`dy/dx = -sin2x`

Explanation:

`f(x)= cos^2x = (cosx)^2`

By chain rule

let `u = cosx` then `y= u^2`

`(du)/dx = -sinx` then `dy/(du) = 2u`

Therefore `dy/dx = (du)/dx . dy/(du)`

`dy/dx=(-sinx)(2u)`

But `u = cosx`

`(-sinx)(2cosx)= -2sinxcosx`

Reminder `-2sinxcosx = -sin2x`

`dy/dx = -sin2x`