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

2 Answers
Feb 4, 2017

#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#

#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#