How do you differentiate f(x) =2sin2x + cos^2x ?

1 Answer
May 20, 2016

4cos2x-sin2x

Explanation:

Both terms here can be differentiated using the color(blue)"chain rule"

d/dx[f(g(x))]=f'(g(x)).g'(x)...........(A)
"-----------------------------------------------------"

Term 1

f(g(x)) = 2sin2xrArrf'(g(x))=2cos2x

g(x) = 2xrArrg'(x)=2

Substitute these values into (A)

rArrd/dx(2sin2x)=2cos2x.2=4cos2x
"--------------------------------------------------------"
Term 2

f(g(x))=cos^2x=(cosx)^2
rArrf'(g(x))=2(cosx)^1=2cosx

g(x)=cosxrArrg'(x)=-sinx

Substitute these values into (A)

rArrd/dx(cos^2x)=2cosx.(-sinx)=-2cosxsinx
"-----------------------------------------------------------------------"

using the trig.identitycolor(red)(|bar(ul(color(white)(a/a)color(black)(sin2x=2sinxcosx)color(white)(a/a)|)))

rArrd/dx(cos^2x)=-sin2x

Now the whole can be put together

f(x)=2sin2x+cos^2x

rArrf'(x)=4cos2x-sin2x