How do you differentiate f(x) =2sin2x + cos^2x ?
1 Answer
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)) = 2sin2x
rArrf'(g(x))=2cos2x g(x) = 2x
rArrg'(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.identity
color(red)(|bar(ul(color(white)(a/a)color(black)(sin2x=2sinxcosx)color(white)(a/a)|)))
Now the whole can be put together
f(x)=2sin2x+cos^2x
rArrf'(x)=4cos2x-sin2x