Question

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

Answer 1

`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)) = 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)|)))`

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

Now the whole can be put together

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

`rArrf'(x)=4cos2x-sin2x`