Question

How do you find the derivative of `cos2x-5cos^2x`?

Answer 1

`3sin2x`

Explanation:

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

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`d/dx(cos2x)=-2sin2x`

`d/dx(-5cos^2x)=10sinxcosx=5sin2x`

`rArrd/dx(cos2x-5cos^2x)`

`=-2sin2x+5sin2x=3sin2x`

Answer 2

`3sin2x`

Explanation:

Given: `d/dx(cos2x-5cos^2x)`.

`=d/dx(cos2x)-d/dx(5cos^2x)`

`=d/dx(cos2x)-5d/dx(cos^2x)`

Using the chain rule,

`d/dx(cos2x)`

`=2*-sin2x`

`=-2sin2x`

For the second part, We find: `d/dx(cos^2x)`.

Let `u=cosx,(du)/dx=-sinx`, `:.f=u^2,(df)/(du)=2u`.

So, `d/dx(cos^2x)=-2usinx`

`=-2cosxsinx`

`=-sin2x` `(because 2cosxsinx=sin2x)`

So, the whole differential becomes:

`=-2sin2x-5(-sin2x)`

`=-2sin2x+5sin2x`

`=3sin2x`