Question

How do you differentiate `sinx+sinxcosx`?

Answer 1

`cosx + cos^2x - sin^2x`

Explanation:

The derivative of `sinx` is `cosx`.

For `sinxcosx`, we need to use the product rule, so let `a,b` be functions of `x`. Then `(ab)' = a'b + ab'`.

`(sinxcosx)' = (sinx)'cosx + sinx(cosx)' = cosx * cosx + sinx * (-sinx)`

`=cos^2x - sin^2x`.

The final derivative is `cosx + cos^2x - sin^2x`.

Answer 2

`cosx+cos2x`

Explanation:

We require to know the following `color(blue)"standard derivatives"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx(sinx)=cosx , d/dx(cosx)=-sinx)color(white)(2/2)|)))`

Also the `color(blue)"trigonometric identity"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(cos2x=cos^2x-sin^2x)color(white)(2/2)|)))`

`sinxcosx" has to be differentiated using the"color(blue)" product rule"`

`"Given "f(x)=g(x).h(x)" then"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=g(x)h'(x)+h(x)g'(x))color(white)(2/2)|)))`

`"here "g(x)=sinxrArrg'(x)=cosx`

`"and "h(x)=cosxrArrh'(x)=-sinx`

`rArrd/dx(sinxcosx)=sinx(-sinx)+cosx(cosx)`

`=cos^2x-sin^2x=cos2x`

`"Thus "d/dx(sinx+sinxcosx)=cosx+cos2x`