Question

How do you find the derivative of `sinx(sinx+cosx)`?

Answer 1

`sin2x+cos2x`

Explanation:

`"differentiate using the "color(blue)"product rule"`

`"Given "y=g(x)h(x)" then"`

`dy/dx=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"`

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

`h(x)=sinx+cosxrArrh'(x)=cosx-sinx`

`rArrdy/dx=sinx(cosx-sinx)+cosx(sinx+cosx)`

`color(white)(rArrdy/dx)=sinxcosx-sin^2x+sinxcosx+cos^2x`

`color(white)(rArrdy/dx)=2sinxcosx+cos^2x-sin^2x`

`color(white)(rArrdy/dx)=sin2x+cos2x`