Question

How do you find the derivative of `y=e^x(sinx+cosx)`?

Answer 1

`dy/dx=2e^xcosx`

Explanation:

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

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

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

`"here "g(x)=e^xrArrg'(x)=e^x`

`"and "h(x)=sinx+cosxrArrg'(x)=cosx-sinx`

`rArrdy/dx=e^x(cosx-sinx)+e^x(sinx+cosx)`

`color(white)(xxxxx)=e^xcosxcancel(-e^xsinx)cancel(e^xsinx)+e^xcosx`

`color(white)(xxxxx)=2e^xcosx`