Question

Find dy/dx of `y=e^x sin x` Help please and show the working? Thanks

Answer 1

The answer is `=e^x(cosx+sinx)`

Explanation:

Reminder the product rule

`(uv)'=u'v+uv'`

Here,

`u(x)=e^x`, `=>`, `u'(x)=e^x`

`v(x)=sinx`, `=>`, `v'(x)=cosx`

Therefore,

`(e^xsinx)'=e^xsinx+e^xcosx=e^x(cosx+sinx)`

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

Answer 2

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

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)=e^xrArrg'(x)=e^x`

`h(x)=sinxrArrh'(x)=cosx`

`rArrdy/dx=e^xcosx+e^xsinx`

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