Question

differentiate y=xe^sinx

Answer 1

we need product rule but first lets look at e^sinx
if we have y=e^sinx
if we take natural log of both sides we have lny=lne^sinx
lny=sinxlne
lny=sinx(1)
lny=sinx
we use chain rule to find(lny)'
y'/y=cosx
y'=ycosx
y'=e^sinx*cosx
so lets go back to y=xe^sinx
we need the product rule
y'=(x)'e^sinx+x(e^sinx)'
y'=1e^sinx+xe^sinx*cosx
y'=e^sinx(1+xcosx)