Question

How do you find the derivative of `sinx/e^x`?

Answer 1

`(cosx-sinx)/e^x`

Explanation:

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

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

`color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))to(A)`

`"here " f(x)=(sinx)/e^x`

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

`"and "h(x)=e^xrArrh'(x)=e^x`

Substituting these values into (A)

`f'(x)=(e^x(cosx)-sinx(e^x))/e^(2x)`

`rArrf'(x)=(e^x(cosx-sinx))/e^(2x)=(cosx-sinx)/e^x`