Question

How do you differentiate `g(x) =e^(1-x)sinx` using the product rule?

Answer 1

`g'(x)=fg'+gf' = e^(1-x) cos x-e^(1-x) sinx`

Explanation:

`f=e^(1-x), g=sin x`
`f'=e^(1-x)*-1 = -e^(1-x), g'=cos x`
`g'(x)=fg'+gf' = e^(1-x) cos x-e^(1-x) sinx`

Answer 2

`e^(1-x)(cosx - sinx)`

Explanation:

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

If g(x)=f(x).h(x) then g'(x) = f(x).h'(x) + h(x).f'(x)
`"-------------------------------------------------------------"`
f(x) = `e^(1-x) rArr f'(x) = e(1-x) d/dx(1-x) = -e^(1-x)`

and h(x)=sinx `rArr h'(x) = cosx`
`"----------------------------------------------------------------"`
substitute these results into g'(x)

`rArr g'(x) =e^(1-x).cosx + sinx.(-e^(1-x) )`

taking out a common factor

`rArr g'(x) = e^(1-x) (cosx - sinx )`