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

2 Answers
Mar 10, 2016

Answer:

#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#

Mar 10, 2016

Answer:

#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 ) #