How do you use the product rule to differentiate #f(x)=e^x/(3-x)#?

2 Answers
Dec 2, 2017

Answer:

#f'(x)= (e^x*(3-x)+e^x)/(3-x)^2#

Explanation:

#f(x)= e^x/(3-x)# Product rule: #d/dx(g*h*)=g'h+h'g#

here #g=e^x:. g'=e^x and h= 1/(3-x)= (3-x)^-1#

#:. h' = -1 (3-x)^-2*(-1)= 1/(3-x)^2#

#:.f(x)= e^x/(3-x):.f'(x)= e^x*1/(3-x)+1/(3-x)^2*e^x# or

#f'(x)= (e^x*(3-x)+e^x)/(3-x)^2# [Ans]

Dec 2, 2017

Answer:

#f'(x)=(e^x(4-x))/(3-x)^2#

Explanation:

#"given "f(x)=g(x)h(x)" then"#

#f'(x)=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"#

#f(x)=e^x/(3-x)=e^x(3-x)^-1#

#g(x)=e^xrArrg'(x)=e^x#

#h(x)=(3-x)^-1#

#rArrh'(x)=-(3-x)^-2(-1)=(3-x)^-2#

#rArrf'(x)=e^x(3-x)^-2+e^x(3-x)^-1#

#color(white)(rArrf'(x))=e^x(3-x)^-2(1+3-x)#

#color(white)(rArrf'(x))=(e^x(4-x))/(3-x)^2#