How do you use the product rule to differentiate #f(x)=e^x/(3-x)#?
2 Answers
Dec 2, 2017
Explanation:
here
Dec 2, 2017
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#