How do you differentiate #e^x /lnx#?
1 Answer
May 9, 2018
Explanation:
#"differentiate using the "color(blue)"quotient rule"#
#"given "y=(g(x))/(h(x))" then"#
#dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#
#g(x)=e^xrArrg'(x)=e^x#
#h(x)=lnxrArrh'(x)=1/x#
#rArrd/dx((e^x)/(lnx))#
#=(e^xlnx-1/x e^x)/(lnx)^2=(e^x(lnx-1/x))/(lnx)^2#
