How do you find the derivative of #ln(e^x+1)#?
1 Answer
Jul 29, 2016
Explanation:
differentiate using the
#color(blue)"chain rule"#
#color(red)(|bar(ul(color(white)(a/a)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(a/a)|)))........ (A)# let
#color(blue)(u=e^x+1)rArr(du)/(dx)=e^x# and
#y=lncolor(blue)(u)rArr(dy)/(du)=1/color(blue)(u)# substitute these values into (A) and convert
#color(blue)(u)# back into x
#rArrdy/dx=1/color(blue)(u).e^x=e^x/(e^x+1)#