Find #dy/dx# for #ln(e^x)#?
1 Answer
Mar 16, 2017
You can simplify this before taking the derivative.
#lne^x = x#
since
#(dy)/(dx) = d/(dx)[x] = color(blue)(1)# .
You can simplify this before taking the derivative.
#lne^x = x#
since
#(dy)/(dx) = d/(dx)[x] = color(blue)(1)# .