How do you use the chain rule to differentiate #y = e^lnx#?

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Answer 1 · 2016-08-22T02:06:01

#dy/dx = e^lnx/x = 1#

Let #y = e^u# and #u = lnx#

Then #y' = e^u xx 1/x = e^lnx xx 1/x = e^lnx/x = 1#

Hopefully this helps!

Answer 2 · 2016-08-22T02:08:06

#f'(x) =1#

#f(x) = e^lnx = x#

#f'(x) = d/dx (x) = 1# (No need of the Chain rule)

However, since the question askes that we use the Chain rule:

#f(x) = e^lnx#
#f'(x) = e^lnx * 1/x# (Standard differentials and Chain rule)

#= x*1/x = 1#