How do you differentiate #d/dx e^ln(x^2)# ?

#d/dx e^ln(x^2)#

2 Answers
VNVDVI
Mar 31, 2018

#2x#

Explanation:

Let us rewrite #e^(ln(x^2))# as #x^2.# We could use the chain rule and logarithmic differentiation rules, but it would be a lot of messy and unnecessary work.

In general, #e^(lnx)=x#:

#x=e^lnx#

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

#lnx=lnxln(e)#

#lnx=lnx# As #lne=1#

#x=x#

So, we really want

#d/dxx^2=2x#, as #d/dxx^n=nx^(n-1).#

Jim G.
Mar 31, 2018

#2x#

Explanation:

#"note that " e^ln(x^2)=x^2#

#"since e and ln are inverse functions"#

#rArrd/dx(e^ln(x^2))=d/dx(x^2)=2x#

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