What is the derivative of #e^(lnx)#?

1 Answer
May 29, 2017

#1#

Explanation:

We can also do this without first using the identity #e^lnx=x#, although we will have to use this eventually.

Note that #d/dxe^x=e^x#, so when we have a function in the exponent the chain rule will apply: #d/dxe^u=e^u*(du)/dx#.

So:

#d/dxe^lnx=e^lnx(d/dxlnx)#

The derivative of #lnx# is #1/x#:

#d/dxe^lnx=e^lnx(1/x)#

Then using the identity #e^lnx=x#:

#d/dxe^lnx=x(1/x)=1#

Which is the same as the answer we'd get if we use the identity from the outset (which is what I recommend you do--this is just a fun way to show that "calculus works".)

#d/dxe^lnx=d/dxx=1#