How do you find the first and second derivative of #ln(x^2e^x)#?

1 Answer
Feb 5, 2017

Answer:

#y' = (2 + x)/x#
#y'' = -2/x^2#

Explanation:

This can be rewritten as

#y = lnx^2 + lne^x#

We know that #y= lnx# and #y = e^x# are inverses, such that #lne = 1#. Using the law #lna^n = nlna#, we can say that #lne^x = x# and that #lnx^2 = 2lnx#.

#y = 2lnx + x#

#y' = 2/x + 1#

This is the first derivative. Differentiate this to get the second derivative.

#y' = 2x^-1 + 1#

#y'' = -2/x^2#

Hopefully this helps!