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

1 Answer
Oct 30, 2016

Answer:

#(ln(x^2e^x))'=(2+x)/x#

#(ln(x^2e^x))''=-2/x^2#

Explanation:

The derivative of the given function is determined by using chain rule and applying some differential properties of the function #lnx# and the product functions.

Let #f(x)=lnx and g(x)=(x^2)(e^x)#

Then #fog(x)=f(g(x))=ln((x^2)(e^x))#

Derivative of a composite function
#color(red)((fog(x))'=f'(g(x))xxg'(x))#

Let us compute #f'(g(x)) and g'(x)#

#f(x)=lnx #
#f'(x)=1/x#

#f'(g(x))=1/(g(x))#

#color(red)(f'(g(x))=1/(x^2e^x))#

#g(x)# is a product of two functions #x^2# and #e^x#

The product rule of differentiation:
#color(green)(u'v+v'u)#

#g'(x)=color(green)((x^2)'e^x+x^2(e^x)')#
#g'(x)=2xe^x+x^2e^x#
#color(red)(g'(x)=xe^x(2+x)#

#color(red)((fog(x))'=f'(g(x))xxg'(x))#
#(ln(x^2e^x))'=1/(x^2e^x)xxxe^x(2+x)#
#(ln(x^2e^x))'=(cancel(xe^x)(2+x))/cancel(x^2e^x)#
#(ln(x^2e^x))'=(2+x)/x#

Let us compute the second derivative by applying quotient rule:

#(ln(x^2e^x))''#

#=((ln(x^2e^x))')'

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

#=(1xxx-1xx(x+2))/x^2#

#=(x-x-2)/x^2#

#=-2/x^2#