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#
