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

1 Answer
Oct 30, 2016

(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