What is the derivative of #f(x)=(x^2-4)ln(x^3/3-4x)#?

1 Answer
Guilherme N.
Dec 30, 2015

We'll need the product rule for #f(x)# and the chain rule for the second term #ln(x^3/3-4x)#

Explanation:

  • Product rule: be #y=f(x)g(x)#, then #(dy)/(dx)=f'(x)g(x)+f(x)g'(x)#
  • Chain rule: #(dy)/(dx)=(dy)/(du)(du)/(dx)#

#(df(x))/(dx)=(2x)ln(x^3/3-4x)+(x^2-4)(1/(x^3/3-4x)(x^2-4))#

#(df(x))/(dx)=2xln(x^3/3-4x)+(x^2-4)^2/((x^3-12x)/3)#

#(df(x))/(dx)=2xln(x^3/3-4x)+(3(x^2-4)^2)/(x(x^2-12))#

Related questions
See all questions in Differentiating Logarithmic Functions with Base e
Impact of this question
1408 views around the world
You can reuse this answer
Creative Commons License