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

1 Answer
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))