What are the first and second derivatives of g(x) = 7x^(1/2) + e^(6x)ln(x)?

1 Answer
Jan 1, 2016

We'll demand some chain rule and also product rule for the second term.

Explanation:

  • Chain rule: (dy)/(dx)=(dy)/(du)(du)/(dx)
  • Product rule: (ab)'=a'b+ab'

(dg(x))/(dx)=(7/2)x^(-1/2)+6e^(6x)ln(x)+(e^(6x))/x

(dg(x))/(dx)=7/(2sqrt(x))+e^(6x)(6ln(x)+x^-1)

Same rules for the second:

(dg(x)^2)/(d^2x)=-7/(4sqrt(x^3))+6e^(6x)(6ln(x)+x^-1)+e^(6x)((6x-1)/x^2)

(dg(x)^2)/(d^2x)=-7/(4sqrt(x^3))+e^(6x)((36xln(x)+6)/x+(6x-1)/x^2)

(dg(x)^2)/(d^2x)=-7/(4sqrt(x^3))+e^(6x)((36x^2ln(x)+12x-1)/x^2)