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

1 Answer
Guilherme N.
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)#

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