Question

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

Answer 1

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)`