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

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Answer 1 · 2016-01-01T20:40:51

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

Chain rule: $\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)(du)/(dx)$
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:

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

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

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

What are the first and second derivatives of g(x) = 7x^(1/2) + e^(6x)ln(x)g(x)=7x12+e6xln(x) g(x) = 7x^(1/2) + e^(6x)ln(x)?