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

Jan 1, 2016

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

Explanation:

• Chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$
• Product rule: $\left(a b\right) ' = a ' b + a b '$

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \left(\frac{7}{2}\right) {x}^{- \frac{1}{2}} + 6 {e}^{6 x} \ln \left(x\right) + \frac{{e}^{6 x}}{x}$

$\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \frac{7}{2 \sqrt{x}} + {e}^{6 x} \left(6 \ln \left(x\right) + {x}^{-} 1\right)$

Same rules for the second:

$\frac{\mathrm{dg} {\left(x\right)}^{2}}{{d}^{2} x} = - \frac{7}{4 \sqrt{{x}^{3}}} + 6 {e}^{6 x} \left(6 \ln \left(x\right) + {x}^{-} 1\right) + {e}^{6 x} \left(\frac{6 x - 1}{x} ^ 2\right)$

$\frac{\mathrm{dg} {\left(x\right)}^{2}}{{d}^{2} x} = - \frac{7}{4 \sqrt{{x}^{3}}} + {e}^{6 x} \left(\frac{36 x \ln \left(x\right) + 6}{x} + \frac{6 x - 1}{x} ^ 2\right)$

$\frac{\mathrm{dg} {\left(x\right)}^{2}}{{d}^{2} x} = - \frac{7}{4 \sqrt{{x}^{3}}} + {e}^{6 x} \left(\frac{36 {x}^{2} \ln \left(x\right) + 12 x - 1}{x} ^ 2\right)$