How do you find the second derivative of $ln (x^{2} + 4)$ $ln(x^2+4)$ ?

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How do you find the second derivative of $ln (x^{2} + 4)$ $ln(x^2+4)$ ?

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Answer 1 · 2016-10-15T18:51:26

$\frac{d^{2} ln (x^{2} + 4)}{{d x}^{2}} = \frac{8 - 2 x^{2}}{{(x^{2} + 4)}^{2}}$ $(d^2ln(x^2 + 4))/dx^2 = (8 - 2x^2)/(x^2 + 4)^2$

Explanation:

The chain rule is:

$\frac{d {f (u (x))}}{d x} = \frac{d f (u)}{d u} (\frac{d u}{d x})$ $(d{f(u(x))})/dx = (df(u))/(du)((du)/dx)$

Let $u (x) = x^{2} + 4$ $u(x) = x^2 + 4$ , then $\frac{d f (u)}{d u} = \frac{d ln (u)}{d u} = \frac{1}{u}$ $(df(u))/(du) =(dln(u))/(du) = 1/u$ and $\frac{d u}{d x} = 2 x$ $(du)/dx = 2x$

$\frac{d ln (x^{2} + 4)}{d x} = \frac{2 x}{x^{2} + 4}$ $(dln(x^2 + 4))/dx = (2x)/(x^2 + 4)$

$\frac{d^{2} ln (x^{2} + 4)}{{d x}^{2}} = \frac{d (\frac{2 x}{x^{2} + 4})}{d x}$ $(d^2ln(x^2 + 4))/dx^2 = (d((2x)/(x^2 + 4)))/dx$

$\frac{d (\frac{2 x}{x^{2} + 4})}{d x} =$ $(d((2x)/(x^2 + 4)))/dx =$

$\frac{2 (x^{2} + 4) - 2 x (2 x)}{{(x^{2} + 4)}^{2}} =$ ${2(x^2 + 4) - 2x(2x)}/(x^2 + 4)^2 =$

$\frac{8 - 2 x^{2}}{{(x^{2} + 4)}^{2}}$ $(8 - 2x^2)/(x^2 + 4)^2$

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How do you find the second derivative of $ln (x^{2} + 4)$ $ln(x^2+4)$ ?

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How do you find the second derivative of $ln (x^{2} + 4)$ $ln(x^2+4)$ ?