What are the first and second derivatives of $g (x) = {(ln x)}^{2} - ln (x^{2})$ $g(x) =(lnx)^2-ln(x^2)$ ?

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Answer 1 · 2015-11-30T09:54:08

$\frac{d}{d x} g (x) = \frac{2 (ln (x) - 1)}{x}$ $d/dxg(x) = (2(ln(x)-1))/x$

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

We will use

$g'(x) = d/dx(ln^2(x) - ln(x^2)) = d/dx(ln^2(x)) - d/dx(ln(x^2))$

$=2ln(x)(d/dxln(x)) - 1/(x^2)(d/dxx^2)$ (by the chain rule)

$= 2ln(x)(1/x) - 1/x^2(2x)$

$= (2ln(x))/x - 2/x$

$= (2(ln(x) - 1))/x$

To find the second derivative, we take the derivative of $g'(x)$ using the quotient rule.

$g''(x) = d/dx(2(ln(x)-1))/x$

$= ((d/dx2(ln(x)-1))x - 2(ln(x)-1)(d/dxx))/x^2$

$=((2(1/x))x - 2(ln(x)-1)(1))/x^2$

$= (2 - 2ln(x) + 2)/x^2$

$= (4-2ln(x))/x^2$

What are the first and second derivatives of g(x) =(lnx)^2-ln(x^2)g(x)=(lnx)2−ln(x2) g(x) =(lnx)^2-ln(x^2)?