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

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Answer 1 · 2016-10-11T22:55:30

$f''(x) = ( 4 - 2x^2 ) / (x^2+2)^2$

$f(x)=ln(x^2+2)$

Using the chain rule we have:
$f'(x)=1/(x^2+2) *(2x)$
$:. f'(x)=(2x)/(x^2+2)$

To find the second derivative we now need to use the product rule. Remember:
$d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2$ ; so

$f''(x) = ( (x^2+2)(2) - (2x)(2x) ) / (x^2+2)^2$
$:. f''(x) = ( 2x^2+4 - 4x^2 ) / (x^2+2)^2$
$:. f''(x) = ( 4 - 2x^2 ) / (x^2+2)^2$

How do you find the second derivative of f(x)=ln (x^2+2)f(x)=ln(x2+2)f(x)=ln (x^2+2)?