How do you find the first and second derivative of $(ln(x^(2)+3))^(3)$ ?

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Answer 1 · 2017-01-31T14:15:26

$f'(x)=(6xln^2(x^2+3))/(x^2+3)$

$f''(x)=((-6x^2+18)ln^2(x^2+3) + 24x ln(x^2+3))/(x^2+3)^2$

The first derivative is:

$f'(x)=3ln^2(x^2+3)*1/(x^2+3)*2x$

$=(6xln^2(x^2+3))/(x^2+3)$

The second derivative is:

$f''(x)=((6ln^2(x^2+3) + 6x*2 ln(x^2+3)* 1/(x^2+3) * 2x)*(x^2+3)-6xln^2(x^2+3)*2x)/(x^2+3)^2$

$=((6ln^2(x^2+3) + (24x ln(x^2+3))/(x^2+3))*(x^2+3)-12x^2ln^2(x^2+3))/(x^2+3)^2$

$=(((6(x^2+3)ln^2(x^2+3) + 24x ln(x^2+3))/cancel(x^2+3))*cancel((x^2+3))-12x^2ln^2(x^2+3))/(x^2+3)^2$

$=(6(x^2+3)ln^2(x^2+3) + 24x ln(x^2+3)-12x^2ln^2(x^2+3))/(x^2+3)^2$

$=((6x^2+18-12x^2)ln^2(x^2+3) + 24x ln(x^2+3))/(x^2+3)^2$

$=((-6x^2+18)ln^2(x^2+3) + 24x ln(x^2+3))/(x^2+3)^2$

How do you find the first and second derivative of (ln(x^(2)+3))^(3) (ln(x^(2)+3))^(3)?