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

Jan 31, 2017

$f ' \left(x\right) = \frac{6 x {\ln}^{2} \left({x}^{2} + 3\right)}{{x}^{2} + 3}$

$f ' ' \left(x\right) = \frac{\left(- 6 {x}^{2} + 18\right) {\ln}^{2} \left({x}^{2} + 3\right) + 24 x \ln \left({x}^{2} + 3\right)}{{x}^{2} + 3} ^ 2$

#### Explanation:

The first derivative is:

$f ' \left(x\right) = 3 {\ln}^{2} \left({x}^{2} + 3\right) \cdot \frac{1}{{x}^{2} + 3} \cdot 2 x$

$= \frac{6 x {\ln}^{2} \left({x}^{2} + 3\right)}{{x}^{2} + 3}$

The second derivative is:

$f ' ' \left(x\right) = \frac{\left(6 {\ln}^{2} \left({x}^{2} + 3\right) + 6 x \cdot 2 \ln \left({x}^{2} + 3\right) \cdot \frac{1}{{x}^{2} + 3} \cdot 2 x\right) \cdot \left({x}^{2} + 3\right) - 6 x {\ln}^{2} \left({x}^{2} + 3\right) \cdot 2 x}{{x}^{2} + 3} ^ 2$

$= \frac{\left(6 {\ln}^{2} \left({x}^{2} + 3\right) + \frac{24 x \ln \left({x}^{2} + 3\right)}{{x}^{2} + 3}\right) \cdot \left({x}^{2} + 3\right) - 12 {x}^{2} {\ln}^{2} \left({x}^{2} + 3\right)}{{x}^{2} + 3} ^ 2$

$= \frac{\left(\frac{6 \left({x}^{2} + 3\right) {\ln}^{2} \left({x}^{2} + 3\right) + 24 x \ln \left({x}^{2} + 3\right)}{\cancel{{x}^{2} + 3}}\right) \cdot \cancel{\left({x}^{2} + 3\right)} - 12 {x}^{2} {\ln}^{2} \left({x}^{2} + 3\right)}{{x}^{2} + 3} ^ 2$

$= \frac{6 \left({x}^{2} + 3\right) {\ln}^{2} \left({x}^{2} + 3\right) + 24 x \ln \left({x}^{2} + 3\right) - 12 {x}^{2} {\ln}^{2} \left({x}^{2} + 3\right)}{{x}^{2} + 3} ^ 2$

$= \frac{\left(6 {x}^{2} + 18 - 12 {x}^{2}\right) {\ln}^{2} \left({x}^{2} + 3\right) + 24 x \ln \left({x}^{2} + 3\right)}{{x}^{2} + 3} ^ 2$

$= \frac{\left(- 6 {x}^{2} + 18\right) {\ln}^{2} \left({x}^{2} + 3\right) + 24 x \ln \left({x}^{2} + 3\right)}{{x}^{2} + 3} ^ 2$