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

Jun 19, 2016

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

#### Explanation:

Here $f \left(x\right) = {\left(g \left(x\right)\right)}^{3}$, where $g \left(x\right) = \ln \left(h \left(x\right)\right)$ and $h \left(x\right) = {x}^{2} + 3$

According to chain rule $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}} \times \frac{\mathrm{dh}}{\mathrm{dx}}$

Hence $\frac{\mathrm{df}}{\mathrm{dx}} = 3 {\left(\ln \left({x}^{2} + 3\right)\right)}^{2} \times \frac{1}{{x}^{2} + 3} \times 2 x$

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