# How do you differentiate f(x) = ln [ x^{2}(x + 3)^{8}(x^2 + 4)^{3} ]?

Apr 29, 2015

I'll leave it to someone else to use $\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$ and the three factor product rule. I'll do it the easy way:

$f \left(x\right) = \ln \left[{x}^{2} {\left(x + 3\right)}^{8} {\left({x}^{2} + 4\right)}^{3}\right]$

$f \left(x\right) = \ln {x}^{2} + \ln {\left(x + 3\right)}^{8} + \ln {\left({x}^{2} + 4\right)}^{3}$

$f \left(x\right) = 2 \ln x + 8 \ln \left(x + 3\right) + 3 \ln \left({x}^{2} + 4\right)$

So
$f ' \left(x\right) = \frac{2}{x} + \frac{8}{x + 3} + \frac{3}{{x}^{2} + 4} \cdot \left(2 x\right)$

$f ' \left(x\right) = \frac{2}{x} + \frac{8}{x + 3} + \frac{6 x}{{x}^{2} + 4}$.