How do you differentiate f(x)=ln((2x+5)^5)^2  using the chain rule?

Nov 14, 2016

Rewrite it first.

Explanation:

$f \left(x\right) = \ln {\left({\left(2 x + 5\right)}^{5}\right)}^{2}$

The convention for reading this is that we find $2 x + 5$, then rais that number to the ${5}^{\text{th}}$ power, then square that result and finally take the natural logarithm of that.

By propeties of exponents, ${\left({u}^{5}\right)}^{2} = {u}^{10}$.

By properties of logarithms, $\ln \left({u}^{10}\right) = 10 \ln u$.

Applying the chain rule, we have

$\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$.

$f \left(x\right) = \ln {\left({\left(2 x + 5\right)}^{5}\right)}^{2} = 10 \ln \left(2 x + 5\right)$

$f ' \left(x\right) = 10 \cdot \frac{1}{2 x + 5} \cdot \frac{d}{\mathrm{dx}} \left(2 x + 5\right)$

$= \frac{20}{2 x + 5}$