# How do I find the derivative of  ln[x(x^2+1)^2/(2x^3-1)^(1/2)] ?

Mar 27, 2016

Use the properties of logarithms and the $\frac{d}{\mathrm{dx}} \ln \left(x\right)$ rule to get $\frac{1}{x} + \frac{4 x}{{x}^{2} + 1} - \frac{3 {x}^{2}}{2 {x}^{3} - 1}$.

#### Explanation:

Begin by using the properties of logs to write:
$\ln \left[\frac{x {\left({x}^{2} + 1\right)}^{2}}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right] = \ln \left(x {\left({x}^{2} + 1\right)}^{2}\right) - \ln {\left(2 {x}^{3} - 1\right)}^{\frac{1}{2}} = \ln \left(x\right) + \ln \left({\left({x}^{2} + 1\right)}^{2}\right) - \frac{1}{2} \ln \left(2 {x}^{3} - 1\right) = \ln \left(x\right) + 2 \ln \left({x}^{2} + 1\right) - \frac{1}{2} \ln \left(2 {x}^{3} - 1\right)$

Now we can take the derivative of this term by term.

Term 1: $\ln \left(x\right)$
This is the easiest, as the derivative of $\ln \left(x\right) = \frac{1}{x}$.

Term 2: $2 \ln \left({x}^{2} + 1\right)$
We need to use the chain rule for this one. We first apply the derivative of $\ln \left(x\right)$ rule to get $2 \cdot \frac{1}{{x}^{2} + 1}$, and then we apply the chain rule to get $2 \cdot \frac{2 x}{{x}^{2} + 1} = \frac{4 x}{{x}^{2} + 1}$.

Term 3: $\frac{1}{2} \ln \left(2 {x}^{3} - 1\right)$
We use the same process as term 2. Use $\ln \left(x\right)$ rule to get $\frac{1}{2} \cdot \frac{1}{2 {x}^{3} - 1}$, then use the chain rule to get $\frac{1}{2} \cdot \frac{6 {x}^{2}}{2 {x}^{3} - 1} = \frac{3 {x}^{2}}{2 {x}^{3} - 1}$.

Put all of these together to get the final result:
$\frac{d}{\mathrm{dx}} \ln \left[\frac{x {\left({x}^{2} + 1\right)}^{2}}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right] = \frac{1}{x} + \frac{4 x}{{x}^{2} + 1} - \frac{3 {x}^{2}}{2 {x}^{3} - 1}$.