# How do you differentiate f(x)=ln(x(x^2+1)/(2x^3-1)^(1/2)) ?

Jul 12, 2017

$\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x \left({x}^{2} + 1\right)}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right)\right) = \frac{3 {x}^{5} - {x}^{3} - 3 {x}^{2} + 1}{x \left({x}^{2} + 1\right) \left(2 {x}^{3} - 1\right)}$

#### Explanation:

Note that in general, using the chain rule:

(1) $\frac{d}{\mathrm{dx}} \ln \left(f \left(x\right)\right) = \frac{f ' \left(x\right)}{f} \left(x\right)$

Using the properties of logarithms we have:

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

and as the derivative is linear:

$\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x \left({x}^{2} + 1\right)}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right)\right) = \frac{d}{\mathrm{dx}} \ln x + \frac{d}{\mathrm{dx}} \ln \left({x}^{2} + 1\right) - \frac{1}{2} \frac{d}{\mathrm{dx}} \ln \left(2 {x}^{3} - 1\right)$

Then, based on (1):

$\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x \left({x}^{2} + 1\right)}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right)\right) = \frac{1}{x} + \frac{2 x}{{x}^{2} + 1} - \frac{1}{2} \frac{6 {x}^{2}}{2 {x}^{3} - 1}$

$\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x \left({x}^{2} + 1\right)}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right)\right) = \frac{1}{x} + \frac{2 x}{{x}^{2} + 1} - \frac{3 {x}^{2}}{2 {x}^{3} - 1}$

$\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x \left({x}^{2} + 1\right)}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right)\right) = \frac{\left(2 {x}^{3} - 1\right) \left({x}^{2} + 1\right) + 2 {x}^{2} \left(2 {x}^{3} - 1\right) - 3 {x}^{3} \left({x}^{2} + 1\right)}{x \left({x}^{2} + 1\right) \left(2 {x}^{3} - 1\right)}$

$\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x \left({x}^{2} + 1\right)}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right)\right) = \frac{2 {x}^{5} - {x}^{2} + 2 {x}^{3} + 1 + 4 {x}^{5} - 2 {x}^{2} - 3 {x}^{5} - 3 {x}^{3}}{x \left({x}^{2} + 1\right) \left(2 {x}^{3} - 1\right)}$

$\frac{d}{\mathrm{dx}} \left(\ln \left(\frac{x \left({x}^{2} + 1\right)}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right)\right) = \frac{3 {x}^{5} - {x}^{3} - 3 {x}^{2} + 1}{x \left({x}^{2} + 1\right) \left(2 {x}^{3} - 1\right)}$