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

Nov 28, 2015

$f ' \left(x\right) = \frac{x \left(4 {x}^{3} - 3 x - 2\right)}{2 {x}^{3} - 1}$

#### Explanation:

This problem look intimidating but if we can rewrite it using logarithm properties, it actually really easy.

*Remember: color(red)(log(a/b) hArr loga- log b ; color(blue)(loga^n = n log a ; $\textcolor{g r e e n}{\ln e = 1}$

Step 1: Rewrite the logarithm expression
$f \left(x\right) = \ln \left(\frac{{e}^{{x}^{2} + 1}}{2 {x}^{3} - 1} ^ \left(\frac{1}{2}\right)\right)$ hArr color(red)(f(x)= ln(e^(x^2+1))-ln(2x^3-1)^(1/2)

$\Leftrightarrow f \left(x\right) = \textcolor{b l u e}{\left({x}^{2} + 1\right) \cdot \left(\ln e\right)} - \frac{1}{2} \ln \left(2 {x}^{3} - 1\right)$

$\Leftrightarrow f \left(x\right) = \textcolor{g r e e n}{\left({x}^{2} + 1\right)} - \frac{1}{2} \ln \left(2 {x}^{3} - 1\right)$

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Step 2: Now can can begin to differentiate the function
*Remember color(red)(g(x)= lnx ; g'(x)= (dx)/x)

$f ' \left(x\right) = \textcolor{g r e e n}{2 x} - \frac{1}{2} \left(\frac{\textcolor{red}{6 {x}^{2}}}{2 {x}^{3} - 1}\right)$ Differentiate

$f ' \left(x\right) = \textcolor{g r e e n}{2 x} - \left(\frac{\textcolor{red}{3 {x}^{2}}}{2 {x}^{3} - 1}\right)$ Simplify

$f ' \left(x\right) = \frac{\textcolor{g r e e n}{2 x} \textcolor{red}{\left(2 {x}^{3} - 1\right)} - 3 {x}^{2}}{2 {x}^{3} - 1}$ Common denominator

$f ' \left(x\right) = \frac{4 {x}^{4} - 2 x - 3 {x}^{2}}{2 {x}^{3} - 1}$ Distributen and simplify

$f ' \left(x\right) = \frac{x \left(4 {x}^{3} - 3 x - 2\right)}{2 {x}^{3} - 1}$ Better answer :)