# What is the second derivative of f(x) = ln (x-x^2) ?

Feb 17, 2016

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

#### Explanation:

To differentiate the natural logarithm function, use the chain rule. The chain rule states that

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

Knowing that $\frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) = \frac{1}{x}$, we can apply this specifically to natural logarithm functions:

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

The given function is $f \left(x\right) = \ln \left(x - {x}^{2}\right)$, so

$f ' \left(x\right) = \frac{1}{x - {x}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(x - {x}^{2}\right)$

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

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

To find the second derivative, use the quotient rule, which states that

$\frac{d}{\mathrm{dx}} \left(\frac{f \left(x\right)}{g \left(x\right)}\right) = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

Applying this to $f ' \left(x\right) = \frac{1 - 2 x}{x - {x}^{2}}$, we see that

$f ' ' \left(x\right) = \frac{\left(x - {x}^{2}\right) \frac{d}{\mathrm{dx}} \left(1 - 2 x\right) - \left(1 - 2 x\right) \frac{d}{\mathrm{dx}} \left(x - {x}^{2}\right)}{x - {x}^{2}} ^ 2$

Simplify through the power rule.

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

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

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