# How do you differentiate  y =- ln( x^2 - x +4)  using the chain rule?

Feb 7, 2016

$y ' = \frac{1 - 2 x}{{x}^{2} - x + 4}$

#### Explanation:

The chain rule, in the case of a natural logarithm function, shows that

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

Applying this to the current question, we see that

$y ' = - \frac{1}{{x}^{2} - x + 4} \cdot \frac{d}{\mathrm{dx}} \left[{x}^{2} - x + 4\right]$

The derivative can be found through the power rule.

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

Simplified, this yields

$y ' = \frac{1 - 2 x}{{x}^{2} - x + 4}$