# How do you differentiate f(x)=2 ln( x^2 - 3x +4)  using the chain rule?

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

#### Explanation:

Given function:

$f \left(x\right) = 2 \setminus \ln \left({x}^{2} - 3 x + 4\right)$

Differentiating above function w.r.t. $x$ using chain rule as follows

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

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

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

$= \setminus \frac{2}{{x}^{2} - 3 x + 4} \left(2 x - 3\right)$

$= \setminus \frac{2 \left(2 x - 3\right)}{{x}^{2} - 3 x + 4}$