# How do you differentiate  y =sqrtln(x^2-3x) using the chain rule?

##### 1 Answer

dy/dx=(2x-3)/(2(x^2-3x)sqrt(ln (x^2-3x))

#### Explanation:

$y = \sqrt{\ln \left({x}^{2} - 3 x\right)}$

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

using formula: $\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

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

using now the formula:

$\frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\ln \left({x}^{2} - 3 x\right)}} \cdot \frac{1}{{x}^{2} - 3 x} \left(2 x - 3\right)$

After simplification, the final answer is

dy/dx=(2x-3)/(2(x^2-3x)sqrt(ln (x^2-3x))