# How do you find the derivative of y= (2x)/ sqrt(x - 1)?

Jun 29, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - 2}{\left(x - 1\right) \sqrt{x - 1}} .$

#### Explanation:

Method I uses the Quotient Rule for Diffn. : $\left(\frac{u}{v}\right) ' = \frac{v \cdot u ' - u \cdot v '}{v} ^ 2$.

Let $y = \frac{u}{v} = \frac{2 x}{\sqrt{x - 1}} .$

$u = 2 x \Rightarrow u ' = 2.$
v=sqrt(x-1)=(x-1)^(1/2) rArr v'=1/2*(x-1)^(1/2-1)*d/dx(x-1)=1/2(x-1)^(-1/2)*1=1/(2sqrt(x-1)

Sub.ing, these values in the Quotient Rule, we have,

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