# How do you differentiate f(x)=sqrt((3x^2)/(2x-3))  using the chain rule?

Oct 18, 2017

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

#### Explanation:

Differentiate the outer function and multiply by the derivative of the inside like so:

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

Evaluating the derivative of the inside function we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(\frac{3 {x}^{2}}{2 x - 3}\right)}^{- \frac{1}{2}} \frac{\left(2 x - 3\right) \left(6 x\right) - \left(3 {x}^{2}\right) 2}{2 x - 3} ^ 2$

Simplify:

dy/dx=(6(x^2-3x))/(2(2x-3)^2sqrt((3x^2)/(2x-3))

Cancel 2's:

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

Technically, it would be proper to rationalize the denominator, but you can do that on your own.