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

Jun 7, 2016

The chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$.

#### Explanation:

First, though we have to break this function up into less complex functions.

Let $y = \sqrt{u} = {u}^{\frac{1}{2}}$

Let $u = \frac{3 x}{2 x - 3}$

We're going to have to differentiate both functions. Let's start with the easiest: y.

By the power rule:

$y ' = \frac{1}{2} {u}^{\frac{1}{2} - 1}$

y' = 1/(2u^(1/2)

Now to $u$. For this function, we must differentiate using the quotient rule.

Let $u \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$. The quotient rule states that $u ' \left(x\right) = \frac{g ' \left(x\right) \times h \left(x\right) - g \left(x\right) \times \left(h ' \left(x\right)\right)}{h \left(x\right)} ^ 2$.

In our function, let $g \left(x\right) = 3 x$ and $h \left(x\right) = 2 x - 3$.

By the power rule $g ' \left(x\right) = 3$ and $h ' \left(x\right) = 2$

$u ' \left(x\right) = \frac{3 \left(2 x - 3\right) - \left(3 x \times 2\right)}{2 x - 3} ^ 2$

$u ' \left(x\right) = \frac{6 x - 9 - 6 x}{4 {x}^{2} - 12 x + 9}$

$u ' \left(x\right) = - \frac{9}{4 {x}^{2} - 12 x + 9}$

Now, substituting:

f'(x) = 1/(2sqrt((3x)/(2x - 3))) xx -9/(4x^2 - 12x + 9

Hopefully this helps!