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

Dec 19, 2016

The first step is always to simplify the function as much as possible before differentiating.

f(x) = sqrt((x^2(3 - x))/(x(1 - 3x))

$f \left(x\right) = \sqrt{\frac{x \left(3 - x\right)}{1 - 3 x}}$

Let's rewrite as $f \left(x\right) = \sqrt{\square}$ where $\square = \frac{- {x}^{2} + 3 x}{1 - 3 x}$. Before applying the chain rule on $f \left(x\right)$, we need to apply the quotient rule to find $\frac{d \square}{\mathrm{dx}}$.

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

$\frac{d}{\mathrm{dx}} \square = \frac{- 2 x + 3 + 6 {x}^{2} - 9 x - 3 {x}^{2} + 9 x}{1 - 3 x} ^ 2$

$\frac{d}{\mathrm{dx}} \square = \frac{3 {x}^{2} - 2 x + 3}{1 - 3 x} ^ 2$

Now, use the chain rule to find the derivative of $f \left(x\right)$.

Let $y = \sqrt{u} = {u}^{\frac{1}{2}}$ and let $u = \square$. Then $y ' = \frac{1}{2 {u}^{\frac{1}{2}}}$ and $u ' = \frac{3 {x}^{2} - 2 x + 3}{1 - 3 x} ^ 2$ (as derived above).

$y ' = \frac{1}{2 {u}^{\frac{1}{2}}} \times \frac{3 {x}^{2} - 2 x + 3}{1 - 3 x} ^ 2$

$y ' = \frac{3 {x}^{2} - 2 x + 3}{2 {\left(1 - 3 x\right)}^{2} \sqrt{\frac{x \left(3 - x\right)}{1 - 3 x}}}$

Hopefully this helps!