# How do you differentiate f(x) = (1-sqrt(3x-1))^2  using the chain rule?

Jun 19, 2016

$y = {u}^{2}$

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

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

We must therefore differentiate both functions.

$y ' = 2 u$

$u$ is a little more complicated: we will have to apply the chain rule to the radical.

The derivative of any constant $c$ is always $0$, so we don't have to worry about the $1$, thankfully.

$u = - {v}^{\frac{1}{2}}$

$v = 3 x - 1$

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

$v ' = 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 \times - \frac{1}{2 {v}^{\frac{1}{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{2 {v}^{\frac{1}{2}}}$

Now that we know the derivative of $u$, we can use the chain rule again to determine the derivative of $f \left(x\right)$.

$f ' \left(x\right) = - \frac{3}{2 {v}^{\frac{1}{2}}} \times 2 u$

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

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

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

$f ' \left(x\right) = \frac{- 6 + 6 \sqrt{3 x - 1}}{2 \sqrt{3 x - 1}}$

f'(x) = (-6(1 - 1sqrt(3x - 1)))/(2sqrt(3x - 1)

f'(x) = (-3(1 - 1sqrt(3x - 1)))/(sqrt(3x - 1)

f'(x) = (-3 + 3sqrt(3x- 1))/(sqrt(3x - 1)

f'(x) = (3sqrt(3x - 1) - 3)/(sqrt(3x - 1)

It's long, but it works!

Hopefully you understand this complicated problem better now!