Given that $y = \frac{1}{(1-3x)^3} \cdot x^{\frac{1}{3}}$, find $\frac{dy}{dx}$. - Scottish Highers Maths - Question 6 - 2022

Next, we will use the product rule for differentiation. If we let:

$u = (1-3x)^{-3} \quad \text{and} \quad v = x^{\frac{1}{3}}$

The product rule states that:

$\frac{dy}{dx} = u'v + uv'$

Now we differentiate $u$ and $v$:

1. For $u = (1-3x)^{-3}$:

   $u' = -3(1-3x)^{-4} \cdot (-3) = 9(1-3x)^{-4}$

2. For $v = x^{\frac{1}{3}}$:

   $v' = \frac{1}{3}x^{-\frac{2}{3}}$

Now substituting into the product rule:

$\frac{dy}{dx} = 9(1-3x)^{-4} \cdot x^{\frac{1}{3}} + (1-3x)^{-3} \cdot \frac{1}{3}x^{-\frac{2}{3}}$

This provides the final expression for the derivative: