Given that y = \frac{1}{(1-3x)^{3}} x^{\frac{1}{3}} find \frac{dy}{dx}. - Scottish Highers Maths - Question 6 - 2019

We will apply the product rule for differentiation. Let:

• u = $(1-3x)^{-3}$
• v = $x^{1/3}$

The product rule states that:

$\frac{dy}{dx} = \frac{du}{dx}v + u\frac{dv}{dx}$

Now, we differentiate u and v:

1. For $u = (1-3x)^{-3}$: $\frac{du}{dx} = -3(1-3x)^{-4}(-3) = 9(1-3x)^{-4}$

2. For $v = x^{1/3}$: $\frac{dv}{dx} = \frac{1}{3}x^{-2/3}$

Now substituting these back into the product rule: $\frac{dy}{dx} = 9(1-3x)^{-4} \cdot x^{1/3} + (1-3x)^{-3} \cdot \frac{1}{3}x^{-2/3}$

This simplifies to: $\frac{dy}{dx} = 9x^{1/3}(1-3x)^{-4} + \frac{1}{3}(1-3x)^{-3}x^{-2/3}$