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

We will apply the product rule for differentiation. The product rule states that if we have two functions $u(x)$ and $v(x)$, then:

$\frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

Let:

• $u = \frac{1}{(1-3x)^{3}}$
• $v = x^{\frac{1}{3}}$

Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$.

First, differentiate $v$:

$\frac{dv}{dx} = \frac{1}{3} x^{-\frac{2}{3}}$

Next, differentiate $u$ using the chain rule:

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

Now we can apply the product rule:

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

Substituting in:

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

This gives:

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