5. (i) Differentiate with respect to $x$ (a) $y = x^2 ext{ln } 2x$ (b) $y = (x + ext{sin } 2x)^3$ Given that $x = ext{cot } y$, (ii) show that $\frac{dy}{dx} = \frac{-1}{1+x^2}$ - Edexcel - A-Level Maths Pure - Question 26 - 2013 - Paper 1

For this function, we will use the chain rule. Let $u = x + \text{sin } 2x$, then $y = u^3$. The derivative using the chain rule is:
$\frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx}$
Now we need to find $\frac{du}{dx}$:
$\frac{du}{dx} = 1 + 2\text{cos } 2x$
Substituting back, we have:
$\frac{dy}{dx} = 3(x + \text{sin } 2x)^2(1 + 2\text{cos } 2x)$.

To solve this, we start with the relationship $x = \text{cot } y$. We will differentiate both sides with respect to $x$:
$\frac{dx}{dy} = -\text{cosec}^2 y$
From this, we find $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{1}{\text{cosec}^2 y}$
Next, using the identity $\text{cosec}^2 y = 1 + \text{cot}^2 y$, we can replace $ext{cosec}^2 y$:
$\frac{dy}{dx} = -\frac{1}{1 + \text{cot}^2 y}$
Since $x = \text{cot} y$, we have $\frac{dy}{dx} = -\frac{1}{1 + x^2}$.
Thus, we have shown the required result.