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 $rac{dy}{dx} = rac{-1}{1+x^2}$ - Edexcel - A-Level Maths Pure - Question 25 - 2013 - Paper 1

To differentiate $y = (x + ext{ sin } 2x)^3$, we use the chain rule.

Let:

• $u = x + ext{ sin } 2x$
  Then: $y = u^3$

Now, applying the chain rule gives: $\frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx}$

Next, we find $\frac{du}{dx}$:

• The derivative of $u$ is: $\frac{du}{dx} = 1 + 2\cos(2x)$

Substituting back, we have: $\frac{dy}{dx} = 3(x + ext{ sin } 2x)^2 \cdot (1 + 2\cos(2x))$

Given $x = \text{ cot } y$, we need to find $\frac{dy}{dx}$. Using implicit differentiation:

1. We know: $\frac{dx}{dy} = -\text{ csc}^2 y$

2. We then express: $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = -\text{ csc}^2 y$

3. Since $\text{ csc}^2 y = 1 + \text{ cot}^2 y$, we can substitute: $\frac{dy}{dx} = -\left(1 + x^2\right)$

Thus, we arrive at: $\frac{dy}{dx} = \frac{-1}{1+x^2}$