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 27 - 2013 - Paper 1

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

• $u = x + ext{sin} 2x$.

Then, the derivative becomes:

$\frac{dy}{dx} = 3u^2 \frac{du}{dx}$

Now we need to differentiate $u$:

$\frac{du}{dx} = 1 + 2 \text{cos} 2x$

Substituting this back into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = 3(x + \text{sin} 2x)^2 (1 + 2 \text{cos} 2x)$

Since $x = \text{cot} y$, we can use implicit differentiation with respect to $x$:

We know from trigonometric identities:

$\frac{dy}{dx} = -\frac{1}{\text{sin}^2 y} \cdot \frac{dy}{dx}$

Also, we can express $x$ in terms of $y$:

$\text{sin}^2 y + x^2 = 1$

Thus:

$\frac{dy}{dx} = -\csc^2 y = -\frac{1}{1+x^2}$

Therefore, we have shown that:

$\frac{dy}{dx} = \frac{-1}{1+x^2}$