5. (i) Differentiate with respect to x (a) y = x² ln 2x (b) y = (x + sin 2x)³ Given that x = cot y, (ii) show that \frac{dy}{dx} = \frac{-1}{1+x²} - Edexcel - A-Level Maths Pure - Question 26 - 2013 - Paper 1

To differentiate y = (x + sin 2x)³, we will apply the chain rule. The chain rule states that if you have a composite function, then:

$\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$

Let:

• f(u) = u³ and g(x) = x + sin 2x

Differentiating f:

• (f'(u) = 3u²)

Now differentiate g:

• (g'(x) = 1 + 2cos 2x)

Therefore,

Applying the chain rule:

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

Given x = cot y, we will first differentiate x with respect to y:

Using the derivative of cot:

• (\frac{dx}{dy} = -csc^{2}y)

Inverting this gives:

• (\frac{dy}{dx} = \frac{-1}{csc^{2}y})

Using the identity (csc^{2}y = 1 + cot^{2}y), we can replace csc²y:

So,

• (\frac{dy}{dx} = \frac{-1}{1 + cot^{2}y})

Since we know that x = cot y,

• (\frac{dy}{dx} = \frac{-1}{1 + x^{2}}) as required.