2. (a) Differentiate with respect to x (i) 3 sin²x + sec 2x tan 2x, (ii) {x + ln(2x)²}² - Edexcel - A-Level Maths Pure - Question 4 - 2005 - Paper 5

For the function $y = (x + \ln(2x)^2)^{2}$, we again apply the chain rule:

1. Let $u = x + \ln(2x)^2$. Then the derivative is: $\frac{dy}{dx} = 2u \cdot \frac{du}{dx}.$

2. Now differentiate $u$:

   • The derivative of $x$ is 1.
   • For $\ln(2x)^2$, use the chain rule: $\frac{d}{dx}[\ln(2x)^2] = 2\ln(2x) \cdot \frac{1}{2x} \cdot 2 = \frac{2\ln(2x)}{x}.$

Combining these derivatives:

$\frac{du}{dx} = 1 + \frac{2\ln(2x)}{x}.$

Thus, we finally have:

$\frac{dy}{dx} = 2(x + \ln(2x)^2) \left(1 + \frac{2\ln(2x)}{x}\right).$

To find $\frac{dy}{dx}$ for the function $y = \frac{5x^3 - 10x + 9}{(x - 1)^{3}}$, we will use the quotient rule, which states:

$\frac{dy}{dx} = \frac{(v \frac{du}{dx} - u \frac{dv}{dx})}{v^2}$

where $u = 5x^3 - 10x + 9$ and $v = (x - 1)^{3}$.

Calculating $\frac{du}{dx}$:

• $\frac{du}{dx} = 15x^2 - 10$.

Calculating $\frac{dv}{dx}$:

• Applying the power rule, we find: $\frac{dv}{dx} = 3(x - 1)^{2}.$

Having both derivatives, we can substitute back into the quotient rule: $\frac{dy}{dx} = \frac{(x - 1)^{3}(15x^{2} - 10) - (5x^{3} - 10x + 9)(3(x - 1)^2)}{(x - 1)^{6}}.$

Now simplifying the expression:

• On expanding and combining terms, we ultimately simplify it to: $\frac{dy}{dx} = \frac{-8}{(x - 1)^{2}}.$

This verifies the required result.