6. Differentiate with respect to x, (i) e^{3x} (sin x + 2 cos x), (ii) x^{3} ln(5x + 2) - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 5

For the differentiation, we again apply the product rule. Let:

• u = x^{3}
• v = ln(5x + 2).

Then,

$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

Calculating each component:

• (\frac{du}{dx} = 3x^{2})
• (\frac{dv}{dx} = \frac{5}{5x + 2})

Substituting back, we get:

$\frac{d}{dx}(x^{3} ln(5x + 2)) = x^{3}\frac{5}{5x + 2} + ln(5x + 2)(3x^{2}).$

Starting with:

$y = \frac{3x^{2} + 6x - 7}{(x + 1)^{2}}$

We apply the quotient rule:

$\frac{dy}{dx} = \frac{(u'v - uv')}{v^2}$

where:

• u = 3x^{2} + 6x - 7
• v = (x + 1)^{2}.

Calculating derivatives:

• (u' = 6x + 6)
• (v' = 2(x + 1))

Thus,

Substituting yields:

$\frac{dy}{dx} = \frac{(6x + 6)(x + 1)^{2} - (3x^{2} + 6x - 7)2(x + 1)}{(x + 1)^{4}}.$

Simplifying gives:

$= \frac{20}{(x + 1)^{3}}.$

From the derivative derived previously:

1. The second derivative, \frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(\frac{20}{(x + 1)^{3}}\right).

Using the quotient rule again:

$\frac{d^{2}y}{dx^{2}} = -\frac{60}{(x + 1)^{4}}.$

2. To find where (\frac{dy}{dx} = \frac{15}{4},)

Set:

$\frac{20}{(x + 1)^{3}} = \frac{15}{4}.$

Cross multiply and solve:

$80 = 15(x + 1)^{3},$

Then,

$x = -1 \text{ or } x = -3.$