Let y = 2x³ - 3x² - 1 - Leaving Cert Mathematics - Question Question 1 - 2013

We can use the product rule for differentiation, where if ( u = 2x^2 + 3x + 1 ) and ( v = x^3 - x + 2 ), then:

[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} ]

Calculating ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):

[ \begin{align*} \frac{du}{dx} & = 4x + 3 \ \frac{dv}{dx} & = 3x^2 - 1 \end{align*} ]

Now, substituting back:

[ \frac{dy}{dx} = (2x^2 + 3x + 1)(3x^2 - 1) + (x^3 - x + 2)(4x + 3) ]

Expanding this will yield the final derivative.

Given ( y = \frac{3x}{2x + 5} ), we can find ( \frac{dy}{dx} ) using the quotient rule:

[ \frac{dy}{dx} = \frac{(2x + 5)(3) - (3x)(2)}{(2x + 5)^2} ]

Substituting ( x = 0 ):

[ \frac{dy}{dx} = \frac{(2(0) + 5)(3) - (3(0))(2)}{(2(0) + 5)^2} = \frac{15}{25} = \frac{3}{5} ]

Therefore, the value is ( \frac{3}{5} ).