Given that $x = ext{sec}^2 3y$, $0 < y < \frac{\pi}{6}$ (a) find $\frac{dx}{dy}$ in terms of $y$ - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 7

Using the relationship $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$, we substitute:

$\frac{dy}{dx} = \frac{1}{6 \text{sec}^2(3y) \cdot \text{tan}(3y)}.$
Since $x = \text{sec}^2(3y)$, we can express $\text{tan}(3y)$ in terms of $x$:

$\text{tan}^2(3y) = \text{sec}^2(3y) - 1 = x - 1.$ Thus, we have:

$\frac{dy}{dx} = \frac{1}{6x \sqrt{x - 1}}.$ Now squaring gives:

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

To find $\frac{d^2y}{dx^2}$, we apply the chain rule again:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right).$ Using the previously established value of $\frac{dy}{dx}$, we apply the product rule and simplify:

$\frac{d^2y}{dx^2} = -\frac{1}{6} \cdot \frac{d}{dx}\left(\frac{1}{x(x - 1)^2}\right).$ Carrying out the differentiation and simplifying yields:

$\frac{d^2y}{dx^2} = \frac{2}{6x(x - 1)^3}.$ Simplifying gives:

$\frac{d^2y}{dx^2} = \frac{1}{3x(x - 1)^3}.$