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

Using the relationship between $\frac{dy}{dx}$ and $\frac{dx}{dy}$, we have:

$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$

Substituting our expression from part (a):

$\frac{dy}{dx} = \frac{1}{6 \sec^2 3y \tan 3y}$

Using the identity $\sec^2 3y = x$, we rewrite it as:

$\frac{dy}{dx} = \frac{1}{6x \tan 3y}$

Now we substitute the identity $\tan 3y = \frac{\sin 3y}{\cos 3y}$ which we can relate back using $x$. After some manipulation, we demonstrate:

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

To find $\frac{d^2y}{dx^2}$, we start with:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

We apply the quotient rule here,

Let $u = 1$ and $v = 6x(x - 1)^2$:

$\frac{d^2y}{dx^2} = \frac{0 \cdot v - u \cdot \frac{d}{dx}(6x(x - 1)^2)}{v^2}$

Calculating $\frac{d}{dx}(6x(x - 1)^2)$:

Using the product rule,

$6[(x - 1)^2 + x \cdot 2(x - 1)] = 6[(x - 1)^2 + 2x(x - 1)]$

Now substituting back:

$\frac{d^2y}{dx^2} = \frac{-6[(x - 1)^2 + 2x(x - 1)]}{[6x(x-1)^{2}]^{2}}$

We can simplify further if necessary. The final result will be in its simplest form based on adjustments here.