8. (a) By writing sec θ = \frac{1}{cos θ}, show that \frac{d}{dθ}(sec θ) = sec θ tan θ (b) Given that \begin{equation} x = e^{sec \, y} \, , \, x > e, \, 0 < y < \frac{π}{2} \end{equation} show that \frac{dy}{dx} = \frac{1}{x \, g(x)} \, , \, x > e where g(x) is a function of ln x. - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 5

Starting with the equation:

$x = e^{sec \, y}$

we take the natural logarithm of both sides:

$ln \, x = sec \, y$

Differentiating both sides with respect to x gives:

$\frac{1}{x} = \frac{dy}{dx} sec \, y$

Next, we rewrite sec y in terms of tan y:

Using the identity: $sec^{2} y - 1 = tan^{2} y$, we know:

$sec \, y = \sqrt{1 + tan^{2} \, y}$

Substituting this into our derivative gives:

$\frac{dy}{dx} = \frac{1}{x sec \, y}$

Now we express g(x):

$g(x) = sec \, y = \sqrt{(ln \, x)^{2} - 1}$

Thus:

$\frac{dy}{dx} = \frac{1}{x g(x)}$

This proves the required relationship.