(a) Given that $$\frac{d}{dx}(\cos x) = -\sin x$$ show that $$\frac{d}{dx}(\sec x) = \sec x \tan x.$$ (b) Given that $$x = \sec 2y$$ find $$\frac{dx}{dy}$$ in terms of $$y$$ - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 4

To find the derivative of $\sec x$, we can use the reciprocal identity:

$\sec x = \frac{1}{\cos x}$.

Using the quotient rule:

$\frac{d}{dx}(\sec x) = \frac{d}{dx}\left( \frac{1}{\cos x} \right)$

Letting $u = \cos x$, we apply the quotient rule:

$\frac{d}{dx}(\sec x) = \frac{0 \cdot u - 1 \cdot \frac{d}{dx}(\cos x)}{(\cos x)^2}$

Substituting the derivative of $\cos x$:

$= \frac{-(-\sin x)}{(\cos x)^2} = \frac{\sin x}{(\cos x)^2}$

Recognizing the identities, we have:

$\frac{\sin x}{(\cos x)^2} = \sec x \tan x.$ This completes the proof.