Given that $$\frac{d}{dx}(\cos x) = -\sin x$$ show that $$\frac{d}{dx}(\sec x) = \sec x \tan x.$$ Given that $$x = \sec 2y$$ (b) find $$\frac{dx}{dy}$$ in terms of y - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 4

To find $\frac{dx}{dy}$, we start by differentiating with respect to y:

Using the chain rule:

$\frac{dx}{dy} = \frac{d}{dy}(\sec 2y) = \sec 2y \tan 2y \cdot \frac{d}{dy}(2y) = 2 \sec 2y \tan 2y.$

To find $\frac{dy}{dx}$, we use the relationship:

$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}.$ Substituting in the expression from part (b):

$\frac{dy}{dx} = \frac{1}{2 \sec 2y \tan 2y}.$ Finally, using the identity $1 + \tan^2 A = \sec^2 A$, we can express this in terms of x:

Substituting $x = \sec 2y$:

$\frac{dy}{dx} = \frac{1}{x^2 \tan 2y}.$