Given $y = 2x(x^2 - 1)^5$, show that dy/dx = g(x)(x^2 - 1)^4 where g(x) is a function to be determined - Edexcel - A-Level Maths Pure - Question 8 - 2017 - Paper 4

To find where $\frac{dy}{dx} > 0$, we analyze the function g(x):

$g(x) = 22x^2 - 2$

Setting $g(x) > 0$ gives:

$22x^2 - 2 > 0 \implies x^2 > \frac{1}{11} \implies x > \pm \frac{1}{\sqrt{11}}$

Since we need to consider the factor $(x^2 - 1)^4$, which is always positive except when $x = \pm 1$, the final intervals where $\frac{dy}{dx} > 0$ are:

$x \in \left(-\infty, -1\right) \cup \left(-\frac{1}{\sqrt{11}}, \frac{1}{\sqrt{11}}\right) \cup \left(1, \infty\right).$

Starting with $x = \text{ln} \left( \text{sec} 2y \right)$, we differentiate using implicit differentiation:

$\frac{dx}{dy} = \frac{2\tan(2y)}{\text{sec}(2y)}$

Rearranging gives:

$\frac{dy}{dx} = \frac{\text{sec}(2y)}{2\tan(2y)}$

This is the required expression for dy/dx as a function of x in its simplest form.