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 9 - 2017 - Paper 4

To find where $\frac{dy}{dx} > 0$, we investigate two factors:

1. $(x^2 - 1)^4 > 0$, which is always positive except at $x = \pm 1$.
2. $g(x) = 2 + 20x^2 > 0$, which is always positive.

Thus, $\frac{dy}{dx} > 0$ for all $x$ except at $x = 1$ and $x = -1$:

The set of values of $x$ for which $\frac{dy}{dx} > 0$ is:

$x \in (-\infty, -1) \cup (-1, 1) \cup (1, \infty)$

Given $x = \ln(\sec(2y))$, we differentiate:

Using implicit differentiation:

1. Differentiate both sides:

2. Rewrite to find $dy/dx$:

3. Substitute $\sec(2y)$:
   • Since $\sec(2y) = e^x$:

4. Expressing in simplest form yields: