# How do you differentiate y=csc^-1(4x^2)?

Nov 12, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2}{x \sqrt{16 {x}^{4} - 1}} \left(= \frac{- 2}{x \sqrt{\left(4 {x}^{2} + 1\right) \left(4 {x}^{2} - 1\right)}}\right)$

#### Explanation:

$y = {\csc}^{-} 1 \left(4 {x}^{2}\right) \iff \csc y = 4 {x}^{2}$

Differentiating implicitly we have:
$\csc y = 4 {x}^{2}$
$- \csc y \cot y \frac{\mathrm{dy}}{\mathrm{dx}} = 8 x$
$- \left(4 {x}^{2}\right) \cot y \frac{\mathrm{dy}}{\mathrm{dx}} = 8 x$
$\cot y \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{x}$ ..... [1]

Now $1 + {\cot}^{2} A \equiv {\csc}^{2} A$
$\therefore 1 + {\cot}^{2} y = 16 {x}^{4}$
$\therefore {\cot}^{2} y = 16 {x}^{4} - 1$
$\therefore \cot y = \sqrt{16 {x}^{4} - 1}$

Substituting into [1] we get:
$\sqrt{16 {x}^{4} - 1} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{x}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \frac{2}{x}}{\sqrt{16 {x}^{4} - 1}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2}{x \sqrt{16 {x}^{4} - 1}} \left(= \frac{- 2}{x \sqrt{\left(4 {x}^{2} + 1\right) \left(4 {x}^{2} - 1\right)}}\right)$