# How do you find (dy)/(dx) given cos(2y)=sqrt(1-x^2)?

Dec 13, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{2 \sin \left(2 y\right) \sqrt{1 - {x}^{2}}}$

or equivalently:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{1 - {x}^{2}}}$

#### Explanation:

$\setminus \setminus \setminus \setminus \setminus \cos \left(2 y\right) = \sqrt{1 - {x}^{2}}$
$\therefore \cos \left(2 y\right) = {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

Differentiating implicitly and applying the chain rule we get:
$- \sin \left(2 y\right) \left(2 \frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{1}{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right)$
$2 \frac{\mathrm{dy}}{\mathrm{dx}} \sin \left(2 y\right) = \frac{x}{\sqrt{1 - {x}^{2}}}$

So we can rearrange to get;
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{2 \sin \left(2 y\right) \sqrt{1 - {x}^{2}}}$

We can also get an explicit expression should we need it;

Using ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ we have:

${\sin}^{2} 2 y + {\cos}^{2} 2 y = 1$
$\therefore {\sin}^{2} 2 y + \left(1 - {x}^{2}\right) = 1$
$\therefore {\sin}^{2} 2 y = {x}^{2}$
$\therefore \sin 2 y = x$

So the earlier solution can be written as:

$\setminus \setminus \setminus \setminus \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{2 x \sqrt{1 - {x}^{2}}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{1 - {x}^{2}}}$