# How do you differentiate sin^2x-sin^2y=x-y-5?

Jun 1, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \sin 2 x}{1 - \sin 2 y}$

#### Explanation:

Write the equation as:

${\sin}^{2} x - x = {\sin}^{2} y - y - 5$

Differentiate both sides with respect to $x$:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \sin 2 x}{1 - \sin 2 y}$

Jun 1, 2017

Answer: $y ' = \frac{\sin \left(2 x\right) - 1}{\sin \left(2 y\right) - 1}$

#### Explanation:

Differentiate ${\sin}^{2} x - {\sin}^{2} y = x - y - 5$

First, we take the derivative of both sides, leaving $\frac{\mathrm{dy}}{\mathrm{dx}}$ as $y '$:
$\frac{d}{\mathrm{dx}} {\sin}^{2} x - {\sin}^{2} y = \frac{d}{\mathrm{dx}} x - y - 5$

$2 \sin \left(x\right) \cos \left(x\right) - 2 \sin \left(y\right) \cos \left(y\right) y ' = 1 - y '$

Solve for $y '$ by adding $2 y ' \sin \left(y\right) \cos \left(y\right)$ to both sides and subtracting $1$ from both sides:
$2 \sin \left(x\right) \cos \left(x\right) - 1 = 2 y ' \sin \left(y\right) \cos \left(y\right) - y '$

$2 \sin \left(x\right) \cos \left(x\right) - 1 = y ' \left(2 \sin \left(y\right) \cos \left(y\right) - 1\right)$

$y ' = \frac{2 \sin \left(x\right) \cos \left(x\right) - 1}{2 \sin \left(y\right) \cos \left(y\right) - 1}$

Note that $\sin \left(2 \theta\right) = 2 \sin \left(\theta\right) \cos \left(\theta\right)$
So, we can write $y '$ as:
$y ' = \frac{\sin \left(2 x\right) - 1}{\sin \left(2 y\right) - 1}$