# Derivatives of Trigonometric, Logarithmic, and Exponential Functions. Implicit Relations. How do I differentiate this implicit function ? Answer: sin(x-y) / sin(x-y) - 1

## y = $\cos \left(x - y\right)$

May 27, 2017

Given: $y = \cos \left(x - y\right)$

Differentiate both sides. The left side is trivial:

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

The right side requires the chain rule:

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

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

Returning to the equation, we substitute into the right side:

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

Distribute the $- \sin \left(x - y\right)$:

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

Move all of the terms containing $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the left:

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

factor out $\frac{\mathrm{dy}}{\mathrm{dx}}$

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

Divide both sides by $\left(1 - \sin \left(x - y\right)\right)$

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

Multiply the right side by 1 in the form of $\frac{- 1}{-} 1$

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