# How do you find dy/dx by implicit differentiation of coty=x-y?

##### 1 Answer
Nov 17, 2016

We start by defining the derivative of $\cot y$. Recall that $\cot x = \frac{1}{\tan} x = \frac{1}{\sin \frac{x}{\cos} x} = \cos \frac{x}{\sin} x$.

$\therefore \cot y = \cos \frac{y}{\sin} y$

$\therefore \left(\cot y\right) ' = \frac{- \sin y \times \sin y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - \left(\cos y \times \cos y\right) \frac{\mathrm{dy}}{\mathrm{dx}}}{\sin y} ^ 2$

$\therefore \left(\cot y\right) ' = \frac{- {\sin}^{2} y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - {\cos}^{2} y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\sin} ^ 2 y$

$\therefore \left(\cot y\right) ' = - \frac{{\sin}^{2} y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + {\cos}^{2} y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\sin} ^ 2 y$

$\therefore \left(\cot y\right) ' = - \frac{\frac{\mathrm{dy}}{\mathrm{dx}}}{\sin} ^ 2 y$

$\therefore \left(\cot y\right) ' = - \frac{\mathrm{dy}}{\mathrm{dx}} {\csc}^{2} y$

Now, the rest of the relation.

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

$- {\csc}^{2} y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + 1 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(- {\csc}^{2} y + 1\right) = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 - {\csc}^{2} y}$

From the relation $1 + {\cot}^{2} y = {\csc}^{2} y$ we see that $1 - {\csc}^{2} y = - {\cot}^{2} y$. We can rewrite the derivative if we wish.

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\cot} ^ 2 y$

If we want, we can use the original function $\cot y = x - y$.

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

Hopefully this helps!