How do you implicitly differentiate -1=x-ycot^2(x-y) ?

Jun 18, 2017

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

Explanation:

Take the derivative, $\frac{d}{\mathrm{dx}}$ of both sides

$\frac{d}{\mathrm{dx}} \left(- 1\right) = \frac{d}{\mathrm{dx}} \left(x\right) - \frac{d}{\mathrm{dx}} \left(y {\cot}^{2} \left(x - y\right)\right)$

$0 = 1 - y \frac{d}{\mathrm{dx}} \left({\cot}^{2} \left(x - y\right)\right) - \frac{\mathrm{dy}}{\mathrm{dx}} {\cot}^{2} \left(x - y\right)$

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

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

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

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

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

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

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