# How do you differentiate 1/cos(x) = x/(x-y^2-y)?

##### 1 Answer
Nov 3, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{- {y}^{2} - y - {\left(x - {y}^{2} - y\right)}^{2} \sin x {\sec}^{2} x}{2 x y + x}$ or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{2} + y + {\left(x - {y}^{2} - y\right)}^{2} \sin x {\sec}^{2} x}{2 x y + x}$

#### Explanation:

$\frac{1}{\cos} x = \frac{x}{x - {y}^{2} - y}$

First put everything on one side

$0 = \frac{x}{x - {y}^{2} - y} - \frac{1}{\cos} x$

Then find the derivative with respect to x let's call it ${f}_{x}$ while holding y constant

${f}_{x} = \frac{\left(x - {y}^{2} - y\right) - x}{x - {y}^{2} - y} ^ 2 + \frac{1}{\cos} ^ 2 x \cdot - \sin x$

$= \frac{x - {y}^{2} - y - x}{x - {y}^{2} - y} ^ 2 - \sin x {\sec}^{2} x$

$= \frac{- {y}^{2} - y - {\left(x - {y}^{2} - y\right)}^{2} \sin x {\sec}^{2} x}{x - {y}^{2} - y} ^ 2$

Next find the derivative with respect to y call it ${f}_{y}$and hold x contant

${f}_{y} = \frac{- x \left(- 2 y - 1\right)}{x - {y}^{2} - y} ^ 2$

$= \frac{2 x y + x}{x - {y}^{2} - y} ^ 2$

To write your final answer use the formula $\frac{\mathrm{dy}}{\mathrm{dx}} = - {f}_{x} / {f}_{y}$

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

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

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