# How do you implicitly differentiate -1=xy+e^ysec(x/y) ?

Oct 6, 2016

I got

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

Since $y = f \left(x\right)$, we must account for that when we take the derivative of $y$ with respect to $x$.

$\frac{d}{\mathrm{dx}} \left[- 1\right] = \frac{d}{\mathrm{dx}} \left[x y + {e}^{y} \sec \left(\frac{x}{y}\right)\right]$

$0 = \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right) + \left({e}^{y} \frac{d}{\mathrm{dx}} \left[\sec \left(\frac{x}{y}\right)\right] + \sec \left(\frac{x}{y}\right) \frac{d}{\mathrm{dx}} \left[{e}^{y}\right]\right)$

$= \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right) + \left({e}^{y} \sec \left(\frac{x}{y}\right) \tan \left(\frac{x}{y}\right) \cdot \frac{d}{\mathrm{dx}} \left[\frac{x}{y}\right] + \sec \left(\frac{x}{y}\right) {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$= x \frac{\mathrm{dy}}{\mathrm{dx}} + y + \left[{e}^{y} \sec \left(\frac{x}{y}\right) \tan \left(\frac{x}{y}\right) \cdot \left(- \frac{x}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{y}\right) + \sec \left(\frac{x}{y}\right) {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

Now we simplify and isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$.

$0 = x \frac{\mathrm{dy}}{\mathrm{dx}} + y + \left[- \frac{x}{y} ^ 2 {e}^{y} \sec \left(\frac{x}{y}\right) \tan \left(\frac{x}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{y} {e}^{y} \sec \left(\frac{x}{y}\right) \tan \left(\frac{x}{y}\right) + \sec \left(\frac{x}{y}\right) {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

$0 = x \frac{\mathrm{dy}}{\mathrm{dx}} + y - \frac{x}{y} ^ 2 {e}^{y} \sec \left(\frac{x}{y}\right) \tan \left(\frac{x}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{y} {e}^{y} \sec \left(\frac{x}{y}\right) \tan \left(\frac{x}{y}\right) + \sec \left(\frac{x}{y}\right) {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$

It's all multiplied out, so separate your variables.

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

$- y - \frac{1}{y} {e}^{y} \sec \left(\frac{x}{y}\right) \tan \left(\frac{x}{y}\right) = \left[x - \frac{x}{y} ^ 2 {e}^{y} \sec \left(\frac{x}{y}\right) \tan \left(\frac{x}{y}\right) + \sec \left(\frac{x}{y}\right) {e}^{y}\right] \frac{\mathrm{dy}}{\mathrm{dx}}$

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

Simplifying this further, we'd get:

$\textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- y - \frac{1}{y} {e}^{y} \sec \left(\frac{x}{y}\right) \tan \left(\frac{x}{y}\right)}{x - {e}^{y} \sec \left(\frac{x}{y}\right) \left(\frac{x}{y} ^ 2 \tan \left(\frac{x}{y}\right) + 1\right)}}$