Originally Posted by

**Bradyns** Take this equation for example.

$\displaystyle x^2 + yx = 2$

It isn't explicitly a function of y, with respect to x, you have to do some manipulation before differentiating normally..

$\displaystyle y = ?$

In this case,

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

If I gave you this originally, you would be able to differentiate this, and the problem is no different.

But, one thing that we do usually forget is that we differentiate both sides.

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}\frac{2 - x^2}{x}$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}2x^{-1}-\frac{\mathrm{d}y}{\mathrm{d} x}x$

So we get,

$\displaystyle 1\cdot \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1$

*Take note of the left hand side.*

Finally we would get:

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1$