How do you differentiate 5x^3-4xy-2y^2=1?

Feb 19, 2015

Hello,

You can differentiate a function, not an equality.

If you want differentiate a 2-variables function $\left(x , y\right) \setminus \mapsto f \left(x , y\right)$, you write

$d f = \setminus \frac{\setminus \partial f}{\setminus \partial x} \mathrm{dx} + \setminus \frac{\setminus \partial f}{\setminus \partial y} \mathrm{dy}$.

If $f \left(x , y\right) = 5 {x}^{3} - 4 x y - 2 {y}^{2}$, you have

$\setminus \frac{\setminus \partial f}{\setminus \partial x} = 15 {x}^{2} - 4 y$ and $\setminus \frac{\setminus \partial f}{\setminus \partial y} = - 4 x - 4 y$

therefore, you finally have

$d f = \left(15 {x}^{2} - 4 y\right) d x - 4 \left(x + y\right) d y$