Math Help - Clear this up for me.

1. Clear this up for me.

$e^y+6-e^-1 =3x^2 + 3y^2$

on the first term:

$e^y$

That uses chain rule? or does it remain the same?

2. Originally Posted by Zanderist
$e^y+6-e^-1 =3x^2 + 3y^2$

on the first term:

$e^y$

That uses chain rule? or does it remain the same?
so what do you want to do?

3. Originally Posted by dedust
so what do you want to do?

Well right now, I'm trying to figure out how it's derivative is.

4. Originally Posted by Zanderist
Well right now, I'm trying to figure out how it's derivative is.

if you differentiate it with respect to y then it remain the same

5. Originally Posted by Zanderist
$e^y+6-e^-1 =3x^2 + 3y^2$

on the first term:

$e^y$

That uses chain rule? or does it remain the same?
Using partial derivative

$e^y+6-e^{-1} =3x^2 + 3y^2$

$e^y+6-e^{-1} -3x^2 - 3y^2 = 0$

$F_x = -6x$

$F_y = e^y -6y$

so

$\frac{dy}{dx} = -\frac{F_x}{F_y} = \frac{6x}{e^y -6y}$

6. Originally Posted by Zanderist
$e^y+6-e^-1 =3x^2 + 3y^2$

on the first term:

$e^y$

That uses chain rule? or does it remain the same?
Originally Posted by dedust
so what do you want to do?
Originally Posted by Zanderist
Well right now, I'm trying to figure out how it's derivative is.

The point was- the derivative with respect to what variable?

The derivative of $e^y$ with respect to y is $e^y$.

The derivative of $e^y$ with respect to x is $e^y \frac{dy}{dx}$.

7. Originally Posted by dedust
if you differentiate it with respect to y then it remain the same
Right, this also means then.

$-e^-1$ Differentiate to.

$-log(e)$