sorry i entered it wrong the first time

e^(x+y)+x-y^2=3

Printable View

- Jun 30th 2008, 08:47 AMgabet16941implicit diff
sorry i entered it wrong the first time

e^(x+y)+x-y^2=3 - Jun 30th 2008, 09:42 AMAryth
All we do for this is differentiate implicitly with respect to x, treating y as a separate variable:

We have:

$\displaystyle e^{x + y} + x - y^2 = 3$

Now we differentiate:

$\displaystyle \frac{d}{dx}[e^{x + y}] + 1 - \frac{d}{dx}[y^2] = 0$

$\displaystyle \frac{d}{dx}[e^x]\frac{d}{dx}[e^y] - 2y\frac{dy}{dx} = -1$

When we differentiate y, we differentiate it just like we would x, but we add a $\displaystyle \frac{dy}{dx}$ after it:

$\displaystyle e^x(e^y\frac{dy}{dx}) - 2y\frac{dy}{dx} = -1$

$\displaystyle \frac{dy}{dx}(e^{x + y} - 2y) = -1$

$\displaystyle \frac{dy}{dx} = -\frac{1}{e^{x + y} - 2y}$

And there you go. - Jun 30th 2008, 09:50 AMo_O
That is

**not**true. The product rule needs to be used if you were to go about it that way: $\displaystyle \frac{d}{dx} e^{x+y} = \frac{d}{dx} \left[e^{x}e^{y}\right] = \left(e^{x}\right)'e^y + e^{x}\left(e^y\right)' = \hdots$

However, it is sufficient to simply know that: $\displaystyle \frac{d}{dx}e^{u(x)} = e^{u(x)} \cdot u'(x)$. Here, we see that u = x + y, so:

$\displaystyle e^{x + y} + x - y^2 = 3$

$\displaystyle (x+y)'e^{x+y} + 1 - 2yy' = 0$

$\displaystyle (1 + y')e^{x+y} - 2yy' = -1$

etc. etc.