implicit diff

**gabet16941** · June 30th 2008, 08:47 AM

sorry i entered it wrong the first time

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

**Aryth** · June 30th 2008, 09:42 AM

All we do for this is differentiate implicitly with respect to x, treating y as a separate variable:

We have:

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

Now we differentiate:

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

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

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

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

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

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

And there you go.

**o_O** · June 30th 2008, 09:50 AM

Originally Posted by Aryth

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

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

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

However, it is sufficient to simply know that: $\frac{d}{dx}e^{u(x)} = e^{u(x)} \cdot u'(x)$ . Here, we see that u = x + y, so:
$e^{x + y} + x - y^2 = 3$
$(x+y)'e^{x+y} + 1 - 2yy' = 0$
$(1 + y')e^{x+y} - 2yy' = -1$

etc. etc.

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