# Implicit differentiation: nasty numbers

If $\frac{x^2}{36} + \frac{y^2}{64} = 1$ and $y(3) = 6.92820323$ find $y'(3)$

Okay so, we get:

$\frac{36\cdot2x}{36^2} + \frac{64\cdot 2y(y')}{64^2} = 0$

$\frac{72x}{1296} + \frac{128y(y')}{4096} = 0$

Where the heck do we go from here?

Okay: Here's what I did. I got really confused with the y(3) and y'(3) stuff. I sometimes don't know what goes where with this added notation (that I'm not used to yet).

$\frac{72\cdot3}{1296} + \frac{128\cdot 6.92820323z}{4096} = 0$

We have to substitute the 3 for the x, and the 6.92820323 for the y, and I'm just going to use z to represent y'.

Then:

$z = -0.769802$

This works because our $x$, $3$ is the same in both cases. When we plug in our y, which we know to be $6.928...$, it is synonymous to plugging the original equation back in (like we'd do if we didn't know what y was).

Thanks!

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How about solving for $y'(x)$ and then substituting in $x=3$ to get the value? – Amzoti Oct 3 '13 at 4:01
Set $x=3$ and solve for $y'(3)$. – lhf Oct 3 '13 at 4:02
I think you have an extra 36 and an extra 64 in your implicit derivative. – DanielV Oct 3 '13 at 4:09
The $y(3)\approx \text{mess}$ does carry some information. There are two values of $y$ corresponding to $x-3$, – André Nicolas Oct 3 '13 at 4:54