1. ## Implicit Differentiation Problem

I've been stuck on this one for quite awhile now.

Find the slope of the tangent line to the curve

at the point (3, 7).

I basically understand implicit differentiation for the most part, I believe I am just stuck on the arithmetic. I can't get y' (dy/dx) to one side.

I rewrote the equation:
(x+2y)^(1/2) + (2xy)^(1/2) = 10.6

I then used the chain rule on both terms:
(1/2)(x+2y)^(-1/2)*(1+2y') + 1/2(2xy)^(-1/2)*(2xy'+2y) = 0

Then I simplified equation by moving ^(-1/2) to denominator:
[(1+2y')/(2sqrt(x+2y))] + (2xy'+2y)/(2sqrt(2xy)) = 0

And them from here I cannot understand the arithmetic to solving for y'. Any help would be appreciated, again, I'm pretty sure I'm correct to this point, I just don't know how to get y' by itself.

2. Originally Posted by kycon90
I've been stuck on this one for quite awhile now.

Find the slope of the tangent line to the curve

at the point (3, 7).

I basically understand implicit differentiation for the most part, I believe I am just stuck on the arithmetic. I can't get y' (dy/dx) to one side.

I rewrote the equation:
(x+2y)^(1/2) + (2xy)^(1/2) = 10.6

I then used the chain rule on both terms:
(1/2)(x+2y)^(-1/2)*(1+2y') + 1/2(2xy)^(-1/2)*(2xy'+2y) = 0

Then I simplified equation by moving ^(-1/2) to denominator:
[(1+2y')/(2sqrt(x+2y))] + (2xy'+2y)/(2sqrt(2xy)) = 0

And them from here I cannot understand the arithmetic to solving for y'. Any help would be appreciated, again, I'm pretty sure I'm correct to this point, I just don't know how to get y' by itself.
$\displaystyle \frac{1+2y'}{2\sqrt{x+2y}} + \frac{xy'+y}{\sqrt{2xy}} = 0$

sub in x = 3 and y = 7 ...

$\displaystyle \frac{1+2y'}{2\sqrt{17}} + \frac{3y'+7}{\sqrt{42}} = 0$

$\displaystyle \frac{1+2y'}{2\sqrt{17}} = -\frac{3y'+7}{\sqrt{42}}$

$\displaystyle \sqrt{42}(1+2y') = -2\sqrt{17}(3y'+7)$

$\displaystyle \sqrt{42} + 2\sqrt{42} y' = -6\sqrt{17} y' - 14\sqrt{17}$

$\displaystyle 2\sqrt{42} y' + 6\sqrt{17} y' = -(\sqrt{42} + 14\sqrt{17})$

$\displaystyle y'(2\sqrt{42} + 6\sqrt{17}) = -(\sqrt{42} + 14\sqrt{17})$

$\displaystyle y' = -\frac{\sqrt{42} + 14\sqrt{17}}{2\sqrt{42} + 6\sqrt{17}}$

3. Thanks skeeter, great job at breaking it down like that. I would have never found the answer at the track I was on.