# Implicit Differentiation Problem

• Feb 12th 2010, 09:49 AM
kycon90
Implicit Differentiation Problem
I've been stuck on this one for quite awhile now.

Find the slope of the tangent line to the curve

https://webwork.math.ohio-state.edu/...c7c70e06a1.png

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.
• Feb 12th 2010, 10:22 AM
skeeter
Quote:

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

https://webwork.math.ohio-state.edu/...c7c70e06a1.png

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 ...

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

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

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

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

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

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

$\displaystyle y' = -\frac{\sqrt{42} + 14\sqrt{17}}{2\sqrt{42} + 6\sqrt{17}}$
• Feb 12th 2010, 10:46 AM
kycon90
Thanks skeeter, great job at breaking it down like that. I would have never found the answer at the track I was on.