Implicit differentation help

**FrustratedCollegeKid** · October 23rd 2011, 02:25 PM

I have to find the slope of the tangent line to the curve at the point (2, 8):

√4x+4y + √2xy = 12

Or also can be written as:

(4x+4y)^(1/2) + (2xy)^(1/2) = 12

I tried differentating got got:

(d/dx)[(4x+4y)^(1/2)] = (1/2)*(4x+4y)^(-1/2)*[4+4(dy/dx)]
(d/dx)[(2xy)^(1/2)] = (1/2)*(2xy)^(-1/2)*[2y+2x*(dy/dx)]

then,

(2+2*(dy/dx))/(4x+4y)^1/2 + (4+x*dy/dx)/(2xy)^(1/2)

I then plugged in numbers and ended up with (4+4sqrt5)/((-2sqrt5)-4). However that answer is wrong. Can anyone give a clue as to what i did wrong?

**SammyS** · October 23rd 2011, 03:40 PM

Originally Posted by FrustratedCollegeKid

I have to find the slope of the tangent line to the curve at the point (2, 8):

√4x+4y + √2xy = 12

Or also can be written as:

(4x+4y)^(1/2) + (2xy)^(1/2) = 12

I tried differentating got got:

(d/dx)[(4x+4y)^(1/2)] = (1/2)*(4x+4y)^(-1/2)*[4+4(dy/dx)]
(d/dx)[(2xy)^(1/2)] = (1/2)*(2xy)^(-1/2)*[2y+2x*(dy/dx)]

then,

(2+2*(dy/dx))/(4x+4y)^1/2 + (4+x*dy/dx)/(2xy)^(1/2)

I then plugged in numbers and ended up with (4+4sqrt5)/((-2sqrt5)-4). However that answer is wrong. Can anyone give a clue as to what i did wrong?

That last expression should be an equation, plus it has a typo.

(2+2*(dy/dx))/(4x+4y)^1/2 + (y+x*dy/dx)/(2xy)^(1/2) = 0

**Soroban** · October 23rd 2011, 07:16 PM

Hello, FrustratedCollegeKid!

It would help if you told us the (alleged) right answer.

Also, there must be typo in the problem.
. . The point (2, 8) is not on the graph.

Find the slope of the tangent line to the curve at the point (2, 8):

. . $\sqrt{4x+4y} + \sqrt{2xy} \:=\: 12$

We have: . $2(x+y)^{\frac{1}{2}} + \sqrt{2}(xy)^{\frac{1}{2}} \;=\;12$

Then: . $2\cdot\frac{1}{2}(x+y)^{-\frac{1}{2}}\left(1 + \frac{dy}{dx}\right) + \sqrt{2}\cdot\frac{1}{2}(xy)^{-\frac{1}{2}}\left(x\frac{dy}{dx} + 1\cdot y\right) \;\;=\;\;0$

. . $\frac{1}{\sqrt{x+y}}\left(1 + \frac{dy}{dx}\right) + \frac{1}{\sqrt{2xy}}\left(x\,\frac{dy}{dx} + y\right) \;\;=\;\;0$

. . $\frac{1}{\sqrt{x+y}} + \frac{1}{\sqrt{x+y}}\,\frac {dy}{dx} + \frac{\sqrt{x}}{\sqrt{2y}}\,\frac{dy}{dx} + \frac{\sqrt{y}}{\sqrt{2x}} \;\;=\;\;0$

. . $\frac{1}{\sqrt{x+y}}\,\frac{dy}{dx} + \frac{\sqrt{x}}{\sqrt{2y}}\,\frac{dy}{dx} \;\;=\;\; -\frac{1}{\sqrt{x+y}} - \frac{\sqrt{y}}{\sqrt{2x}}$

. . $\left(\frac{1}{\sqrt{x+y}} + \frac{\sqrt{x}}{\sqrt{2y}}\right)\frac{dy}{dx} \;\;=\;\; -\left(\frac{1}{\sqrt{x+y}} + \frac{\sqrt{y}}{\sqrt{2x}}\right)$

. . $\left(\frac{\sqrt{2y} + \sqrt{x}\sqrt{x+y}}{\sqrt{2y}\sqrt{x+y}}\right) \frac{dy}{dx} \;\;=\;\; -\left(\frac{\sqrt{2x} + \sqrt{y}\sqrt{x+y}}{\sqrt{2x}\sqrt{x+y}}\right)$

. . $\frac{dy}{dx} \;\;=\;\; -\left(\frac{\sqrt{2x} + \sqrt{y}\sqrt{x+y}}{\sqrt{2x}\sqrt{x+y}}\right) \left(\frac{\sqrt{2y}\sqrt{x+y}}{\sqrt{2y} + \sqrt{x}\sqrt{x+y}}\right)$

. . $\frac{dy}{dx} \;\;=\;\; -\frac{\sqrt{y}\left(\sqrt{2x} + \sqrt{y}\sqrt{x+y}\right)}{\sqrt{x}\left(\sqrt{2y} + \sqrt{x}\sqrt{x+y}\right)}$

. . $\frac{dy}{dx} \;\;=\;\; -\frac{\sqrt{2xy} + y\sqrt{x+y}}{\sqrt{2xy} + x\sqrt{x+y}}$

**mr fantastic** · October 24th 2011, 04:16 AM

Originally Posted by Soroban

Hello, FrustratedCollegeKid!

It would help if you told us the (alleged) right answer.

Also, there must be typo in the problem.
. . The point (2, 8) is not on the graph.

We have: . $2(x+y)^{\frac{1}{2}} + \sqrt{2}(xy)^{\frac{1}{2}} \;=\;12$

Then: . $2\cdot\frac{1}{2}(x+y)^{-\frac{1}{2}}\left(1 + \frac{dy}{dx}\right) + \sqrt{2}\cdot\frac{1}{2}(xy)^{-\frac{1}{2}}\left(x\frac{dy}{dx} + 1\cdot y\right) \;\;=\;\;0$

. . $\frac{1}{\sqrt{x+y}}\left(1 + \frac{dy}{dx}\right) + \frac{1}{\sqrt{2xy}}\left(x\,\frac{dy}{dx} + y\right) \;\;=\;\;0$ *

. . $\frac{1}{\sqrt{x+y}} + \frac{1}{\sqrt{x+y}}\,\frac {dy}{dx} + \frac{\sqrt{x}}{\sqrt{2y}}\,\frac{dy}{dx} + \frac{\sqrt{y}}{\sqrt{2x}} \;\;=\;\;0$

. . $\frac{1}{\sqrt{x+y}}\,\frac{dy}{dx} + \frac{\sqrt{x}}{\sqrt{2y}}\,\frac{dy}{dx} \;\;=\;\; -\frac{1}{\sqrt{x+y}} - \frac{\sqrt{y}}{\sqrt{2x}}$

. . $\left(\frac{1}{\sqrt{x+y}} + \frac{\sqrt{x}}{\sqrt{2y}}\right)\frac{dy}{dx} \;\;=\;\; -\left(\frac{1}{\sqrt{x+y}} + \frac{\sqrt{y}}{\sqrt{2x}}\right)$

. . $\left(\frac{\sqrt{2y} + \sqrt{x}\sqrt{x+y}}{\sqrt{2y}\sqrt{x+y}}\right) \frac{dy}{dx} \;\;=\;\; -\left(\frac{\sqrt{2x} + \sqrt{y}\sqrt{x+y}}{\sqrt{2x}\sqrt{x+y}}\right)$

. . $\frac{dy}{dx} \;\;=\;\; -\left(\frac{\sqrt{2x} + \sqrt{y}\sqrt{x+y}}{\sqrt{2x}\sqrt{x+y}}\right) \left(\frac{\sqrt{2y}\sqrt{x+y}}{\sqrt{2y} + \sqrt{x}\sqrt{x+y}}\right)$

. . $\frac{dy}{dx} \;\;=\;\; -\frac{\sqrt{y}\left(\sqrt{2x} + \sqrt{y}\sqrt{x+y}\right)}{\sqrt{x}\left(\sqrt{2y} + \sqrt{x}\sqrt{x+y}\right)}$

. . $\frac{dy}{dx} \;\;=\;\; -\frac{\sqrt{2xy} + y\sqrt{x+y}}{\sqrt{2xy} + x\sqrt{x+y}}$

As a sidenote, when required to get the value of dy/dx at a given point, I find it much easier NOT to first make dy/dx the subject. Rather, I'd substitute the point into the line marked * and then simplify and make dy/dx the subject. Less algebra, less chance of error, less time taken.

**Soroban** · November 4th 2011, 05:52 AM

Hello, mr F!

Absolutely right . . . excellent advice!

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