Implicit Differentiation (Square Roots)

**jestrick** · March 9th 2010, 01:57 PM

I have tried and tried to do this and I need help. Please show your work so I can figure out where I messed up.

Find $\frac {dy}{dx}$ of $\sqrt {2x + 4y} + \sqrt {4xy} = 18$

**skeeter** · March 9th 2010, 03:40 PM

Originally Posted by jestrick

I have tried and tried to do this and I need help. Please show your work so I can figure out where I messed up.

Find $\frac {dy}{dx}$ of $\sqrt {2x + 4y} + \sqrt {4xy} = 18$

... show your attempt so one may see where you "messed up".

**jestrick** · March 9th 2010, 05:00 PM

Originally Posted by skeeter

... show your attempt so one may see where you "messed up".

I will show my best attempt... I have a feeling that I am starting wrong

I start by taking the derivative of both sides...

$\frac {d}{dx} \sqrt {2x + 4y} + \sqrt {4xy} = \frac {d}{dx} 18$

leaving $\frac {d}{dx} 18 = 0$ so $\frac {d}{dx} \sqrt {2x + 4y} + \sqrt {4xy} = 0$

and this is how find the derivative of the first term...

$\frac {1}{2}(2x + 4y)^\frac {-1}{2} (2 + 4y')$

$\frac {1}{2}(4x + 8y)^\frac {-1}{2} + \frac {1}{2}(8xy' + 16yy')^\frac {-1}{2}$

$\frac {4}{2}(x + 2y)^\frac {-1}{2} + \frac {8}{2}(x + 2y)^\frac {-1}{2}y'$

I will leave that there and show the second term...

$\frac {1}{2}(4xy)^\frac {-1}{2} (4y + 4xy')$

$\frac {1}{2}(16xy^2)^\frac {-1}{2} + \frac {1}{2}(16x^2yy')^\frac {-1}{2}$

$\frac {16}{2}(xy^2)^\frac {-1}{2} + \frac {16}{2}(x^2y)^\frac {-1}{2}y'$

from here I put the two terms together to get...

$\frac {16}{2}(x^2y)^\frac {-1}{2}y' + \frac {8}{2}(x + 2y)^\frac {-1}{2}y' + \frac {16}{2}(xy^2)^\frac {-1}{2} + \frac {4}{2}(x + 2y)^\frac {-1}{2} = 0$

$\frac {16}{2}(x^2y)^\frac {-1}{2}y' + \frac {8}{2}(x + 2y)^\frac {-1}{2}y' = -\frac {16}{2}(xy^2)^\frac {-1}{2} - \frac {4}{2}(x + 2y)^\frac {-1}{2}$

$y'(\frac {16}{2}(x^2y)^\frac {-1}{2} + \frac {8}{2}(x + 2y)^\frac {-1}{2}) = -\frac {16}{2}(xy^2)^\frac {-1}{2} - \frac {4}{2}(x + 2y)^\frac {-1}{2}$

$y' = \frac {dy}{dx} = \frac {-\frac {16}{2}(xy^2)^\frac {-1}{2} - \frac {4}{2}(x + 2y)^\frac {-1}{2}}{\frac {16}{2}(x^2y)^\frac {-1}{2} + \frac {8}{2}(x + 2y)^\frac {-1}{2}}$

This was my best attempt but I know it's wrong. If someone could explain if I am starting wrong or if my algebra is wrong and maybe push me in the right direction I would greatly appreciate it.

**HallsofIvy** · March 10th 2010, 03:24 AM

Originally Posted by jestrick

I will show my best attempt... I have a feeling that I am starting wrong

I start by taking the derivative of both sides...

$\frac {d}{dx} \sqrt {2x + 4y} + \sqrt {4xy} = \frac {d}{dx} 18$

leaving $\frac {d}{dx} 18 = 0$ so $\frac {d}{dx} \sqrt {2x + 4y} + \sqrt {4xy} = 0$

and this is how find the derivative of the first term...

$\frac {1}{2}(2x + 4y)^\frac {-1}{2} (2 + 4y')$

$\frac {1}{2}(4x + 8y)^\frac {-1}{2} + \frac {1}{2}(8xy' + 16yy')^\frac {-1}{2}$

I think what you have done here write this as
$\frac{1}{2}(2x+4y)^\frac{-1}{2}(2)+ \frac{1}{2}(2x+4y)^\frac{-1}{2}(4y')$
which is correct, and then take terms inside the square root. But you did that incorrectly. $2= \sqrt{4}= \frac{1}{4^\frac{-1}{2}}$ so
$\frac{1}{2}$ $(2x+ 4y)^\frac{-1}{ 2}$ $= \frac{1}{2}(\frac{2x+4y}{4})^\frac{-1}{2}$ .

It would be much better to just cancel that "2" with the initial "1/2"!

And it gets worse if you try to take "4y'" inside the square root! "implicit differentiation" always gives you and equation that is linear in y'. Don't make things complicated by changing y' to $\sqrt{y'}$ !

$\frac {4}{2}(x + 2y)^\frac {-1}{2} + \frac {8}{2}(x + 2y)^\frac {-1}{2}y'$

I will leave that there and show the second term...

$\frac {1}{2}(4xy)^\frac {-1}{2} (4y + 4xy')$

$\frac {1}{2}(16xy^2)^\frac {-1}{2} + \frac {1}{2}(16x^2yy')^\frac {-1}{2}$

$\frac {16}{2}(xy^2)^\frac {-1}{2} + \frac {16}{2}(x^2y)^\frac {-1}{2}y'$

from here I put the two terms together to get...

$\frac {16}{2}(x^2y)^\frac {-1}{2}y' + \frac {8}{2}(x + 2y)^\frac {-1}{2}y' + \frac {16}{2}(xy^2)^\frac {-1}{2} + \frac {4}{2}(x + 2y)^\frac {-1}{2} = 0$

$\frac {16}{2}(x^2y)^\frac {-1}{2}y' + \frac {8}{2}(x + 2y)^\frac {-1}{2}y' = -\frac {16}{2}(xy^2)^\frac {-1}{2} - \frac {4}{2}(x + 2y)^\frac {-1}{2}$

$y'(\frac {16}{2}(x^2y)^\frac {-1}{2} + \frac {8}{2}(x + 2y)^\frac {-1}{2}) = -\frac {16}{2}(xy^2)^\frac {-1}{2} - \frac {4}{2}(x + 2y)^\frac {-1}{2}$

$y' = \frac {dy}{dx} = \frac {-\frac {16}{2}(xy^2)^\frac {-1}{2} - \frac {4}{2}(x + 2y)^\frac {-1}{2}}{\frac {16}{2}(x^2y)^\frac {-1}{2} + \frac {8}{2}(x + 2y)^\frac {-1}{2}}$

This was my best attempt but I know it's wrong. If someone could explain if I am starting wrong or if my algebra is wrong and maybe push me in the right direction I would greatly appreciate it.

**jestrick** · March 10th 2010, 07:54 AM

I figured it out.

I used $\frac {-Fx}{Fy}$ where

$-Fx =$ the derivative treating y as a constant

$Fy =$ the derivative treating x as a constant

after doing that I ended up with

$\frac {-(2x + 4y)^\frac {-1}{2} + (4xy)^\frac{-1}{2} (2y)}{2(2x + 4y)^\frac {-1}{2} + (4xy)^\frac{-1}{2} (2x)} = \frac {-Fx}{Fy} = \frac {dy}{dx}$

**Miss** · March 10th 2010, 07:57 AM

Originally Posted by jestrick

I figured it out.

I used $\frac {-Fx}{Fy}$ where

$-Fx =$ the derivative treating y as a constant

$Fy =$ the derivative treating x as a constant

after doing that I ended up with

$\frac {-(2x + 4y)^\frac {-1}{2} + (4xy)^\frac{-1}{2} (2y)}{2(2x + 4y)^\frac {-1}{2} + (4xy)^\frac{-1}{2} (2x)} = \frac {-Fx}{Fy} = \frac {dy}{dx}$

You did not say that is a multivariable calculus problem.

**jestrick** · March 10th 2010, 08:14 AM

Originally Posted by Miss

You did not say that is a multivariable calculus problem.

well technically I am not supposed to know how to use $\frac {-Fx}{Fy}$ yet because I am in CALC 1 and I have been told it's CALC 2 material... but our teacher showed us the method and a classmate reminded me of it.

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