Another Implicit Differentiation Problem that's driving me nuts

[Vulcan]

\(\displaystyle \mbox{Differentiate }\, \dfrac{2x\, +\, 3y}{x^2\, +\, y^2}\, =\, 1\, \mbox{ and find }\, \dfrac{dy}{dx}.\)

I used the Quotient Rule:

. . . . .\(\displaystyle \dfrac{\dfrac{d}{dx}\, (2x\, +\, 3y)\, \cdot\, (x^2\, +\, y^2)\, -\, (2x\, +\, 3y)\, \cdot\, \dfrac{d}{dx}\, (x^2\, +\, y^2)}{(x^2\, +\, y^2)^2}\, =\, \dfrac{d}{dx}\, 1\)

. . . . .\(\displaystyle \Rightarrow\, \dfrac{\left(2\, +\, 3\, \dfrac{dy}{dx}\right)\, \cdot\, (x^2\, +\, y^2)\, -\, (2x\, +\, 3y)\, \cdot\, \left(2x\, +\, 2y\, \dfrac{dy}{dx}\right)}{(x^2\, +\, y^2)^2}\, =\, 0\)

. . . . .\(\displaystyle \Rightarrow\, \dfrac{2\, (x^2\, +\, y^2)\, +\, 3\, \dfrac{dy}{dx}\, (x^2\, +\, y^2)\, -\, 2x\, (2x\, +\, 3y)\, +\, 2y\, \dfrac{dy}{dx}\, (2x\, +\, 3y)}{(x^2\, +\, y^2)^2}\, =\, 0\)

Not sure where to go next! Multiplying out doesn't really help, and the denominator prevents me using the RHS of the equals sign.

Or am I completely on the wrong track?

Is there an easier way to show the working that using screen print?

Any assistance would be much appreciated.

 

[Deleted member 4993]

Or am I completely on the wrong track?

Is there an easier way to show the working that using screen print?

Any assistance would be much appreciated.

An easier process would be:

2x + 3y = x² + y² .........x² + y² > 0

2 + 3*(dy/dx) = 2x + 2y(dy/dx)

continue.....

 

[stapel]

\(\displaystyle \mbox{Differentiate }\, \dfrac{2x\, +\, 3y}{x^2\, +\, y^2}\, =\, 1\, \mbox{ and find }\, \dfrac{dy}{dx}.\)

I used the Quotient Rule:

. . . . .\(\displaystyle \dfrac{\dfrac{d}{dx}\, (2x\, +\, 3y)\, \cdot\, (x^2\, +\, y^2)\, -\, (2x\, +\, 3y)\, \cdot\, \dfrac{d}{dx}\, (x^2\, +\, y^2)}{(x^2\, +\, y^2)^2}\, =\, \dfrac{d}{dx}\, 1\)

. . . . .\(\displaystyle \Rightarrow\, \dfrac{\left(2\, +\, 3\, \dfrac{dy}{dx}\right)\, \cdot\, (x^2\, +\, y^2)\, -\, (2x\, +\, 3y)\, \cdot\, \left(2x\, +\, 2y\, \dfrac{dy}{dx}\right)}{(x^2\, +\, y^2)^2}\, =\, 0\)

. . . . .\(\displaystyle \Rightarrow\, \dfrac{2\, (x^2\, +\, y^2)\, +\, 3\, \dfrac{dy}{dx}\, (x^2\, +\, y^2)\, -\, 2x\, (2x\, +\, 3y)\, +\, 2y\, \dfrac{dy}{dx}\, (2x\, +\, 3y)}{(x^2\, +\, y^2)^2}\, =\, 0\)

...the denominator prevents me using the RHS of the equals sign.

What do you mean by this comment? The square of the sum of squares will only cause problems at (x, y) = (0, 0), so put the origin off to the side as an exception (to check at the end, if needed) and then multiply to get:

. . . . .\(\displaystyle 2\, (x^2\, +\, y^2)\, +\, 3\, \dfrac{dy}{dx}\, (x^2\, +\, y^2)\, -\, 2x\, (2x\, +\, 3y)\, +\, 2y\, \dfrac{dy}{dx}\, (2x\, +\, 3y)\, =\, 0\)

. . . . .\(\displaystyle 3\, \dfrac{dy}{dx}\, (x^2\, +\, y^2)\,+\, 2y\, \dfrac{dy}{dx}\, (2x\, +\, 3y)\, =\, 2x\, (2x\, +\, 3y)\, -\, 2\, (x^2\, +\, y^2)\)

...and so forth.

 

[Ishuda]

\(\displaystyle \mbox{Differentiate }\, \dfrac{2x\, +\, 3y}{x^2\, +\, y^2}\, =\, 1\, \mbox{ and find }\, \dfrac{dy}{dx}.\)

I used the Quotient Rule:

. . . . .\(\displaystyle \dfrac{\dfrac{d}{dx}\, (2x\, +\, 3y)\, \cdot\, (x^2\, +\, y^2)\, -\, (2x\, +\, 3y)\, \cdot\, \dfrac{d}{dx}\, (x^2\, +\, y^2)}{(x^2\, +\, y^2)^2}\, =\, \dfrac{d}{dx}\, 1\)

. . . . .\(\displaystyle \Rightarrow\, \dfrac{\left(2\, +\, 3\, \dfrac{dy}{dx}\right)\, \cdot\, (x^2\, +\, y^2)\, -\, (2x\, +\, 3y)\, \cdot\, \left(2x\, +\, 2y\, \dfrac{dy}{dx}\right)}{(x^2\, +\, y^2)^2}\, =\, 0\)

. . . . .\(\displaystyle \Rightarrow\, \dfrac{2\, (x^2\, +\, y^2)\, +\, 3\, \dfrac{dy}{dx}\, (x^2\, +\, y^2)\, -\, 2x\, (2x\, +\, 3y)\, \pmb{\huge{+}} \, 2y\, \dfrac{dy}{dx}\, (2x\, +\, 3y)}{(x^2\, +\, y^2)^2}\, =\, 0\)

Not sure where to go next! Multiplying out doesn't really help, and the denominator prevents me using the RHS of the equals sign.

Or am I completely on the wrong track?

Is there an easier way to show the working that using screen print?

Any assistance would be much appreciated.

I realize staple was just typing out what the original poster had but am I missing something? Shouldn't that \(\displaystyle \pmb{\huge{+}}\)above be a \(\displaystyle \pmb{\huge{-}}\)?

 

[Vulcan]

An easier process would be:

2x + 3y = x² + y² .........x² + y² > 0

2 + 3*(dy/dx) = 2x + 2y(dy/dx)

continue.....

Can't believe I never thought of that! so much easier.

Thank you!

 

[Vulcan]

What do you mean by this comment? The square of the sum of squares will only cause problems at (x, y) = (0, 0), so put the origin off to the side as an exception (to check at the end, if needed) and then multiply to get:

. . . . .\(\displaystyle 2\, (x^2\, +\, y^2)\, +\, 3\, \dfrac{dy}{dx}\, (x^2\, +\, y^2)\, -\, 2x\, (2x\, +\, 3y)\, +\, 2y\, \dfrac{dy}{dx}\, (2x\, +\, 3y)\, =\, 0\)

. . . . .\(\displaystyle 3\, \dfrac{dy}{dx}\, (x^2\, +\, y^2)\,+\, 2y\, \dfrac{dy}{dx}\, (2x\, +\, 3y)\, =\, 2x\, (2x\, +\, 3y)\, -\, 2\, (x^2\, +\, y^2)\)

...and so forth.

I do not understand why the denominator has disappeared as it is part of the Quotient Rule!

 

[Vulcan]

What do you mean by this comment? The square of the sum of squares will only cause problems at (x, y) = (0, 0), so put the origin off to the side as an exception (to check at the end, if needed) and then multiply to get:

. . . . .\(\displaystyle 2\, (x^2\, +\, y^2)\, +\, 3\, \dfrac{dy}{dx}\, (x^2\, +\, y^2)\, -\, 2x\, (2x\, +\, 3y)\, +\, 2y\, \dfrac{dy}{dx}\, (2x\, +\, 3y)\, =\, 0\)

. . . . .\(\displaystyle 3\, \dfrac{dy}{dx}\, (x^2\, +\, y^2)\,+\, 2y\, \dfrac{dy}{dx}\, (2x\, +\, 3y)\, =\, 2x\, (2x\, +\, 3y)\, -\, 2\, (x^2\, +\, y^2)\)

...and so forth.

How do you get the equations to appear like you do? I see some sort of code written when I reply?

 

[stapel]

I do not understand why the denominator has disappeared as it is part of the Quotient Rule!

What did you get when you multiplied through by the denominator? Please reply showing your steps and results. Thank you!

 

[Ishuda]

How do you get the equations to appear like you do? I see some sort of code written when I reply?

The code is called LaTex. See
ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf
for example for an introduction.

In this forum, you put your code between a [t e x][/t e x] tag pair (without the spaces). For example

[tex]\frac{101}{2}[/tex]

produces \(\displaystyle \frac{101}{2}\)

Other forums may use different tag pairs to indicate a beginning and end of the LaTex code.