MHB What is the Differentiation Challenge?

[Theia]

Let's have a snack challenge for a while. ^^

Let $$x$$ and $$y$$ be real numbers (with restrictions $$y \ne 0, \ y \ne -x$$) and $$\frac{x - y}{x + y} = \frac{x + y}{y}$$.

Find $$\frac{\mathrm{d}y}{\mathrm{d}x}$$ in whatever form you like most. I mean, for example forms $$\frac{\mathrm{d}y}{\mathrm{d}x} = f(x, y)$$ or $$\frac{\mathrm{d}y}{\mathrm{d}x} = g(x)$$ are equally good at this time.

Please remember to use spoiler tags when you run to post the answer! :)

 

[HallsofIvy]

How is this a "challenge"?

Spoiler

Once you have multiplied on both sides by y(x+ y), it is a straight forward Calculus I "implicit differentiation".

 

[Theia]

How is this a "challenge"?

Spoiler

Once you have multiplied on both sides by y(x+ y), it is a straight forward Calculus I "implicit differentiation".

I said this is 'a snack', not 'an afwully complicated thing you need to think about seven years'...

Spoiler

Anyway, then you can show me your answer, right? ^^

 

[MarkFL]

I said this is 'a snack', not 'an afwully complicated thing you need to think about seven years'...

Spoiler

Anyway, then you can show me your answer, right? ^^

We appreciate challenge problems of virtually any level of difficulty. My advice to everyone is that if you find a posted problem "too easy" to solve, then leave it for those who will be challenged by it and enjoy posting a solution. :D

 

[I like Serena]

My solution:

Spoiler

$\d y x = \text{Undefined}$

Yummy snack! ;)

 

[Theia]

My solution:

Spoiler

$\d y x = \text{Undefined}$

Yummy snack! ;)

Spoiler

That's right! Can you give a short explanation too, please? :D

 

[I like Serena]

Spoiler

That's right! Can you give a short explanation too, please? :D

Sure! (Bandit)

Spoiler

$x$ and $y$ are given to be real numbers with $y\ne 0$ and $y\ne x$.

We have:
$$\frac{x-y}{x+y}=\frac {x+y}{y} \quad\Rightarrow\quad
(x-y)y=(x+y)^2 \quad\Rightarrow\quad
2y^2+xy+x^2=0 \quad\Rightarrow\quad
y = \frac 14 (-x\pm \sqrt{x^2 - 8x^2}) =\frac 14 x(-1\pm i\sqrt 7)
$$
Since $y\ne 0$, there are no real numbers $x,y$ that satisfy the equation.

Therefore $\d y x$ is not defined anywhere. (Cake)

 

[Theia]

Spoiler

Well done! Thank you! ^^

And now, I'd think I try to be more careful when it comes to 'implicit differentiation'... :D