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The reason I asked this question is that I am trying to differentiate the relation $x^2 + y^2 = 25$. In an attempt to understand what's going on I gave the expression $x^2 + y^2x$ a name. "S" is essentially a function of two variables; it takes every point(x,y) on the plane and associates it with a number.

For points on this circle, that number happens to be $25$.

What it means to take a derivative of this expression, a derivative of S, is to consider a tiny change to both these variables, some tiny change dx to x, and some tiny change $dy$ to $y$.

The key point is that when you restrict yourself to steps along this circle, you're essentially saying you want to ensure that this value S doesn't change; it starts at a value of $25$, and you want to keep it at a value of $25$; that is, $dS$ should be $0$. So setting the expression $2xdx + 2ydy$ equal to $0$ is the condition under which a tiny step stays on the circle.

What I don't understand is this is the solution my textbook gave me...

$x^2 + y^2 = 25$

$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}25$

Remembering that y is a function of x and using the chain rule, we have

$\frac{d}{dx}(y^2) = \frac{d}{dx} (y^2) \frac{dy}{dx} = 2y\frac{dy}{dx}$

and now we solve this equation for $\frac{dy}{dx}$ which equals $-\frac{x}{y}$.

YUKITERU_AMANO
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    Right after "you have" there are two copies of the same term. I think you made a typo and the second one should be $\frac{d}{dy}(y^2)\frac{dy}{dx}.$ Please check that and fix. – coffeemath Oct 08 '21 at 09:21
  • Thank you so much for fixing this problem for me and the exponents look cool. I'll need to learn how to do that. – YUKITERU_AMANO Oct 08 '21 at 09:31

2 Answers2

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Informally, $dx$ and $dy$ are just small changes to $x$ and $y$, so we can treat them like numbers and do division. Thus from your $$2x\,dx+2y\,dy=0$$ we could just divide both sides by $dx$ and get $$2x+2y\frac{dy}{dx}=0.$$ Actually, this is not very rigorous since $dx$ and $dy$ are not really real numbers. But this is good enough for our purposes. You could also replace $dx$ and $dy$ by some finite change $\Delta x$ and $\Delta y$, to be slightly more rigorous.

Note that your solution and the given solution are basically equivalent.

YiFan Tey
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  • Thank you for taking the time to answer my question. But I actually made an error and forgot to include this question. "Remembering that y is a function of x and using the chain rule" why would y be a function of x? – YUKITERU_AMANO Oct 08 '21 at 09:28
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    If you have $x^2 + y^2 = 25$ then $y = \pm \sqrt{25-x^2}$ right? – Saegusa Oct 08 '21 at 09:32
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As ODEs, the equations $$M(x,y)dx+N(x,y)dy=0 \tag{1}$$ and $$M(x,y)+N(x,y)\dfrac{dy}{dx}=0 \tag{2}$$

are not the same things, but almost identical. The only difference between them is their solutions. A solution of $(2)$ must be a function of $x$. Meanwhile, for $(1)$, the role of $x$ and $y$ are equal, so a solution of $(2)$ maybe a function of $y$ with respect to $x$ (that is, $y=y(x)$), or a function of $x$ with respect to $y$ (that is, $x=x(y)$), or an implicit relation of variables $x$ and $y$ (that is, $\Phi(x,y)=0$).

As an example, $2xdx+2ydy=0$ has solutions $x^2+y^2=C^2$; and $2x+2y\dfrac{dy}{dx}=0$ has solutions $y=\pm\sqrt{C^2-x^2}$.

Black Mild
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