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I've been watching a video series called "The Essence of Calculus" on the YouTube channel $3$Blue$1$Brown, and I'm familiar with taking derivates, even moderately advanced derivatives, but on this occasion I come across a use of the notation I hadn't seen before. In this particular case, the following was said:

"The derivative of the equation $x^2+y^2=5^2$ can be taken by the following means: the derivative of $x^2$ is given as $2x~dx$ and the derivative of $y^2$ is given as $2y~dy$, therefore $2x~dx+2y~dy=0$ [etc]"

Now, I'm familiar with the chain rule, but only very briefly familiar with implicit differentiation. I understand how, when differentiation something like the equation for a circle, you end up with $$2x+2y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=-\frac{x}{y}$$

but even so, I'm unsure as to where the individual $dx$ and $dy$ tags came from. Provided we follow a slower approach, and we consider $x^2+y^2$ to be a multivariable function, $S$, of $x$ and $y$, then $$S=x^2+y^2$$ $$S+dS=(x+dx)^2+(y+dy)^2$$ $$dS=x^2+2x~dx+dx^2+y^2+2y~dy+dy^2-x^2-y^2$$ $$dS=2x~dx+dx^2+2y~dy+dy^2$$ we know that, since this is the derivative of a circle equation, that $dS=0$, since the radius of the circle will remain the same everywhere, but even we set the current expression to equal $0$, i.e. $0=2x~dx+dx^2+2y~dy+dy^2$, this I'm still far from where I want to be, which stokes a few questions. One is, is the $dx$ and $dy$ a notational trick, or is there more than this behind it? And what is it that we are even differentiating with respect to? Given the situation, this question has led me to believe that $x$ and $y$ are in fact functions of some other variable, as if I wasn't lost before!

Having followed the symbolic (non-intuitive) method of differentiating implicit functions has left me a little stumped by this situation.

Also, I'm aware that my question and this question are very similar, however I have not managed to find an answer that has explained my particular situation in a satisfactory way to myself in it.

Any help is appreciated, thank you.

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The concept you're looking for is the differential. If you look it up, this whole calculation will make sense. I didn't watch that video; but if it actually says that

the derivative of $x^2$ is given as $2x\,dx$

then technically speaking it's incorrect. The derivative of $f(x)=x^2$ is $\displaystyle f'(x)=\frac{df}{dx}=2x$, while the differential of $f(x)=x^2$ is $df=f'(x)\,dx=2x\,dx$.

In fact, you've pretty much correctly calculated the differential of the left-hand side yourself! One correction: where you have

$$dS=2x\,dx+dx^2+2y\,dy+dy^2$$

it actually must be $$\Delta S=2x\,dx+dx^2+2y\,dy+dy^2.$$ The differential is, informally speaking, the linear part of the change of a function. (You can look up a rigorous definition e.g. here.) Its linear part is obtained by discarding the higher degree terms in infinitesimal increments $dx$ and $dy$. So $dx^2$ and $dy^2$ go away, and we end up with $$dS=2x\,dx+2y\,dy.$$

$dx$ and $dy$ are not just a notational trick. They are also differentials; but for an independent variable, say $x$, its differential is defined to be equal to its change: $dx=\Delta x$.

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  • $\begingroup$ The typical format for a differential is given by $$dy=f'(x)dx$$ as far as my own reading is concerned. Given that $dx$ refers to an infinitesimally small change in $x$, then the shape formed between the point on the function and the $x$ axis will be a rectangle. To me, this means that the differential itself gives an equation for the area of this rectangle, i.e. $dy$ is the area, $f'(x)$ is the height of the rectangle, and $dx$ is the width. I don't understand why this works in this way yet, but is this correct? $\endgroup$ – joshuaheckroodt Nov 13 '17 at 2:26
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This is a standard technique but based on multi variable calculus. If $f(x, y) =0$ defines $y$ as a function of $x$ in implicit manner then we have $f_{x} dx+f_{y} dy=0$ or $\dfrac{dy} {dx} =-\dfrac{f_{x}} {f_{y}} $ where $f_{x}, f_{y} $ denote partial derivatives of $f(x, y) $ with respect to $x$ and $y$ respectively. These ideas are not difficult but do require some understanding of the concepts of multi variable calculus. If you are dealing with single variable calculus then one does not need to worry about these techniques. The usual rules of implicit differentiation work with almost equal ease and lead to the same result.

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