What exactly does $\frac{\mathrm{d}y}{\mathrm{d}x}$ calculate in implicit differentiation? (details given)

Question

Does it calculate the derivative of the given equation?
Because in implicit differentiation we don't have $y= equation$ so I don't know what we are differentiating using the $\frac{\mathrm{d}y}{\mathrm{d}x}$

Do we consider/assume the given equation lets say $x^2+y^2=36$ as $y$ when we say calculate $\frac{\mathrm{d}y}{\mathrm{d}x}$?
Does it mean we just have to differentiate the given equation and $\frac{\mathrm{d}y}{\mathrm{d}x}$ is its derivative or the gradient function?
This question might seem redundant but I swear nobody asks this question, its a really dumb thing that I don't know or can't seem to understand.

Answer 1

An equation of the form $f(x,y)=0$ describes a set of points in the plane $\mathbb R^2.$ Usually (but not always) this set consists of some curves. At a point on such a curve, we can, unless the tangent is vertical, find a neighborhood in which there are no distinct points with the same value of $y$. The curve segment in that neighborhood is the graph of some function $y=g(x).$ What we calculate with implicit differentiation is the derivative of that function: $\frac{dy}{dx}=g'(x).$

Here's a quadrifolium, the solution to $(x^{2}+y^{2})^{3}=a^{2}(x^{2}-y^{2})^{2}.$ It isn't the graph of any function $x\mapsto y$ globally, but locally, for example inside the red circle, it can be written as $y=g(x).$ (We might not be able to find an explicit expression for $g(x)$ though.) That part of the graph is smooth and using implicit differentiation we can find $g'(x).$

Answer 2

It's the same thing it always means, the slope of the tangent line to the curve that is the graph of the equation.

"Implicit" differentiation just has to do with how you differentiate the expressions. The meaning of the derivative is the same.

You need to be careful about how you say it -- you aren't taking the derivative of the equation, you're differentiating the expressions on each side of the equation.

For example, you can rewrite the equation $y=f(x)$ as $\frac{f(x)}{y}=1$ and use implicit differentiation to find $dy/dx$: $$\require{cancel}\frac{f(x)}{y}=1$$ $$\frac{d}{dx}\Bigg(\frac{f(x)}{y}\Bigg) = \frac{d}{dx}\Bigg(1\Bigg)$$ $$\frac{y\cdot\frac{d}{dx}[f(x)] - f(x)\cdot\frac{d}{dx}[y]}{y^2} = 0$$ $$\frac{y\cdot f'(x) - f(x)\cdot\frac{dy}{dx}}{y^2}= 0$$ $$y\cdot f'(x) - f(x)\cdot\frac{dy}{dx}= 0$$ $$\frac{dy}{dx}= \frac{\cancel{y}\cdot f'(x)}{\cancel{f(x)}}=f'(x)$$

Answer 3

Is just the slope of the tangent line. It's closely tied to the gradient.

Suppose $f(x,y)=0$, e.g. $f(x,y)=x^2+y^2 -R^2 =0$.

Then $\nabla f = 2x \hat{i} + 2y \hat{j}$

Let $d\vec{s} =dx \hat{i} + dy\hat{i}$

Then the chain rue tells us $df=\nabla f \cdot d\vec{s} =\frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy =2xdx + 2ydy . $

And since $f\equiv 0, df\equiv 0. $

$2xdx + 2ydy =0 \implies \frac{dy}{dx}=\frac{-x}{y}$

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You can see this is the slope in another way. Parameterize the curve.

Suppose we hae $x^2+y^2-9=0$.

Then we can let $x=3\cos t . y= 3 \sin t$.

$\frac{dx}{dt}=-3\sin t=-y$

$\frac{dy}{dt}=3\cos t = x$

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{x}{-y}=\frac{-x}{y}$

Here $t$ is just a parameter, you can think of it as time and then $x$ and $y$ give you position with respect to time, but what is the geometric meaning of $dx/dt$? Geometrically, not much, at least not directly, yet the ratio of y and x velocity gives you the slope of the tangent line, a solidly geometric object.