Undergrad Questions about implicit differentiation?

[EchoRush]

TL;DR

Quick questions about implicit differentiation.

I am new to calculus. I am doing well in my class. I just have a few questions about implicit differentiation. First, why do we call it "implicit" differentiation?

Also, when we do it, why when we differentiate a term with a "y" in it, why do we have to multiply it by a dY/dX? What does that actually do? What is the point of doing it? Why would it be wrong to not do it?

Here is an example in the attached image:

 

[fresh_42]

Implicit differentiation means we have a function $f\, : \,\mathbb{R}\longrightarrow \mathbb{R}$, say $y \longmapsto f(y)$. The derivative is $\dfrac{df}{dy}$ and nothing implicit happened up to here. Now what if $y=g(x)$ is itself a function $g\, : \,\mathbb{R}\longmapsto\mathbb{R}$. Then we have $x \longmapsto g(x) = y \longmapsto f(y)$ and $\dfrac{df}{dy}$ doesn't tell us how $f$ changes with $x$. So we do some quotient algebra: $\dfrac{df}{dx} = \dfrac{df}{dy} \cdot \dfrac{dy}{dx} = \dfrac{df}{dy}\cdot \dfrac{dg}{dx}$. The first factor is the derivative of $f$ as we started with, according to the variable $y$, and the second factor is the derivative of $y=g(x)$ according to the variable $x$. Together we have $\dfrac{df}{dx}=f\,'(y)\cdot g'(x)$. It is called implicite, since the function $y=g(x)$ is nested in $f$. The variable of $f$, which is $y$, depends implicitly on the variable $x$ via the function $g$.

 

[Mark44]

First, why do we call it "implicit" differentiation?

Because we have an equation in which one variable, say y, is not explicitly given as a function of another variable, say x. For example, in the equation $x^2 + y^2 = 4$ we can treat y as if it were a function of x, even though we have no equation that directly relates x and y; i.e., we don't have a formula that gives a y value for a given x value. That would be an explicit ("explained") relationship between the two variables rather than the implicit ("implied") relationship in the equation I wrote.

Also, when we do it, why when we differentiate a term with a "y" in it, why do we have to multiply it by a dY/dX? What does that actually do?

Because of the chain rule.
If I use implicit differentiation with respect to x on the equation I wrote above, we have $\frac d{dx}(x^2 + y^2) = \frac d{dx}4$, or $2x + \frac d{dx}(y^2) = 0$.
To get the remaining derivative, use the chain rule like so: $\frac d{du}u^2 \cdot \frac {du}{dx}$, where $u = y^2$.
We end up with this equation: $2x + 2y\frac{dy}{dx} = 0$. If necessary, we can solve for $\frac{dy}{dx}$ algebraically.

I've skipped a few steps, but I hope you are able to follow my explanation.

 

[Delta2]

It is called implicit because we are not given an explicit equation $y=f(x)$ which we are able to explicitly differentiate with respect to x. Instead we are given a perplexed equation $f(x,y)=0$ (like $x^2+y^3x=0$ for example) which we can explicitly differentiate the terms involving $x$ (differentiation with respect to $x$ always) and implicitly differentiate the terms involving $y=y(x)$ using the chain rule as follows $$\frac{dg(y(x))}{dx}=\frac{dg(y)}{dy}\frac{dy}{dx}$$
where $g(y)$ is any function of y like $g(y)=y^3$. I believe the above answers your second question also as to where the term $\frac{dy}{dx}$ comes from. It comes from the application of the chain rule of differentiation.