Implicit Differentiation Explained: In-Depth Examples, Equations, and Expert Insights

Using implicit differentiation, we have that the derivative of $4x^{2}y-<!--\; -\; -->3/y=0$ is
$\frac{dy}{dx}=-<!--\; -\; -->\frac{8x{y}^3}{4x^2{y}^2+3}.$
But, if we multiply both sides of the original equation by y, we have the following equation $4x^2{y}^2-<!--\; -\; -->3=0$ which is seemingly equivalent for $y\ne 0$. Taking the derivative of this new function yields $-<!--\; -\; -->y/x$ What is causing the difference here. Is it a rule of implicit differentiation or because the 0 in the original equation that allows us to rewrite it in such a form? I am new to implicit differentiation and any help is appreciated.

For finding the derivative, $\frac{\mathrm{d}y}{\mathrm{d}x}$ of an implicit function, there is this formula,
When $f(x,y)=0$,
$\frac{\mathrm{d}y}{\mathrm{d}x}=-<!--\; -\; -->\frac{\text{Differentiation of }f\text{ w.r.t to }x\text{ keeping }y\text{ constant}}{\text{Differentiation of }f\text{ w.r.t to }y\text{ keeping }x\text{ constant}}$I want to understand how this method came about.
Is there a clear proof for this formula?

How do you differentiate ${\displaystyle 5x^3-4xy-2y^2=1}$

Use the process of implicit differentiation to find ${\displaystyle \frac{dx}{dy}}$
given that $\sin <!--\; \; -->(x)+\sin <!--\; \; -->(3y)=1$

Given $S=\pi x(x+2y)$ and $V=\pi x^{2}y$,I'm to show that $dS/dx=2\pi (x-<!--\; -\; -->y)$ when $V$ is constant, and $dV/dx=-<!--\; -\; -->\pi x(x-<!--\; -\; -->y)$ when $S$ is constant.
I just don't know where to start.

What does dy/dx really mean in the context of implicit differentiation
The problem is that I can't relate it back to the limit definition. If you can't write y explicitly as a function of x, often y approaches multiple values as x approach some value, or at some x, y has multiple values hence the limit does not exist and therefore the derivative will not exist

Today I learnt about implicit differentiation using this:
$\frac{df}{dy}\times <!--\; \times \; -->\frac{dy}{dx}$
I don't understand when doing implicit differentiation how the d/dy part works for y terms:
$\frac{d(y^2)}{dx}=\frac{d(y^2)}{dy}\times <!--\; \times \; -->\frac{dy}{dx}=2y\frac{dy}{dx}$
Why is the derivative of y2 with respect to $y$, $2y$? Do you assume it's a function of something else? And similarly if you apply this same formula to an $x$ term (even though its not needed) you get:
$\frac{d(x^2)}{dx}=\frac{d(x^2)}{dy}\times <!--\; \times \; -->\frac{dy}{dx}$
How does that simplify to the $2x$ I know it is?

Find the equation of the tangent line to the curve (a lemniscate) $2(x^2+y^2{)}^2=25(x^2-y^2)$ at the point (3,1). Write the equation of the tangent line in the form $y=mx+b.$
Every time that we do it we get a ridiculous number for the slope (150/362)

Alright so my issue is that i get stuck at this point and do not know what i should do to isolate $\frac{dy}{dx}$ since it is asking for implicit differentiation and this is what i have so far
$\ln <!--\; \; -->(y)\ast <!--\; \ast \; -->y^x\ast <!--\; \ast \; -->\frac{dy}{dx}=\ln <!--\; \; -->(x)\ast <!--\; \ast \; -->x^y\ast <!--\; \ast \; -->\frac{dy}{dx}$

Use implicit differentiation to determine $\frac{\mathrm{\partial <!--\; \partial \; -->}z}{\mathrm{\partial <!--\; \partial \; -->}x}$ in $yz=ln(x+z)$ and $\sin <!--\; \; -->(xyz)=x+2y+3z$.
Here is my answer:
$yz=ln(x+z)$
$y{z}^{\prime }=(1+z^{\prime })\frac{1}{x+z}$
$z^{\prime }=\frac{1}{yx+yz-<!--\; -\; -->1}$
and
$\sin <!--\; \; -->(xyz)=x+2y+3z$
$y(z+x{z}^{\prime })\cos <!--\; \; -->(xyz)=1+3z^{\prime }$
$yz\cos <!--\; \; -->(xyz)+xy{z}^{\prime }\cos <!--\; \; -->(xyz)=1+3z^{\prime }$
$yz\cos <!--\; \; -->(xyz)-<!--\; -\; -->1=z^{\prime }(3-<!--\; -\; -->xy\cos <!--\; \; -->(xyz))$
$z^{\prime }=\frac{yz\cos <!--\; \; -->(xyz)-<!--\; -\; -->1}{3-<!--\; -\; -->xy\cos <!--\; \; -->(xyz)}$
is that right?

How do you use implicit differentiation to find ${\displaystyle y\prime }$ for ${\displaystyle \sin \left(xy\right)=1}$?

How do you find dy/dx for ${\displaystyle y+2x-3x{y}^3=4}$

Given is $F(x,y)=y{e}^{3x}-<!--\; -\; -->2x^2=0$
I was asked to calculate $y^{\prime }$ using implicit differentiation.
I know that $y^{\prime }=-<!--\; -\; -->\frac{Fx}{Fy}=-<!--\; -\; -->\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}.$
So I obtained: $y^{\prime }(x)=-<!--\; -\; -->\frac{3y{e}^{3x}-<!--\; -\; -->4x}{e^{3x}}=\frac{-<!--\; -\; -->3y{e}^{3x}+4x}{e^{3x}}.$
But, in the solution manual I found another approach:
$y=f(x)$
i.e. $f(x)e^{3x}-<!--\; -\; -->2x^2=0$, We use the product rule
$\Rightarrow <!--\; \Rightarrow \; -->f^{\prime }(x)e^{3x}+f(x)3e^{3x}-<!--\; -\; -->4x=0$
$f^{\prime }(x)e^{3x}+f(x)3e^{3x}=4x$
$f^{\prime }(x)e^{3x}=4x-<!--\; -\; -->(f(x)3e^{3x})$
$f^{\prime }(x)=\frac{4x-<!--\; -\; -->(3e^{3x}f(x)}{e^{3x}}$
And finally
$f^{\prime }(x)=y^{\prime }(x)=\frac{4x-<!--\; -\; -->3e^{3x}y}{e^{3x}}=\frac{-<!--\; -\; -->3y{e}^{3x}+4x}{e^{3x}}$
So, the result is the same.

Are both approaches valid? Is there any difference between them? Which one would you recommend to use?

I have seen how people implicitly differentiate the equation $x^2+y^2=c$.
$d/dx(x^2)+d/dx(y^2)=d/dx(c)$
treating "$y$" as "$f(x)$" and using the chainrule we get
$2x+2y(y^{\prime })=0$
and solving for $y^{\prime }$
$y^{\prime }=-<!--\; -\; -->2x/2y$
The problem is that I just don´t understand implicit differentiation, I do know the rules but they don´t make any sense to me. The fact that it is valid to differentiate both "$x$" and "$y$" on the same side of the equation is what´s bothering me and even if I see "$y$" as a function of "$x$" I just end up imagining
$x^2+(-<!--\; -\; -->x^2+c)=c$
which doesn´t help me. I also don´t know very much about partial derivatives but I´m willing to learn about them if that helps me understand implicit differentiation.

Could someone explain chain rule when it comes to implicit differentiation?
so basically I want to know why when we have something like:
$v(x)=x-<!--\; -\; -->y+1$
If we take the derivative with respect to x, it yields:
$v^{\prime }(x)=1-<!--\; -\; -->\frac{dy}{dx}$
Now I still don't understand why when it comes to implicit differentiation, we need to tag a $y^{\prime }$ or $\frac{dy}{dx}$ after every time we take the derivative of a y term.

Find ${\displaystyle \frac{dy}{dx}}$ by implicit differentiation.
$\tan <!--\; \; -->(x+y)=x$
So far, I got to
${\displaystyle {\sec }^2<!--\; \; -->(x+y)(1+\frac{dy}{dx})=1}$
but then im lost.. can someone please help and explain? I would really appreciate it!!

Using implicit differentiation, find ${\displaystyle \frac{dy}{dx}}$
$x^3{y}^6=(x+y{)}^9$
My Attempt:
$x^3{y}^6=(x+y{)}^9$
Differentiating both sides with respect to $x$, we get:
${\displaystyle \frac{d}{dx}}(x^3{y}^6)={\displaystyle \frac{d}{dx}}((x+y{)}^9)$
$x^3.{\displaystyle \frac{d}{dx}}{y}^6+y^6.{\displaystyle \frac{d}{dx}}{x}^3=9(x+y{)}^8{\displaystyle \frac{d}{dx}}(x+y)$
$6x^3{y}^5.{\displaystyle \frac{dy}{dx}}+3x^2{y}^6=9(x+y{)}^8(1+{\displaystyle \frac{dy}{dx}})$
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{9(x+y{)}^8-<!--\; -\; -->3x^{2}y}{6x^3{y}^5-<!--\; -\; -->9(x+y{)}^8}}$
But the answer is
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{y}{x}}$

I am trying to understand implicit differentiation; I understand what to do (that is no problem), but why I do it is another story. For example:
$3y^2=5x^3$
I understand that, if I take the derivative with respect to x of both sides of the equation, I'll get:
$\frac{d}{dx}(3y^2)=\frac{d}{dx}(5x^3)$
$6y\frac{d}{dx}(y)=15x^2\frac{d}{dx}(x)$
$6y\frac{dy}{dx}=15x^2\frac{dx}{dx}$
$6y\frac{dy}{dx}=15x^2$
$\frac{dy}{dx}=\frac{15x^2}{6y}$
Unless I made some sort of error, this is what I am suppose to do. But why? Specifically, on the second line, I utilize the chain rule for the "outer function" and get 6y, but I still need to utilize the chain rule for the "inner function" which is the y. So why don't I go ahead and take the derivative of y and get 1? I know that I am not suppose to, but I don't really "get it." It seems to me that I only use the chain rule "halfway". Why isn't it an all or nothing? If it's all done with respect to x, it would seem to me that the 3y^2 should remain unchanged entirely. This is my problem.

How do you show that ${\displaystyle f(x,y)=x^4+y^4}$ has a saddle point at (x,y) = (0,0)?

How do you differentiate ${\displaystyle y\sin x=y}$