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

Use Implicit differentiation to prove
$y^{″}=\frac{-<!--\; -\; -->9}{y^2}$ if $4x^2-<!--\; -\; -->2y^2=9$

Consider the curve given by the equation xy - 3x + 2y = -2. Find the equation of the tangent in the point (2,1).
So through implicit differentiation I've found that (if I'm correct) y' = (1-y)/(x-2)
How do I proceed to find the tangent line since there are still 2 variables in the function?

I'm sure this question has been asked before and yet I can't find it.
Is there a proof of implicit differentiation or is it simply an application of the chain rule? If it's the former, could you give or point me to the proof? If the latter, could you explain exactly how conceptually it works (or point to a link that does so)?

I stumbled upon this calculus implicit differential question: find $\frac{du}{dy}$ of the function $u=\sin <!--\; \; -->(y^2+u)$.
The answer is $\frac{2y\cos <!--\; \; -->(y^2+u)}{1-\cos <!--\; \; -->(y^2+u)}$. I understand how to get the answer for the numerator, but how do we get the denominator part?
And can anyone point out the real intuitive difference between chain rule and implicit differentiation? I can't seem to get my head around them and when or where should I use them.

$Y^3+XY+X^2=0$
and I was suppose to find dY/dX and substitute the value 10 for X after differentiating. Weirdly enough, this equation does not let me get rid of the variable Y after solving and hence I couldn't find a valid answer.

Would request you to help me approach the problem. Also, I would like to do so without finding the roots of Y and solving, as someone pointed out earlier in class.

I am doing implicit differentiation and I want to know the fastest way to simplify this and find y'. My online algebra calculator fails to ever solve problems the easy way. Hoping a math pro on here could show me "the easy way".
$\frac{x+y+y^{\prime }}{xy}=e^{7x-<!--\; -\; -->y}(7-<!--\; -\; -->y^{\prime })$
$=$
$\frac{1}{x}+\frac{y^{\prime }}{y}=7e^{7x-<!--\; -\; -->y}-<!--\; -\; -->y^{\prime }{e}^{7x-<!--\; -\; -->y}$
$y^{\prime }=?$

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

An equation is defined as $x+y+x^5-<!--\; -\; -->y^5=0$.
$a$. Show that the equation determines $y$ as a function of $x$ in a neighbourhood of the origin (0,0).
$b$. Denote the function from $(a)$ by $y=\mathrm{\phi <!--\; \phi \; -->}(x)$. Find ${\mathrm{\phi <!--\; \phi \; -->}}^{(5)}(0)$ and ${\mathrm{\phi <!--\; \phi \; -->}}^{(2004)}(0)$.
For the first part,I simply needed to prove that the equation is $C^1$ in the neighbourhood of the origin and that the first derivative w.r.t.y does not vanish at that point. These are the sufficient conditions for the existence of an implicit function in the neighbourhood of the origin.
For the second part, I could do implicit differentiation (in a usual way) had it been for lower orders. But I need to learn the case for higher orders like the question asks. Please give me some hints.

Given $e^L+KL=K{e}^K$, we are being asked to find $\frac{dL}{dK}$. I think O need to use implicit differentiation, but I am really not sure how to do it!

$\frac{dz}{dx}=?$
$cos(zy)+z{x}^2=(1+y)e^{(}x-<!--\; -\; -->z)$
For the left side through implicit differentiation I have found $(-<!--\; -\; -->sin(zy))(y\ast <!--\; \ast \; -->(dz/dx))+2xz+(dz/dx)x^2$. I am completely unsure how to approach the right side, however.

I'm given $x^3+y^3=6xy$. It's stated that $y$ is a function of $x$ and I'm tasked to differentiate with respect to $x$.The implicit differentiation is:
$3x^2+3y^2{y}^{\prime }=6x{y}^{\prime }+6y$
Now simplify and express in terms of $y^{\prime }$. I'm going to number these steps.
1. $x^2+y^2{y}^{\prime }=2x{y}^{\prime }+2y$
2. $y^{\prime }=\frac{2x{y}^{\prime }+2y-<!--\; -\; -->x^2}{y^2}$
3. $y^{\prime }=\frac{2x{y}^{\prime }}{y^2}+\frac{2y-<!--\; -\; -->x^2}{y^2}$
4. $y^{\prime }=y^{\prime }\ast <!--\; \ast \; -->\frac{2x}{y^2}+\frac{2y-<!--\; -\; -->x^2}{y^2}$
5. $\frac{y^{\prime }}{y^{\prime }}=\frac{2x}{y^2}+\frac{2y-<!--\; -\; -->x^2}{y^2}.$ We know that $\frac{y^{\prime }}{y^{\prime }}=1$ and therefore I've made a mistake in my algebra. But I'm not sure what is wrong.Here is the correct simplification, taking it from step 1 again:
1. $x^2+y^2{y}^{\prime }=2x{y}^{\prime }+2y$
2. $y^2{y}^{\prime }-<!--\; -\; -->2x{y}^{\prime }=2y-<!--\; -\; -->x^2$
3. $y^{\prime }(y^2-<!--\; -\; -->2x)=2y-<!--\; -\; -->x^2$
4. $y^{\prime }=\frac{2y-<!--\; -\; -->x^2}{y^2-<!--\; -\; -->2x}$
This makes complete sense. What was wrong with my simplification? If it's not wrong, how can I arrive at the correct final expression of $y^{\prime }$?

How to find
$dy/dx$
from
$x^3{y}^6=(x+y{)}^9$
using implicit differentiation? I tried solving but I ended up with solution that does not agree with my textbook answer. How can I get
$dy/dx=y/x$

Use implicit differentiation to find $\frac{dc}{dv}$ when $v=\sqrt{c^2+v^2}$

I know I can solve this using normal implicit methods however I was wondering, why can I not square this so it becomes $v^2=c^2+v^2$ and then simplify so $c^2=0⟹<!--\; ⟹\; -->c=0$ then the derivative of this is just $c^{\prime }=0$, what is flawed?

Consider the curve $R^2$ given by the equation:
$x^2-<!--\; -\; -->y^2=1$
a/ Use the method of implicit differentiation to find all points on the curve at which that tangent has a slope of $\frac{5}{3}$
b/ Explain why there are no points on the curve at which the tangent is horizontal.
So far, I have found the implicit differentiation, which is $\frac{dy}{dx}=\frac{x}{y}$. Can you guys help me what should I do next. Any help is appreaciate

How do you differentiate ${\displaystyle x\cos y+y\cos x=1}$

Using implicit differentiation, verify that $u=f(x-<!--\; -\; -->tu)$ satisfies $\frac{\mathrm{\partial <!--\; \partial \; -->}u(x,t)}{\mathrm{\partial <!--\; \partial \; -->}t}+u(x,t)\frac{\mathrm{\partial <!--\; \partial \; -->}u(x,t)}{\mathrm{\partial <!--\; \partial \; -->}x}$.
Could someone explain how implicit differentiation works with a pde?

Consider the curve defined by $x^2+2y^2+4\mathrm{\beta <!--\; \beta \; -->}xy=K$ with $K>0$ and where $\mathrm{\beta <!--\; \beta \; -->}$ is a (sufficiently small) parameter. Assuming that the above can be used to define a function $x=G(y)$, use implicit differentiation to find out what is the largest value that $x$ can take on along this curve.

My attempt: Since $x=G(y)$, the largest value of $x$ is obtained when $G(y)$ is maximized, or when $G^{\prime }(y)=0$. However, the given equation of the curve isn't $G(y)$ itself. How do I express the curve equation as $G(y)$, or is there another way of finding the largest value of $x$ along this curve?

I was just wondering where the y'/(dy/dx) in implicit differentiation comes from
$x^2+y^2=25$
$(d/dx)x^2+(d/dy)y^2\ast <!--\; \ast \; -->\ast <!--\; \ast \; -->(dy/dx)\ast <!--\; \ast \; -->\ast <!--\; \ast \; -->=25(d/dx)$
$2x+2y(dy/dx)=0$
$(dy/dx)=-<!--\; -\; -->x/y$
Where does the bold part come from?

How do you find the second derivative for ${\displaystyle v=\frac{xy}{x-y}}$

Show that the sum of the x and y intercepts of any tangent line to the curve $x^{1/2}+y^{1/2}=4$ is equal to 16
So far I have found the derivative, $-<!--\; -\; -->\frac{\sqrt{y}}{\sqrt{x}}$, but am having trouble as to how I would found the point of tangency as as that is what I would think you do next?