Implicit Differentiation Proof 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)?
Thank you.
 A: Implicit differentiation is simply the use of the chain rule to differentiate a function. Often this makes it possible to differentiate a function that is difficult or impossible to separate into the form $y = f(x)$.
For example, consider the function $y = e^{xy}$. There is no way to separate $y$, so it is impossible to take an 'explicit' derivative. However, we can find an 'implicit' derivative in terms of $x$ and $y$ using the chain rule.
How do we do this? Well, simply take the derivative of both sides:
$$
\frac{d}{dx} y = \frac{d}{dx} e^{xy}
$$
The left hand side is simply $dy/dx$, or $y'$. The right hand side looks strange to derive, since the equation depends on both y and x. However, a few applications of the chain rule will clear things right up. First, let the 'inside function' be $u = xy$ and the outside function $e^u$. The derivative is, of course, 
$$
\frac{d}{dx} e^u = \frac{du}{dx} \frac{d}{du} e^u = \frac{du}{dx} e^u
$$
Now, using the product rule and then the chain rule, 
$$
\frac{du}{dx} = \frac{d}{dx} xy = (\frac{d}{dx} x)y + x (\frac{d}{dx} y) = (1)y + x(\frac{dy}{dx} \frac{d}{dy}y) = y + x(\frac{dy}{dx}*1) = y + xy'
$$
Finally, our implicit derivative is
$$
y' = (y + xy')e^{xy}
$$
and we can easily solve for $y'$.
So you see that implicit differentiation is the result of giving up on separating variables to get a derivative in terms of $x$ only, and cleverly using the chain rule to instead get a derivative in terms of $x$ and $y$.
A: It doesn't make any more sense to "prove implicit differentiation" than it does to "prove numbers," but I assume you're asking why implicit differentiation is valid i.e. preserves the truth of equations.
Implicit differentiation is just an application of the chain and other derivative rules to both sides of an equation, with (in the usual case) $y$ an abridgment of $f(x)$. Observe:
$$\frac{d}{dx}g(f(x),x)=\frac{\partial g}{\partial f}(f(x),x)\cdot\frac{df}{dx}(x)+\frac{\partial g}{\partial x} \tag{1}$$
$$\left(g(y,x)\right)'=\frac{\partial g}{\partial y}y'+\frac{\partial g}{\partial x} \tag{2}$$
The first is how you would take the derivative of a function of both $y=f(x)$ and $x$, but notice the difference between $d/dy$ and $\partial/\partial y$, while the latter equation is the same thing in fewer symbols courtesy of implicit differentiation. You always use ID in contexts where you have a function of both dependent and independent variables, and it's valid because of the above analogy - and it indeed generalizes to arbitrarily many dependent (or independent!) variables. For example, if we understand $u$ and $v$ to be functions of $t$ then we can differentiate
$$u^2+tv^2=1\quad\longrightarrow\quad 2uu'+(1\cdot v^2+t\cdot2vv')=0.$$
A: If you see, for example,
$$
\cdots + x^3 y^4 + \cdots
$$
remember that $y$ is a function of $x$, so it's as if you have
$$
\cdots + x^3 \big(f(x)\big)^4 + \cdots.
$$
Then
$$
\frac{d}{dx}\left( \cdots + x^3 \big(f(x)\big)^4 + \cdots \right) = \cdots + \left(\frac{d}{dx} x^3\right) \big(f(x)\big)^4 + x^3 \frac{d}{dx} \left(\big(f(x)\big)^4\right) + \cdots
$$
(that's the product rule)
$$
= \cdots + 3x^2 \big(f(x)\big)^4 + x^3 4\big( f(x)\big)^3 \frac{d}{dx} f(x) + \cdots
$$
That's the chain rule.
Now write it with $y$ wherever $f(x)$ is (and that means $\dfrac{dy}{dx}$ wherever $\dfrac{d}{dx} f(x)$ is).
A: I had a similar problem to firmly understand implicit differentiation, mostly because all explanations I had seen didn't make clear enough why the so called implicitly defined function qualifies the clause from the function definition (namely that for each element of its domain there is only one corresponding element from its range).
When we have an equation $x^2 + y^2 = 1$ it's said that it implicitly defines a function, let me call it $z$. But:


*

*it's not a function of x (if it was, it wouldn't satisfy the clause mentioned above)

*it cannot be plotted on the plane, on which you see the plot of the equation itself, so don't be deceived
The domain of this function is the set of pairs of x and y, where each y (in the domain, not on the x,y plane we use to plot the equation) is one of the several functions of x on itself (in the case of this particular equation we have two such functions: $y = \sqrt{1 - x^2}$ and $y = -\sqrt{1 - x^2}$ )
If we abstract all such functions as a variable y or $f(x)$ we will get the equation of the implicitly defined function $z$:
$$
z([y, x]) = x^2 + y^2 - 1
$$
Then (to me it's obvious, so sorry if I am not logical enough here) one can differentiate it as any other function. But since y is a function of x and we want to differentiate against x, it's required to use the chain rule.
That is why after finding $z'$ one have to supply a pair of x and y (not any y, but one that lies on the curve of the equation, that is, belongs to the domain of the implicit function) to get the correct value of m for the tangent line at that point (x, y).
