# 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'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)?
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sodoendev7
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 $f\left(x\right)$. Observe:
$\begin{array}{}\text{(1)}& \frac{d}{dx}g\left(f\left(x\right),x\right)=\frac{\mathrm{\partial }g}{\mathrm{\partial }f}\left(f\left(x\right),x\right)\cdot \frac{df}{dx}\left(x\right)+\frac{\mathrm{\partial }g}{\mathrm{\partial }x}\end{array}$
$\begin{array}{}\text{(2)}& {\left(g\left(y,x\right)\right)}^{\prime }=\frac{\mathrm{\partial }g}{\mathrm{\partial }y}{y}^{\prime }+\frac{\mathrm{\partial }g}{\mathrm{\partial }x}\end{array}$
The first is how you would take the derivative of a function of both $y=f\left(x\right)$ and x, but notice the difference between $d/dy$ and $\mathrm{\partial }/\mathrm{\partial }y$, 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
${u}^{2}+t{v}^{2}=1\phantom{\rule{1em}{0ex}}⟶\phantom{\rule{1em}{0ex}}2u{u}^{\prime }+\left(1\cdot {v}^{2}+t\cdot 2v{v}^{\prime }\right)=0.$
###### Did you like this example?
KesseTher12
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
$\cdots +{x}^{3}\left(f\left(x\right){\right)}^{4}+\cdots .$
Then
$\frac{d}{dx}\left(\cdots +{x}^{3}\left(f\left(x\right){\right)}^{4}+\cdots \right)=\cdots +\left(\frac{d}{dx}{x}^{3}\right)\left(f\left(x\right){\right)}^{4}+{x}^{3}\frac{d}{dx}\left(\left(f\left(x\right){\right)}^{4}\right)+\cdots$
(that's the product rule)
$=\cdots +3{x}^{2}\left(f\left(x\right){\right)}^{4}+{x}^{3}4\left(f\left(x\right){\right)}^{3}\frac{d}{dx}f\left(x\right)+\cdots$
That's the chain rule.
Now write it with y wherever $f\left(x\right)$ is (and that means $\frac{dy}{dx}$ wherever $\frac{d}{dx}f\left(x\right)$ is).