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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. And I apologize if I got some of the terminology wrong.

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    $\begingroup$ "Why don't I ... take the derivative of $y$ and get $1$?" You do take the derivative of $y$ but you don't get $1$. If you were differentiating with respect to $y$, then you'd get $dy/dy=1$, but you're differentiating with respect to $x$. (If you really wanted to differentiate the left side of the equation with respect to $y$, then you'd have to also differentiate the right side with respect to $y$. You can't differentiate one side with respect to $x$ and the other with respect to $y$. You have to do the same thing to both sides of the equation.) $\endgroup$ – Andreas Blass Aug 23 '15 at 21:20
  • $\begingroup$ You could continue to get $y' y' + y y'' = 5 x$ - but it is unclear what your problem actually is... $\endgroup$ – johannesvalks Aug 23 '15 at 21:22
  • $\begingroup$ @John, actually $y$ is not a function of $x$ - or at least it is multi-valued. $\endgroup$ – johannesvalks Aug 23 '15 at 21:23
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    $\begingroup$ @johannesvalks, fair enough, I've removed my comment in order not to cause too much confusion. $\endgroup$ – John_dydx Aug 23 '15 at 21:25
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    $\begingroup$ @user263961, well $\frac{dz}{dx} = \frac{dz}{dy} \frac{dy}{dx}$ put in $z = 3y^2$ so $\frac{dz}{dy} = 6 y$ and so you get $\frac{dz}{dx} = 6 y \frac{dy}{dz}$. $\endgroup$ – johannesvalks Aug 23 '15 at 21:30
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$\frac{\mathrm d y}{\mathrm d x}$ may equal $1$ in special cases, but it is not generally so.   The very point of implicit derivation is that you don't know what it is.

So, the Chain Rule is used to separate the derivative of a function of $y$ wrt $x$ into two differential terms, one you can resolve, and one unknown.

To be clear, here's the process with the stage added and highlighted. Does this help? $$\begin{align} 3y^2&=5x^3 \\[1ex] \frac{\mathrm d(3y^2)}{\mathrm d x} &= \frac{\mathrm d(5x^3)}{\mathrm dx} & \text{Take the derivative w.r.t. }x \\[1ex] \color{blue}{\frac{\mathrm d(3y^2)}{\mathrm d y}\frac{\mathrm d y}{\mathrm d x}}&=\frac{\mathrm d(5x^3)}{\mathrm dx} & \text{Apply the Chain Rule to the L.H.S.} \\[1ex] 6y\frac{\mathrm d y}{\mathrm d x}&=15x^2 & \text{Evaluate the polynomial derivatives }\frac{\mathrm d(c\,z^n)}{\mathrm d z}=c\,n\,z^{n-1} \\ && \text{where }n\in\Bbb N,\text{ and }c\text{ is constant} \\[1ex] \frac{\mathrm d y}{\mathrm d x}&=\frac{5x^2}{2y} & \text{Use arthimetic rearrangement} \end{align}$$


Alternatively, we might use explicit differentiation, as follows: $$\begin{align} 3y^2&=5x^3 \\[1ex] y & = x^{3/2}\,\sqrt{\frac 5 3\;} & \star \\[1ex] \frac{\mathrm d y}{\mathrm d x} & = \frac{\mathrm d x^{3/2}}{\mathrm d x}\;\sqrt{\frac 5 3\;} \\[1ex] ~ & = \frac{3 x^{1/2}\sqrt 5}{2\sqrt 3} \\[1ex] ~ & = \frac{3 x^{1/2}\sqrt 5}{2\sqrt 3}\times\frac{x^{3/2}\sqrt 5}{y\sqrt 3} & \text{ re: }\star \\[1ex] ~ & = \frac{5x^2}{2y} \end{align}$$

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Change $y$ to $y(x)$ and everything should be clear. $y$ is not a constant, but a function of $x$.

$$\frac{dy}{dx}=\frac{d}{dx}y(x)=y'(x)$$

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For a concrete example, let $y(x)=x^2$. Clearly $y'=2x$. However, per the question, lets consider $y^2=x^4$.

Say you wanted to find $y'$. Then:

$$\frac{d}{dx} y^2=4x^3$$ $$2yy'=4x^3$$ $$y'=\frac{4x^3}{2y}=\frac{4x^3}{2x^2}=2x$$

Which is good! However:

$$2y=4x^3$$

is patently not true and even worse, doesn't contain $y'$, which is the entire point of this!

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