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EDIT: I forgot to add that I desperately need clarification on this as I will be sitting an exam shortly.

Sorry in advance if this does not comply with guidelines -- I'm still new to this site. I'll try my best to use correct formatting.

I am having a little difficulty with implicit differentiation, and I cannot seem to find an answer to my question anywhere.

My question: Is implicit differentiation ONLY used when taking the derivative of both sides with respect to the INDEPENDENT variable, or does the implicit differentiation process of $\frac {d}{da}[b^n]\Rightarrow \frac{d}{db}[b^n] * \frac {d}{da}[b]$ still work if $a$ is a dependent upon $b$?

I know the above process works if $a$ is the independent variable, and $b$ is a function of $a$ (i.e. $b(a)$), because we have to use the chain rule.

Basically, my question is, if we had (e.g.) $x^3 + y = 2y-3x^4$, such that $y=f(x)$, how would we/can we even simplify: $$\frac{d}{dy}[x^3+y]=\frac{d}{dy}[2y-3x^4]$$ Would it be:$$\frac{d}{dy}[x^3] +\frac{d}{dy}[y]=\frac{d}{dy}[2y]-\frac{d}{dy}[3x^4]$$ $$\color{red}{3x^2 * \frac{dx}{dy}}+1=2\color{red}{-12x^3\frac{dx}{dy}}$$

Or what if instead of $\frac{d}{dy}$, we took $\frac{d}{dm}$ of both sides -- i.e. the derivative of both sides with respect to a completely random variable. Could we still simplify as I have done in the working in red?

I am SO confused and I feel like I am going in circles. I'm so sorry if this doesn't make a great deal of sense. Any help would be GREATLY appreciated (although may I please request it be explained in such a way a numpty like me can understand).

Thank you in advance.

user642965
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  • Have you read the accepted answer to https://math.stackexchange.com/questions/379569/can-someone-give-me-a-deeper-understanding-of-implicit-differentiation – ancient mathematician May 10 '19 at 06:47
  • Yes I've read that answer, and it was actually extremely helpful. I suppose I was just confused because the logic behind the "classic" implicit differentiation process is based on the fact that the "b" in my working above is dependent on "a" -- just like y is generally dependent on x. And so I am confused as to why this process would work on an "x" term when differentiating with respect to y, as x is not dependent on y? (sorry if I'm over-complicating things) – user642965 May 10 '19 at 07:05
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    You should just forget for a bit all this "dependent" and "independent" stuff. The specific relation that you quote, has a graph which is a curve, and at almost every point on the graph, if we look at points close by, we see that we have $y=f(x)$ for some $f$ and $x=g(y)$ for some $g$. – ancient mathematician May 10 '19 at 07:09
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    Thank you so much for your help :) So I suppose that "classic" implicit differentiation process works even for d/dy of an "x" term, because x CAN be expressed in terms of y (which is the value we are differentiating with respect to)? – user642965 May 10 '19 at 07:16

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