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We know that $\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}$.

Under what conditions is this valid? Obviously, one such condition is that $y$ is differentiable wrt $t$ and $t$ is differentiable wrt $x$.

Are there any other necessary conditions? I think not. I remember reading that if $\lim\limits_{n\to\infty}a_n=l_1$ and $\lim\limits_{n\to\infty}b_n=l_2$, then $\lim\limits_{n\to\infty}a_nb_n=l_1l_2$.

Also, what if one or both of $l_1$ and $l_2$ are $\infty$? How does this become invalid? Could someone illustrate with an example?

Thanks in advance!

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    $\begingroup$ This is where Leibniz notation can become deceiving. In case this is what you were doing, those two cases $\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}$, and $\lim_{n \to \infty} a_n b_n$ are false parallels. Remember that the first is functional composition, so it is better expressed as: if $f : X \to Y$ $g: Y \to Z$, then $(g \circ f )'(x) = g'(f(x)) f'(x)$. $\endgroup$ – Ryder Bergerud Feb 24 '14 at 6:09
  • $\begingroup$ All I'm doing is simplifying two separate expressions, and seeing if they're equal. $\endgroup$ – algebraically_speaking Feb 24 '14 at 6:16
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Like Ryder says, this is better expressed in functional notation as $(f\circ g)'=g'*f'\circ g$. When looking at it this way, we see that all we really need for this to work is for the derivatives to be defined for both the interior and exterior function. No extra conditions necessary.

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