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I have a feeling this question is going to have an obvious answer, but I'm left a bit puzzled. I have the following equation:

$$\tag 1 \frac{1}{S(x)} = \frac{W}{12}\cdot(1 - U(x)^2)$$, where $W$ is a constant

If I take a derivative of that equation with respect to $x$ as is, I come with the following results (solving for $S'(x)$):

$$\tag 2 S'(x) = \frac{W}{6}(S(x)^2)\cdot U(x)\cdot U'(x)$$

However, if I solve for $S(x)$ first and then take the derivative $(S(x) = (\frac{12}{W})/(1 - U(x)^2))$, I get the following result (again, solving for $S'(x)$):

$$\tag 3 S'(x) = \frac{24}{W}\cdot\frac{U(x)}{(1 - U(x)^2)^2}\cdot U'(x)$$

Shouldn't I arrive at the same answer regardless of the form of the equation? Or am I missing something fundamental here? Both $S(x)$ and $U(x)$ are continuous.

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  • $\begingroup$ Have you tried to use your original equation $(1)$ to see what you obtained in $(2)$ and $(3)$ is the same? $\endgroup$
    – Pedro
    May 10, 2013 at 20:33
  • $\begingroup$ What makes you think these are different? $\endgroup$
    – Erick Wong
    May 10, 2013 at 20:34

1 Answer 1

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Use that $$S^2=\frac{12^2}{W^2}\frac{1}{(1-U^2)^2}$$ to prove what you get is the same

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  • $\begingroup$ Ah, damn. I knew it was something obvious that I was missing. Clearly I need more coffee. Thanks! $\endgroup$
    – Axles
    May 10, 2013 at 20:40

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