Simple Derivative paradox
Suppose I define $y(x)=x^3$
$${dy(x) \over dx} = 3x^2$$
$${dy(x) \over dy} = 1 = 3x^2 \frac{dx}{dy} = 0\text{ since }x \neq f(y)$$
$1 \neq 0$
If you take the differential $d()$ where $dy(x)$ then there is no issues since the derivative definition isn't explicitly used and it results in a $\dfrac{dx}{dy(x)}$. Why does this paradox exist? Thank you!!
$\endgroup$ 32 Answers
$\begingroup$The derivative $\frac{dx}{dy}$ isn't $0$; since $x = y^{1/3}$, we have $\frac{dx}{dy} = \frac{1}{3} y^{-2/3} = \frac{1}{3x^2}$. The same sort of argument holds for arbitrary $y(x)$ (modulo being careful about the uniqueness of inverses, which isn't relevant to the paradox you present).
$\endgroup$ $\begingroup$Why would you think $\frac{dx}{dy}=0$?
$x$ can be made dependent on $y$ by simple rearrangement, to give $x = y^{\frac 13}$, so everything should work out nicely.
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