Derivative of $s = u^2 \iff\sqrt s = u$ Suppose we have $s = u^2 \iff\sqrt s = u$.
Then taking the derivatives result in different things even thus $s$ takes on the same value for every $u$.
 A: Actually, they do:
$$
s=u^2\\
\sqrt s=u
$$
Differentiating:
$$
ds=2u\,du\\
\frac{ds}{2\sqrt s}=du
$$
Now, we can substitute $s=u^2$ in the second equation:
$$
du=\frac{ds}{2\sqrt u}=\frac{ds}{2u}
$$
Thus, we get $ds=2u\,du$ again.
This is true for any function $f(x)$ with inverse $g(x)=f^{-1}(x)$.
(using the chain-rule)
$$
f'(g(x))g'(x)=\frac{d}{dx}f(g(x))=\frac{dx}{dx}=1
$$
Solving for $g'(x)$ gives
$$
g'(x)=\frac1{f'(g(x))}
$$
A: Taking the differental
$$s=u^2\Rightarrow ds=2udu$$
and
$$\sqrt s= u\Rightarrow \frac{ds}{2\sqrt s}=du\iff \frac{ds}{2u}=du\iff ds=2udu$$
so we find the same result.
A: First yields $ds=2u\,du$.
Second yields $\displaystyle\frac{ds}{2\sqrt s}=du$. Substituting $\sqrt s=u$ will give the same as the first one.
A: Well, $$1 = \frac{d}{ds} s = \frac{d}{ds} u^2 = 2u \frac{du}{ds} \implies \frac{du}{ds} = \frac{1}{2u}= \frac{1}{2s^{1/2}}$$
On the other hand, $$\frac{du}{ds} = \frac{d}{ds} s^{1/2} = \frac{1}{2s^{1/2}}$$ as before.
A: Because $s=u^2\iff \sqrt{s}=|u|$, so your assumption is just wrong.
Edit, using Ragnar's answer:
From differentiation of 
$$
s=u^2\quad\text{vs}\quad
\sqrt s=u
$$
we have
$$
ds=2u\,du\quad\text{vs}\quad
\frac{ds}{2\sqrt s}=du.
$$
Substititing $s=u^2$ in the last equation yields
$$
du=\frac{ds}{2\sqrt{ u^2}}=\frac{ds}{2|u|}
$$
Thus, we don't get $ds=2u\,du$.
