Question: It seems to me that derivatives are treated differently in certain places, but I do not understand why.

Here is an example, if

\begin{align}
\frac{d}{dx} (\sqrt{1 + 4x^2}) & = \frac{1}{2\sqrt{1 + 4x^2}} \cdot \frac{d}{dx} (1 + 4x^2) \\
& = \frac{4 \cdot \frac{d}{dx}x^2}{2\sqrt{1 + 4x^2}} \\
& = \frac{4x}{\sqrt{1 + 4x^2}}
\end{align}

then why

$\frac{d}{dx} (1 + 4x^2) = 8x$?

Seems to me that $\frac{d}{dx} (1 + 4x^2)$ has two different answers

Answer: 
> 
> ...it turns out there are non-trivial high dimensional smooth knots $S^n\subset S^{n+2}$ 
>  such that $\pi\_1$ of the knot complement $X$ is
>  infinite cyclic. The inclusion of the meridian $S^1 \subset X$ is a
>  homology isomorphism, a $\pi\_1$ isomorphism but is never a weak
>  equivalence since by a result of Levine if it were then the knot would
>  be trivial (we should assume $n \ge 3$ here).
> 
> 
> 

[John Klein](https://mathoverflow.net/users/8032/john-klein) @ <https://mathoverflow.net/a/104534/1556>
You know already that the answer for your question is no, nevertheless, there is a generalisation which you might like:

> 
> Suppose that $X$, $Y$ are path-connected, simple CW-complexes. If $f:X\rightarrow Y$ induces isomorphism in $H\_n$ for all $n\geq 0$, then $f$ is a homotopy equivalence.
> 
> 
>