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: 
The answer is no. An easy example is provided in Allen Hatcher's Algebraic Topology, Example 4.35.

<http://www.math.cornell.edu/~hatcher/AT/ATch4.pdf>

There, a CW-complex $X$ is formed by attaching an appropriate $(n+1)$-cell to the wedge $S^1 \vee S^n$ of a circle and an $n$-sphere. The inclusion of the $1$-skeleton $S^1 \to X$ induces an isomorphism on integral homology and on homotopy groups $\pi\_i$ for $i < n$ but not on $\pi\_n$.

That example is also worked out as Problem 1 here:

<http://www.home.uni-osnabrueck.de/mfrankland/Math527/Math527_HW10_sol.pdf>
> 
> ...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>