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>
There are two finite 2-dimensional complexes $A, B$ which are not homotopy-equivalent but are homology-equivalent, i.e., there exists a continuous map
$$
f: A\to B
$$
inducing isomorphism of fundamental groups and homology groups. See the last paragraph (page 522) of [this paper](http://www.ams.org/journals/proc/1986-098-03/S0002-9939-1986-0857954-1/S0002-9939-1986-0857954-1.pdf), the actual example is due to Dunwoody.

Edit 1: The homology equivalence part follows from Dyer's paper, see the link above (read the very last paragraph of his paper). Dyer's results are by no means obvious: He proves that equality of Euler characteristics plus one extra condition ($m=1$, whatever $m$ is) imply homology equivalence of 2-d complexes with isomorphic fundamental groups. Then Dyer verifies his conditions in the case of Dunwoody's example (the only nontrivial condition is $m=1$, since equality of Euler characteristics is clear). 

Edit 2: The correct version of the "homological Whitehead's theorem" is indeed requires a cohomology isomorphisms with sheaf coefficients, see the discussion and reference [here](https://math.stackexchange.com/questions/717018/cohomological-whitehead-theorem?rq=1).