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: 
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).
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.
> 
> 
>