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>
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.
> 
> 
>