Clarification on Rules Differentiation and First Principles Derivatives My grade 11 class has just started differential calculus, the one area seemingly glazed over in our book. We have covered some simple rules of differentiation, like 
f(x) = n x^(n-1), and have applied these with subsequent rules like the constant rule, constant multiple rule, and sum or difference rule, child's play to you, I am sure. Anyway, we also covered the derivative, defined as:

The limit as h approaches zero = (f(x+h)-f(x))/h...

We were told by our teacher that tossing any equation into that formula will result in the spitting out of our derivative function...
We were also told that using the rules of differentiation will break things down into a derivative function; therefore, should using these to methods of derivation not produce the exact same results?

Apparently not:
Take x^2 + 1

From the rules of differentiation we get 2 x right?

But from the above formula we get 2 x+h

Now, clearly just by the presence of the h, which cannot magically come into the formula via the rules of differentiation, the rules and the formula do/produce different things... 

Does one produce the function of the tangent line?
Does one produce the function of the gradients of possible tangents?

Which one of the above is the derivative? The specific tangent? Or the function to produce the slopes of said tangents? Which is yielded by which?

Sorry, I understand how stupid and trivial this all seems but I want to set a proper calculus foundation for the future and cannot risk misunderstanding something, and if I do, then I need rectification!

Thank you to any and all who may answer I will certainly appreciate the help.
The problem is with your calculation of the limit:

$$\begin{align*}
\lim_{h\to 0}\frac{f(x+h)-f(x)}h&amp;=\lim_{h\to 0}\frac{\big((x+h)^2+1\big)-\big(x^2+1\big)}h\\\\
&amp;=\lim_{h\to 0}\frac{(x^2+2xh+h^2+1)-(x^2+1)}h\\\\
&amp;=\lim_{h\to 0}\frac{2xh+h^2}h\\\\
&amp;=\lim_{h\to 0}(2x+h)\\\\
&amp;=2x\;,
\end{align*}$$

not $2x+h$. In fact the formula comes from applying the definition to the general case, so it cannot give a different result.
The formula does not give 2x + h. 
It is the limit as h goes to 0 of (f(x+h))-f(x))/h yes....
so f(x) = x^2 + 1 and the limit is

limit as h goes to 0 of ((x+h)^2 +1) - (x^2 + 1))/h

If you actually do that limit you will get 2x...