Why I can't understand the derivative (with example). Can someone help me understand? The derivative of a function is said to be $\lim_{\Delta{x}\to 0} \frac{f(\Delta{x}+x)-f(x)}{\Delta{x}}$. This I believe I understand. But, I always have trouble believing something that is so theoretical without walking through a concrete example. So, I took the derivative of the function $f(x)=x^2+2x+4$ and I, of course, got $f'(x)=2x+2$. So, this says that for a marginal change in $x$, there will be a change in $y$ of $2(1)+2=4$.
I evaluate the original function at $x=1$ and I find that $y=7$. Then I evaluate the original function at $x=2$ and I find that $y=12$. In this case $\frac{\Delta{y}}{\Delta{x}}=5$, not the $4$ predicted by our derivative function. Where is the mistake in my thinking? Can someone provide me with a concrete example of the derivative that will show me that it does work?
 A: $$
\begin{align*}
f'(1) &= \lim_{\Delta x\to 0}\frac{f(\Delta x + 1) - f(1)}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{(\Delta x + 1)^{2} + 2(\Delta x + 1) + 4 - 7}{\Delta x}\\
&=\lim_{\Delta x\to 0}\frac{(\Delta x)^{2} + 2\Delta x + 1 + 2\Delta x + 2 + 4 - 7}{\Delta x}\\
&=\lim_{\Delta x \to 0}\frac{(\Delta x)^{2} + 4\Delta x}{\Delta x}\\
&=\lim_{\Delta x\to 0}\Delta x + 4\\
&=4. 
\end{align*}
$$
A: You did the derivative right.  But graph the function (it's a parabola).  Now draw a tangent line at $x=1;y=7$.  Notice that at the very instant the tangent line touches the curve, both the function and the line are increasing, at that moment, at a rate (slope) of $4$.
But the parabola is not a straight line.  Even though it is increasing at a rate of 4 at that point, it is constantly "accelerating" and "pulls away from the tangent line and increases even faster.
That's what marginal change means; the rate of change at a moment.  The difference between $f (1) $ and $f (2) $ is $5$, which is only $25\% $ off what we expected. That's actually pretty good as going from $x=1$ to $x=2$ is ... a pretty large margin.
Try a smaller margin.  Say from $x=1$ to $x=1.01$.  Then the difference is $\frac {[(1.01)^2 + 2 (1.01)+4]-[1^2+2+4]}{.01}=\frac {[1.0201+2.02+4]-7}{.01}=\frac {.0401}{.01}=4.01$.
Even this small value isn't perfect (it's off by .25%).  Because the function is constantly changing and its rate change is constantly changing.
