# 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?

• Your understanding of a derivative is a bit off. The derivatives only says that marginal change is 4 when $\Delta x \to 0$. If you take values of $x$ much closer to 1, you will see that $\dfrac{\Delta y}{\Delta x}$ gets much closer to 4. Commented Apr 8, 2017 at 3:49
• It is just the limit of the slopes of the secants. Commented Apr 8, 2017 at 5:04
• Going from x=1 to x=2 is not a marginal change. It's huge change. Try it for $\Delta x$ is smaller. Say .01. y then equals 1+.02+.0001+2.02 + 4=7.0401. A difference of .0401. The discepency is that even as we take a difference of .01 the function got steeper and the marginal rate of change got slightly larger. You need to wrap your mind around the idea that calculus is about INSTANTANEOUS change. Even a change of .01 is pretty large. A change of one is just plain out of the ballpark. Commented Apr 8, 2017 at 5:24

\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*}
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 $4$.
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 $\frac {[(1.01)^2 + 2 (1.01)+4]-[1^2+2+4]}{.01}=\frac {[1.0201+2.02+4]-7}{.01}=\frac {.0401}{.01}=4.01$.