# 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.

• In short, you are right, you will obtain $2x + h$. However, the point of the definition is that you take $h \to 0$---that is, you consider the limit of the function as $h$ goes to 0. In that way, the $h$ that appeared in the derivation process disappears in the end result. There are some technicalities to deal with, but that is roughly what is going on. To add some intuition to this, you can think of the $h$ as the deviation from being a tangent. That is to say, $h$ is the distance to the auxilary point in which you take the secant. The point is to take $h \to 0$ to get the tangent. Nov 24, 2013 at 22:34
• math.stackexchange.com/questions/1205760/…
– user117644
Dec 31, 2015 at 21:50

The problem is with your calculation of the limit:

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

• In this equation is h the value that x approaches? Nov 24, 2013 at 22:39
• @William: No, $x$ is the fixed value, and $x+h\to x$. Nov 24, 2013 at 22:43
• What is h then? I asked my teacher and she said it is something that just 'is' and doesn't matter which helped me not at all. Nov 24, 2013 at 22:47
• @William: $\frac{f(x+h)-f(x)}h$ is the slope of the secant line between the point $\langle x,f(x)\rangle$ at which you want the tangent and a nearby point $\langle x+h,f(x+h)\rangle$. The idea is to let the separation $h$ between $x$ and $x+h$ get smaller and smaller; if the curve actually has a tangent line at $\langle x,f(x)\rangle$, these secant lines will get closer and closer to the tangent line, and their slopes will get closer and closer to the slope of the tangent line. Nov 24, 2013 at 22:50
• That last comment aided immeasurably in my understanding of all this, much more than hours of my teacher could ever, thank you so much. Nov 24, 2013 at 22:57

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...

• Does solving using only the Power Rule and associated rules assume an h-value of zero? Nov 24, 2013 at 22:41
• You clearly don't understand the definition of the derivative. There are plenty of resources online for learning basic calculus..I recommend... tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx Nov 24, 2013 at 22:43
• Thanks. But if solving this problem using the rules of differentiation and solving using this first-principles formula yield the same result doesn't that make the rules of differentiation applicable with an assumed h->0? As if h->1 in the formula the result would be different. I am not trying to be cheeky I just wish to learn from the wiser. Nov 24, 2013 at 22:49
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