I'm currently taking maths at A Level and I have found it strange that it is not explained why the differential of $x^n$ is $nx^{n-1}$, for example. I can see that it works through observation and first principle, but how can it be derived? And what about for an unknown function, not based on trigonometry, $e^x$ or polynomials? Is there some sort of intuition or derivation that can lead to a general answer, other than making observations? Another thing that I do not quite understand is why higher derivatives of some function is denoted by $\frac{d^ny}{dx^n}$.

Sorry if this sounds really basic but I would appreciate any explanations/links to good resources (I've had a look online and couldn't find too much other than some nice explanations special cases etc). Thanks :)

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    $\begingroup$ Do you know how the derivate of a function is defined? Have you tried to calculate $(x^n)'$ and where did you get stuck? $\endgroup$
    – user339727
    Feb 20, 2017 at 23:34
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    $\begingroup$ Your teacher is supposed to show all these geometric interpretations and motivations on class. $\endgroup$
    – Filburt
    Feb 20, 2017 at 23:36
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    $\begingroup$ The other is just notation. $\endgroup$
    – copper.hat
    Feb 20, 2017 at 23:40
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    $\begingroup$ Try plugging $x^2$ into the definition of the derivative. Simplify it out as much as you can, it's pretty simple to get the derivative through the definition. Try it again with $x^3$ and you should start to see the repeating pattern. Of course, know why the definition works, which you can find on Google pretty easy. $\endgroup$
    – Kaynex
    Feb 20, 2017 at 23:47
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    $\begingroup$ I recommend watching Herbert Gross's series "Calculus Revisited" from MIT's open courseware available on youtube. He specifically goes over the proof of this result in this video at 4min30sec in, though it may help you to go over a few of the earlier videos as well in the series. $\endgroup$
    – JMoravitz
    Feb 20, 2017 at 23:48

5 Answers 5


Firstly I'll go over the definition of the derivative, and why the derivative of $x^n$ is $n x^{n-1}$. Then I'll try to explain what the derivative is trying to capture, and why the definition makes sense.

For a function $f(x)$, its derivative is defined as being $$f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$ If you haven't met limits before (as I suspect an A-level student may not have), the idea is that the limit tries to capture what a function looks like near a point, in particular here what $\frac{f(x+h)-f(x)}{h}$ is like near $0$, even though at $0$ it's not defined.

For $f(x) = x^n$, we have

$$f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$ $$f'(x) = \lim_{h\to0} \frac{(x+h)^n-x^n}{h}$$ $$f'(x) = \lim_{h\to0} \frac{x^n + n h x^{n-1} + \frac{n(n-1)}{2} h^2 x^{n-1} + ... + h^n -x^n}{h}$$ $$f'(x) = \lim_{h\to0} (n x^{n-1} + \frac{n(n-1)}{2} h x^{n-1} + ... + h^{n-1})$$ Now you can see that each term except the first contains an $h$, so for really small $h$ they're $0$, but the first term is independent of $h$. This gives that $$f'(x) = n x^{n-1}$$.

Now let's talk about the way you should think about the derivative of a function. The derivative of a function tries to say how much the value of the function changes when you change the input by a tiny amount - and you'll consider the ratio of the change, in the same way as you'd consider a percentage change. In particular, you might want to know how much $x^2$ changes when you move from $x=2$ to $x=2.01$, for instance. Another way of thinking about this is the tangent line to the function at a point, so we might want to know what the slope of the tangent to $y=x^2$ is when $x=2$, and it doesn't take much thought to see why these are both the same idea. To look at the tangent line, it makes sense to instead consider the point $(2, 2^2)$ and another point really close to it, say $(2.01, 2.01^2)$, or $(2.0001, 2.0001^2)$, or $(2+h, (2+h)^2)$, where $h$ is really close to $0$ (possibly negative), and look at what the slope of the line connecting these two is for tiny $h$. This starts to look like the definition of the derivative, because the slope of the line connecting $(2+h, (2+h)^2)$ and $(2, 2^2)$ is $$\frac{(2+h)^2 - 2^2}{2+h-2} = \frac{(2+h)^2 - 2^2}{h}$$

and just like above, you can compute the limit of this expression to be $4$.

This easily generalises to any point $x$ (instead of $2$), and to any (differentiable) function (instead of $x^2$), to give a limit as defined above.

It's important to realise that the derivative is another function in itself, and the meaning of $f'(x) = 2x$ is that at the point $t$, the slope of the tangent line to $y=f(x)$ at $t$ is $2t$.

The derivative of any differentiable function could, in theory, be computed directly from the limit, but that's often tedious. So, we use tricks like linearity (sometimes called the sum rule), the product and quotient rules, and the chain rule. It might be instructive to try to prove linearity and the product rule yourself, directly from the definition of the limit.

  • $\begingroup$ Thanks for your explanation, we have used this method of first principle before (I take further maths too) but it did not give a general result, only an output of say 2x. $\endgroup$ Feb 20, 2017 at 23:48
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    $\begingroup$ Is it actually possible that derivatives are taught before limits? $\endgroup$
    – ChatterOne
    Feb 21, 2017 at 7:49
  • $\begingroup$ @ChatterOne: my first week of university, in a context where no calculus was taught in highschool. Class is Physics (Mechanics). First lecture: vectors. Second lecture: derivatives. Third lecture: integrals. $\endgroup$ Feb 21, 2017 at 8:28
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    $\begingroup$ @ChatterOne In the UK A-level syllabus, limits aren't even specifically included - a good teacher would go over them but the course itself just goes straight into derivatives, a lot of students aren't familiar with the definition of the derivative at all, only how to compute it. $\endgroup$
    – B. Mehta
    Feb 21, 2017 at 11:45
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    $\begingroup$ @ChatterOne The US system teaches limits before derivatives. But in international terms I don't think this is usual. And while pedagogically it makes sense conceptually to teach limits first, most applied users of mathematics (in science/engineering/economics) only need to understand how to calculate a derivative and its meaning as a rate of change - being able to formally prove e.g. that the derivative of $e^{kx}$ is $k e^{kx}$ may not be so practically relevant. Of course the disadvantage of teaching it all in terms of practical skills, is that "what's going on behind it all?" goes unseen $\endgroup$
    – Silverfish
    Feb 21, 2017 at 13:16

Geometrically, the derivative is the slope of the best linear approximation of a function $f$ at a particular point $x_0$.

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Recall that a linear function over the real numbers has the form $L(x) = ax+b$ for some constants $a$ and $b$. In this equation, $a$ is the slope of the line and $b$ is the $y$-intercept. If we were to choose the best line to approximate a function $f$ near the point $x_0$, then we'd probably want for them to both the pass through the point $(x_0,f(x_0))$. One way to ensure this is to write $L$ as $L(x) = a(x-x_0) + f(x_0)$. Then when $x=x_0$, we'd get $L(x_0) = f(x_0)$. Note that $L$ is still a linear function because $a(x-x_0) + f(x_0) = ax + (-ax_0+f(x_0))$ where $-ax_0 + f(x_0)$ is a constant.

Also if $L$ really is the best linear approximation to $f$ at $x_0$, then $L$ should probably have the same slope as $f$ at $x_0$. Here's the good thing, though: while the slope of $f$ may be constantly changing, $L$ is a line and so it has the same slope everywhere -- that slope being $a$. We know that $f'(x_0)$ should be the slope of $f$ at $x_0$ so we just plug that in for $a$. So the best linear approximation for $f$ near $x_0$ should be $$\bbox[5px,border:2px solid blue] {L(x) = f'(x_0)(x-x_0)+f(x_0)}$$

Now, here's how we use this. Imagine we could rewrite $f$ everywhere in some small interval around $x_0$ as $$f(x) = L(x) + r(x-x_0)$$ where $r$ is an error function that gives the error between $f$ and $L$ at each point. Note that at $x_0$, $r$ should just be $0$ because $L$ and $f$ pass through the same point. As it turns out, we can say something even stronger about $r$. The error function $r$ should go to $0$ faster than $x-x_0$ at $x_0$. I.e. $$\lim_{x\to x_0} \frac{r(x-x_0)}{x-x_0} = 0\tag{*}$$ In fact this can be used to uniquely define the derivative in this way:

If, for any $x$ near $x_0$, we can write the function $f$ as $f(x) = f(x_0) + B(x-x_0) + r(x-x_0)$ where $r$ satisfies $(*)$, then $B$ must equal $f'(x_0)$.

I provide a proof in this answer (but depending on your level of mathematical maturity you might just want to take it as a fact for now).

With all that out of the way, we can use this to figure out that the derivative of $f(x)=x^n$ at $x_0$ is $n{x_0}^{n-1}$. We just have to see if we can somehow write $f(x)$ in the form ${x_0}^n + B(x-x_0) + r(x-x_0)$ where $B$ is a constant and $r$ is some function that satisfies the limit condition in $(*)$. Then $B$ will be the derivative at $x_0$. To do so we'll use the binomial theorem and, just to make it a little easier, we'll rewrite $x$ as $x_0 + \Delta x$. Then we see that:

$$\begin{align}f(x) = x^n &= (x_0 + \Delta x)^n = {x_0}^n + n{x_0}^{n-1}\Delta x + \cdots + nx_0\Delta x^{n-1} + \Delta x^n \\ &= {x_0}^n + n{x_0}^{n-1}\left[(x_0 +\Delta x) - x_0\right] + \cdots + nx_0\left[(x_0 +\Delta x) - x_0\right]^{n-1} + \left[(x_0 +\Delta x) - x_0\right]^n \\ &= {x_0}^n + n{x_0}^{n-1}(x - x_0) + \cdots + nx_0(x - x_0)^{n-1} + (x - x_0)^n\end{align}$$

Notice that every term after $n{x_0}^{n-1}(x - x_0)$ has at least a power of $(x - x_0)^2$ in it. So these go to $0$ faster than $x - x_0$ (in the sense of $(*)$). So this is really saying $$f(x) = x^n = {x_0}^n + n{x_0}^{n-1}(x - x_0) + r(x-x_0)$$ where $\lim_{x\to x_0}\frac{r(x-x_0)}{x-x_0} = 0$. Thus $$\bbox[5px,border:2px solid red] {f'(x_0) = n{x_0}^{n-1}}$$


definition of a derivative

Plug in x^n to the formula, you get:

$$\frac{dy}{dx}x^{n}=\lim_{h\to 0}\frac{(x+h)^n-x^n}{h}$$

$$=\lim_{h\to 0}\frac{x^{n}+\binom{n}{1}hx^{n-1}+\binom{n}{2}h^{2}x^{n-2}+...+\binom{n}{n-1}h^{n-1}x^1+h^n-x^n}{h}$$ This simplifies to $$=\lim_{h\to 0}{\binom{n}{1}x^{n-1}+\binom{n}{2}hx^{n-2}+...+\binom{n}{n-1}h^{n-2}x^1+h^{n-1}}$$ Now you can plug in 0 for h and every term containing h dissapears leaving you with $$\frac{dy}{dx}x^{n}=\binom{n}{1}x^{n-1}$$ Since $$\binom{n}{1}=n$$ $$\frac{dy}{dx}x^{n}=nx^{n-1}$$


Let me try to answer you some of your questions.

  1. The derivate of a function is the slope of the tangent line to the graph of the function. Have a look here.
  2. To compute the derivate of $x\mapsto x^n$ you need to know that \begin{align*} \frac{x^n-x_0^n}{x-x_0}=\sum_{j=0}^{n-1}x^jx_0^{n-1-j} \end{align*} for $x\neq x_0$. (You can check the formula by multiplying both sides with $x-x_0$ and computing the right hand side.) Therefore you get \begin{align*} \lim\limits_{x\to x_0}\frac{x^n-x_0^n}{x-x_0} = \lim\limits_{x\to x_0}\sum_{j=0}^{n-1}x^jx_0^{n-1-j} = \sum_{j=0}^{n-1}x_0^jx_0^{n-1-j}=nx_0^{n-1}, \end{align*} which confirms the well known rule $(x^n)' =nx^{n-1}$.
  3. Since differentiating is linear, i.e. $(f+g)' =f' +g'$ and $(cf)'=cf'$ you can apply part 2 on every polynomial of the type $p(x)= \sum_{j=0}^n a_j x^j$ to get $$ p'(x) = \sum_{j=1}^n a_j j x^{j-1} .$$
  4. The definition of higher derivates is just a definition. If you like you can write $f^{(n)}$ for the $n$-th derivate aswell. An other common notation is $D^n f$.
  5. And what about for an unknown function? Lets call the function $f$. First we have to check if the function is differentiable (at a point $x_0$) or not, i.e. you have to check if the limit $$ \lim\limits_{x\to x_0}\frac{f(x) -f(x_0)}{x-x_0}$$ exists or not. If the limit exists we call him (by definition) $f'(x_0)$. A typical example of a function which is not in every point differentiable is $x\mapsto |x|$ since one can show that the limit $$ \lim\limits_{x\to 0}\frac{|x|}{x}$$ does not exist.

Concerning your question of why higher derivatives are denoted by $$\frac{d^ny}{dx^n}$$ There is a reason for this notation. Now $$\begin{align}\frac{d^2y}{dx^2} &= \lim_{h \to 0} \frac{y'(x + h) - y'(x)}{h}\\&=\lim_{h \to 0}\frac{\lim_{h_1 \to 0} \frac{y(x + h + h_1) - y(x + h)}{h_1} - \lim_{h_2 \to 0} \frac{y(x + h_2) - y(x)}{h_2}}h\\&=\lim_{h \to 0}\lim_{h_1 \to 0}\lim_{h_2 \to 0} \frac{ \frac{y(x + h + h_1) - y(x + h)}{h_1} - \frac{y(x + h_2) - y(x)}{h_2}}h\end{align}$$

Now assuming both the iterated and the combined limit exist, we can set the three values $h, h_1, h_2$ to be equal without changing the value. So $$\begin{align}\frac{d^2y}{dx^2} &=\lim_{h \to 0} \frac{ \frac{y(x + 2h) - y(x + h)}{h} - \frac{y(x + h) - y(x)}{h}}h\\&=\lim_{h \to 0} \frac{ y(x + 2h) - 2y(x + h) + y(x)}{h^2}\end{align}$$

The expression on the top is the "2nd difference". It is what you get from taking a difference of a difference, so it is the taking of differences that is squared in the short-hand notation: $d^2y$. But the denominator is just $h^2$, which is the square of a single difference in $x$. Thus $dx^2$ (which is intended to mean $(dx)^2$, not $d(x^2)$).


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