# Mathematics Engineering: How do you prove the power rule?

Consider the real-valued function $f(x)=x^r$ where $r$ is a real number.

(1) For what values of $x$ and $r$, is this function differentiable?

(2) How do you prove the power rule, when $f(x)$ is differentiable?

It would be nice if the proof is accessible to a 12th grade student.

EDIT: Perhaps, I have not been very clear about my question. I am not satisfied with the "textbook proofs" I have seen. Please established the existence (of the derivative) first before proving the power rule. That is, please consider first question first. This is the reason why I call the question a mathematical engineering question. (If we can we bring high level math down to Grade 12 level, without loosing precision, then I call it good mathematical engineering.)

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Have you checked the Wikipedia proof at en.wikipedia.org/wiki/Power_rule? – Ross Millikan Oct 9 '11 at 0:24
@RossMillikan: I just looked at it. The generalization part is vague. – Sony Oct 9 '11 at 0:30
It is differentiable for: positive $x$ and all $r$; negative $x$ and $r$ rational with an odd denominator; zero $x$ with $r$ positive rational with an odd denominator; and if you define $0^0=1$ also zero $x$ with zero $r$. – Henry Oct 9 '11 at 0:33
"negative $x$ and $r$ rational with an odd denominator" - depends on the branch of the logarithm taken... – J. M. Oct 9 '11 at 0:38
How does your hypothetical Grade 12 student define derivative? Unless he has a precise definition, you're not going to be able to get any kind of "proof" across to him, because the very notion of derivative is not clear. The Inverse Function Theorem is easy enough to get across, but products and compositions are not. – Arturo Magidin Oct 9 '11 at 1:16

For positive integer $r$, we can define $x\mapsto x^r$ for all $r$, and the formula follows from the definition of derivative and the Binomial Theorem: \begin{align*} \frac{d}{dx} x^r &= \lim_{h\to 0}\frac{(x+h)^r - x^r}{h} = \lim_{h\to 0}\frac{x^r + rx^{r-1}h + h^2(\text{some factors}) - x^r}{h}\\ &= \lim_{h\to 0}\frac{rx^{r-1}h + h^2(\text{some factors})}{h} = \lim_{h\to 0}\left(rx^{r-1} + h(\text{some factors})\right)\\ &= rx^{r-1} + 0 = rx^{r-1} \end{align*} It is continuous for all $x$, since it is differentiable for all $x$.

For negative integers $r= -n$ with $n$ a positive integer, continuity at all $x\neq 0$ follows because $x^n$ is continuous at all $x$. For differentiability, we know it is differentiable at all places where it is defined, because it is the quotient of two differentiable functions (the constant function $1$, and the function $x\mapsto x^n$, which we just proved is differentiable). To find the formula, we use the Product Rule: \begin{align*} 0 &= \frac{d}{dx} 1 = \frac{d}{dx}x^{-n}x^n = (x^{-n})'x^n + (x^{-n})(x^n)'\\ &= (x^{-n})'x^n + x^{-n}(nx^{n-1}) = (x^{-n})' + nx^{-1}. \end{align*} Solving for $(x^{-n})'$ we obtain $$\frac{d}{dx} x^{-n} = -\frac{nx^{-1}}{x^n} = -nx^{-n-1} = rx^{r-1},$$ yielding the power rule. (We could also use the Quotient Rule to get differentiability of $\frac{1}{x^n}$).

For rational $r=\frac{p}{q}$ with $p$ and $q$ in reduced terms, the definition is that $x^r = (x^p)^{1/q}$. For $q$ odd, this is defined for all $x$ and is continuous (it's the composition of $x^p$ and the inverse of $x^q$, which is continuous on all $x$), but differentiable only at $x\neq 0$; differentiability of $x\mapsto x^{1/q}$ follows by the Inverse Function Theorem, and differentiability of $x^{p/q}$ now follows because it is a composition of differentiable functions, so the proof of the Chain Rule shows that it is differentiable; for $q$ even, this is defined for all $x\geq 0$ and is continuous there, but only differentiable at $x\gt 0$ (the tangent at $x=0$ is vertical). Differentiability elsewhere again follows by the Inverse Function Theorem and the Chain Rule (which in fact proves that the composition is differentiable).

For $x^{1/q}$ with $q$ and integer, we use the Chain Rule: $$1 = \frac{d}{dx}x = \frac{d}{dx}(x^{1/q})^q = q(x^{1/q})^{q-1}(x^{1/q})',$$ so solving for $(x^{1/q})'$ we get $$\frac{d}{dx}x^{1/q} = \frac{1}{q}x^{(1-q)/q} = \frac{1}{q}x^{(1/q) - 1}.$$

Then for $x^{p/q}$ we have, again by the Chain Rule, $$\frac{d}{dx}x^{p/q} = \frac{d}{dx}(x^p)^{1/q} = \frac{1}{q}(x^p)^{(1/q)-1}(x^p)' = \frac{p}{q}(x^{p})^{(1/q)-1}x^{p-1} = \frac{p}{q}x^{(p/q)-p+p-1} = \frac{p}{q}x^{(p/q)-1}.$$

For irrational exponent $r$ we only define $x^r$ for $x\gt 0$, and we define it $x^r = \exp(r\ln x)$. For $x\gt 0$ the function $x\mapsto \ln x$ is differentiable; hence $x\mapsto r\ln x$ is differentiable. The exponential is differentiable everywhere, so the commposition $x\mapsto \exp(r\ln x)$ is differentiable everywhere that it is defined, again by the Chain Rule. To get the formula, we can just use the Chain Rule again: $$\frac{d}{dx} x^r = \frac{d}{dx}\exp(r\ln x) = \exp(r\ln x)(r(\ln x)') = \frac{r}{x}\exp(r\ln x) = \frac{r}{x}(x^r) = rx^{r-1}.$$

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@ArthroMagidin: I like your proof. Let us assume that the hypothetical Grade 12 student has a very good understanding what limit of a function (at a point) is. – Sony Oct 9 '11 at 1:37
I followed your proof up to rational powers. In the real case, you use "exp function is differentiable" and "ln function is differentiable" How do you establish one or the other at Grade 12 level? – Sony Oct 9 '11 at 10:16
@Sony: That the logarithm function is differentiable follows because it's the inverse of the exponential: inverse of a differentiable function is differentiable, provided the original function does not have derivative equal to $0$ (think about the graph of the inverse: it is obtained as a reflection of the original graph about the line $x=y$. Draw a graph and draw the tangent: if the tangent has slope $m$, the reflection has slope $\frac{1}{m}$; unless the original tangent was horizontal, in which case the inverse will have a vertical tangent, and not be differentiable). – Arturo Magidin Oct 9 '11 at 19:25
@Sony: You cannot prove things without assumptions and basic definitions. There is a vast difference between "getting idea across" and "proving". One can get the idea of the Chain Rule across pretty easily, and one can even provide an argument using limits which is almost right but not really a proof, but actually proving it is more complicated. Likewise for the inverse function (which can be gotten across reasonably well using geometric arguments). But if you want proofs, then you need a lot more, and you are vacillating wildly between "proving" and "getting across". (cont) – Arturo Magidin Oct 9 '11 at 20:58
@Sony: In my opinion, you are being quite muddled and unclear about just what it is you want or what it is you want to assume or not assume, and as such your "suspicion" is likely little more than a reinforcement of that muddleness, born out of itself. – Arturo Magidin Oct 9 '11 at 20:59

Let's prove the power rule for an arbitrary real number. So $r$ can be an integer, rational, irrational. We make use of the fact that $e^{r\ln x}=x^r$.
\begin{align*} \frac{\text{d}}{\text{dx}} (x^r) & = \frac{\text{d}}{\text{dx}} \left(e^{r\ln x}\right)\\ &= e^{r\ln x} \frac{\text{d}}{\text{dx}}\left(r\ln x \right)\\ & = e^{r\ln} x \left(\frac{r}{x}\right)\\ & =rx^{r-1}. \end{align*} where we have used the fact that $\frac{d}{dx}\ln x =\frac{1}{x}$. ofcourse this works if $x>0$.

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You use $\ln x$ in $e^{r\ln X}$. How do you know $x>0$? By definition, (Grade 12 level) a real-valued function is differentiable at $x_0$ if the derivative exists at $x_0$. Therefore, existence of the derivative must be established first. Hence, the question 1. – Sony Oct 9 '11 at 0:40
@Sony: The Chain Rule is not just a formula: the proof of the Chain Rule actually shows that if $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$, then $f\circ g$ is differentiable at $x$ and $(f(g(x))' = f'(g(x))g'(x)$. So to know that $e^{r\ln x}$ is differentiable wherever it is defined, you just need to know that the exponential function is differentiable everywhere, and that the logarithm is differentiable for all $x\gt 0$ (because it is the inverse of a function that is differentiable everywhere and whose derivative is never $0$). – Arturo Magidin Oct 9 '11 at 1:13

That it is true of nonnegative integer values of $r$ can be proved by any of several methods.

• One of those uses mathematical induction and the power rule: $$\frac{d}{dx} x^{r+1} = \frac{d}{dx} (x^r\cdot x) = x\frac{d}{dx} x^r + x^r \frac{d}{dx} x = x(rx^{r-1}) + x^r\cdot 1 = (r+1)x^r.$$
• Another uses the definition of derivative directly: $$\frac{d}{dx} x^r = \lim_{w\to x} \frac{w^r-x^r}{w-x} = \lim_{w\to x}\frac{(w-x)(w^{r-1}+w^{r-2}x + w^{r-3}x^2 + \cdots + x^{r-1})}{w-x}$$ $$= \lim_{w\to x} (w^{r-1}+w^{r-2}x + w^{r-3}x^2 + \cdots + x^{r-1})$$ $$= \underbrace{x^{r-1}+\cdots+x^{r-1}}_{r\text{ terms}}.$$

• Another uses the binomial theorem and the definition of the derivative. There's a bit of overkill in that one, since the binomial expansion gives you all $r+1$ coefficients where you need only the first two.

For positive $x$ and real $r$ (non-integer and not necessarily positive), you can use logarithmic differentation, provided you know the chain rule and how to differentiate the natural exponential and logarithmic functions: $$\frac{d}{dx} x^r = \frac{d}{dx} e^{r\log_e x} = e^{r\log_e x} \frac{r}{x} = x^r\cdot \frac{r}{x}.$$

For negative integer values of $r$, you can use the reciprocal rule: $$\frac{d}{dx}x^r = \frac{d}{dx} \frac{1}{x^{-r}} = \frac{-\frac{d}{dx}x^{-r}}{x^{-2r}} = \frac{rx^{-r-1}}{x^{-2r}} = rx^{r-1}.$$

It's also true of arbitrary real values of $r$ even when $x$ is not positive. I'll think about the quickest way to prove that. Except that it is not true when $x=0$ and $r<0$, since that involves dividing by $0$.

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Edited to include absolute values.

Write $y=x^r$. Then $\log|y|=r\log|x|$. Then implicit differentiation gives $\displaystyle\frac{y'}{y}=r\frac{1}{x}$, or $\displaystyle y'=r\frac{y}{x} = r\frac{x^r}{x} = rx^{r-1}$. This works for all real numbers $r$, and so $f(x)=x^r$ is differentiable for all $r$ and for all $x$ for which $x^r$ is defined, unless $r-1<0$ in which case $f$ is not differentiable at $0$.

Of course, this assumes you already know that $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x} \log(x) = \frac{1}{x}$.

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To apply log (I presume log base 10) $y$ has to be positive. How do you know that $y$ is positive? – Sony Oct 9 '11 at 0:32
@Sony: I mean log base e. You are right to be concerned about $y$ being positive. Maybe you can adapt my method to take this into account. – wckronholm Oct 9 '11 at 0:35
@Sony: Arbitrary exponentials are only defined for positive bases; so $x\mapsto x^r$ for arbitrary $r$ is only defined if $x\gt 0$. For rational exponents which, in reduced form have an odd denominator, you can establish the Power Rule by considering $(x^{p/q})^q$, using the Chain Rule, and the Power Rule for positive integral exponents. – Arturo Magidin Oct 9 '11 at 0:36
@Sony: Don't assume base 10. "$\log x$" in this kind of context normally means $\ln x$ i.e. $\log_e x$. – Michael Hardy Oct 9 '11 at 0:45
@Sony: To echo Michael Hardy, $\log$ usually denotes whichever logarithm is more commonly used in the particular field. In mathematics (outside Calculus classes, at any rate) it almost invariably means the natural logarithm. In other areas it may mean something else (e.g., in Computer Science, particularly in analysis of algorithms, it often means logarithm base 2). – Arturo Magidin Oct 9 '11 at 2:03