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I've been using it for a while but still don't really understand why it works.

For integer exponents greater or equal to 2, its easy to intuitively understand it using the geometric interpretation of the product rule: http://web.mit.edu/wwmath/calculus/differentiation/products.html

But could someone give me an intuition about why it works (not a proof)? An intuition that extends to non-integer and negative powers?

Shaun
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dfg
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    For exponents larger than two, you just iterate: $x^3=x\cdot x\cdot x$! – Mariano Suárez-Álvarez Jan 04 '14 at 19:19
  • For negative exponents, there is no choice, really: if you apply the product rule to $x\cdot x^{-1}=1$ you find that there is no choice for the derivative of $x^{-1}$. – Mariano Suárez-Álvarez Jan 04 '14 at 19:22
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    @MarianoSuárez-Alvarez I don't have a problem understanding exponents larger than 2, I explicitly said that. And I don't understand your second comment. – dfg Jan 04 '14 at 19:23
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    I would argue that this question is sort of backwards: intuition for the power rule is from the integer case (the product rule for more than two factors), and that it turns out to hold for other cases like negative integer exponents or non-integral exponents is not intuitive and is precisely why a proof is needed. Your question is somewhat analogous to asking for intuition behind rules of exponents like $(a^x)^y = a^{xy}$; it is precisely the integer case where the rules are intuitive, and that it holds more broadly is the reason for a careful proof (esp. if the exponents are irrational). – KCd Jan 04 '14 at 19:25
  • I think your question is a good illustration of von Neumann's remark that in math we don't understand things but just get used to them. Be grateful that at least the integer exponent case has an intuitive explanation. (Note that positive integer exponents are precisely when exponents have a concrete meaning involving repeated multiplication; other exponents do not, so intuition truly lies in the case of positive integer exponents.) – KCd Jan 04 '14 at 19:27
  • Indeed, what is your intuition about $x^r$ for non-integer and negative $r$ in the first place? –  Jan 04 '14 at 19:32

2 Answers2

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Do you genuinely understand how or why $x'=1$? If so, then, through an effort of imagination, try to picture this x, which you regard as mono-dimensional, as being in reality a multi-dimensional entity itself, and write it as a product of n equal quantities, namely $x=y^n\iff y=\sqrt[n]x$ . Now, apply the integer-dimensional product rule, which you genuinely seem to grasp, to x, but this time viewing it as a composite n-dimensional object, and y as the new unit. What would we get ? $$x'=(y^n)'=n\cdot y^{n-1}\cdot y'\iff y'=\dfrac{x'}{n\cdot y^{n-1}}\iff\Big(\sqrt[n]x\Big)'=\dfrac1{n\cdot\Big(\sqrt[n]x\Big)^{n-1}}$$ For instance, we could represent various three-dimensional volumes using mono-dimensional line segments, e.g., $1$ cm $\equiv$ $1$ m$^3$, or $1$ inch $\equiv$ $1$ gallon, etc. So just because we use a mono-dimensional representation for something, that does not mean that that certain something is uncompounded.

Lucian
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I would not call the product rule thing an intuition, it is more of a proof.

If you want an intuitive explanation for the integer case, you can for example consider $r^n$ as the volume of an $n$-ball of radius $r$ (up to a constant). As you change the radius, the volume changes and the derivative of $r^n$ tells you how fast it changes. Now it should be obvious that the larger the surface of tthe sphere, the faster the volume will grow, and the surface is $r^{n-1}$ (again, up to a constant factor)

Mariano Suárez-Álvarez
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