# Why does the power rule work?

If $$f(x)=x^u$$ then the derivative function will always be $$f'(x)=u*x^{u-1}$$ I've been trying to figure out why that makes sense and I can't quite get there.

I know it can be proven with limits, but I'm looking for something more basic, something I can picture in my head.

The derivative should be the slope of the function. If $$f(x)=x^3=x^2*x$$ then the slope should be $x^2$. But it isn't. The power rule says it's $3x^2$.

I understand that it has to do with having variables where in a more simple equation there would be a constant. I'm trying to understand how that exactly translates into the power rule.

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What if you write $x^3 = x*x*x$ and use the product rule. What do you get? What do you get with the product rule for $x^3 = x^2*x$? Do they all agree? – Amzoti Mar 17 '13 at 22:34
As per Amzoti's suggestion, if you want intuition think of the case where $u$ is an integer and use the product rule. So in the case of $x^3$, use the product rule and take the derivative of $x\cdot x\cdot x$. This is symmetric, so you will get $3\cdot x^2$. – Lepidopterist Mar 17 '13 at 22:37
Since the definition of $f'(x)$ involves limits, you can't hope for "something more basic" that proves things about derivatives. – Henning Makholm Mar 17 '13 at 22:39
"If $f(x)=x^3=x^2\cdot x$ then the slope should be $x^2$" - Why? (You are perhaps confusing it with rules that work to find slopes of lines of the form $y=ax+b$; you simply take the $a$ in front of the $x$ then.) – anon Mar 17 '13 at 22:52
@Amzoti, so that is just restating the question as.... why does the product rule work? – Pacerier Jan 6 '14 at 13:44

First let's try to understand why the derivative of the function $f$ given by $f(x) = x^2$ is equal to $2x$ and not to $x$. (The product rule and the power rule are both generalizations of this.)

Imagine that you have a square whose sides have length $x$. Now imagine what happens to its area if we increase the length of each side by a small amount $\Delta x$. We can do this by adding three regions to the picture: two thin rectangles measuring $x$ by $\Delta x$ (say one on the right of the square and another on the top) and one small square measuring $\Delta x$ by $\Delta x$ (say added in the top right corner.) So the change in the area $x^2$ is equal to $2x \cdot \Delta x + (\Delta x)^2$. If we divide this by $\Delta x$ and take the limit as $\Delta x$ approaches zero, we get $2x$.

So geometrically what is happening is that the small square in the corner is too small to matter, but you have to count both rectangles. If you only count one of them, you will get the answer $x$; however, this only tells you what happens when you lengthen, say, the horizontal sides and not the vertical sides of your square to get a rectangle. This is a different problem than the one under consideration, which asks (after we put it in geometrical terms) how the area varies as we lengthen all the sides.

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That's a really cool way to explain it. Like the OP, I too used to understand the math of the limits, but couldn't picture it. My high school calculus teacher explained it the same way as @Trevor, and it really helped me get my head around the concept visually. – David John Welsh Mar 18 '13 at 3:21

For positive integers $n$, we can use the Binomial Theorem. Let $f(x)=x^n$. We want to find the slope of the tangent line to $y=f(x)$ at $x=a$. So take a very small $h$, and calculate $$\frac{(a+h)^n-a^n}{h}.$$ By the Binomial Theorem, $(a+h)^n=a^n+na^{n-1}h +\binom{n}{2}a^{n-2}h^2+\cdots$.

Since $h$ is tiny, $h^2$, $h^3$, and so on are negligible compared to $h$. Thus $$\frac{(a+h)^n-a^n}{h}\approx na^{n-1}.$$

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Ok, this explanation is way too mathy – Pacerier Jan 6 '14 at 13:58
True, it is rather symbol-laden. – André Nicolas Jan 6 '14 at 15:55

If you know the product rule, you can derive this for when $u$ is a positive integer, which should give you a basic intuitive understanding.

Suppose, as an example, that $f(x) = x^2$. Equivalently, $f(x) = x*x$. By using the product rule, we have $$f(x) = (1)(x) + (x)(1)$$ $$= 2x$$

More generally, suppose that $f(x) = x^u$. Suppose for now that $u$ is a positive integer, which allows us to expand like this: $$f(x) = \underbrace{(x)(x)...(x)(x)}_{u\text{ terms}}$$ Using the product rule again, we can say $$f(x) = \underbrace{\underbrace{(1)(x)...(x)(x)}_{u \text{ terms}} + (x)(1)...(x)(x) + ... + (x)(x)...(1)(x) + (x)(x)...(x)(1)}_{u\text{ terms}}$$ which simplifies to $$f(x)=ux^{u-1}$$

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+1 I had never seen this before. Thanks for posting it! – templatetypedef Mar 18 '13 at 5:16

This may be too advanced for you right now, but knowing about the derivative of the log function can be very helpful.

The basic idea is that $(\ln(x))' = 1/x$, where $\ln$ is the natural log.

Applying the chain rule, $(\ln(f(x))' = f'(x)/f(x)$.

For this case, set $f(x) = x^n$. Then $\ln(f(x)) = \ln(x^n) = n \ln(x)$. Taking derivatives on both sides, $(\ln(x^n))' = f'(x)/f(x) = f'(x)/x^n$ and $(\ln(x^n))' = (n \ln(x))' = n/x$, so $f'(x)/x^n = n/x$ or $f'(x) = n x^{n-1}$.

More generally, if $f(x) = \prod a_i(x)/\prod b_i(x)$, then $\ln(f(x)) = \sum \ln(a_i(x))-\sum \ln(b_i(x))$ so

\begin{align} f'(x)/f(x) &= (\ln(f(x))'\\ &= \sum (\ln(a_i(x))'-\sum (\ln(b_i(x))'\\ &=\sum (a_i(x))'/a_i(x)-\sum (b_i(x))'/b_i(x)\\ \end{align},

so $f'(x) = f(x)\left(\sum (a_i(x))'/a_i(x)-\sum (b_i(x))'/b_i(x)\right)$.

Note that this technique, called logarithmic differentiation, generalizes both the product and quotient rules for derivatives.

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The exponent "comes out in front" as a coefficient because higher polynomial powers represent a greater rate of change. That is to say, $x^5$ grows faster than $x^4$ and so on.

Let's just look at $y = x$ though. If you integrate $x$ from $0$ to $1$, what do you get? Obviously $1/2$ because the $y = x$ line cuts the $((0,0) (1,1))$ square in half. So the integral must be $\frac{x^2}{2}$. If you want the integral to be $x^2$, you have to integrate $2x$, not $x$.

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