What is the intuitive reasoning behind taking the exponent from a variable and moving it to the front when taking a derivative?

Question

For example in the function $x^2$ if you want to find the slope at a point you would take the derivative of the function at that point which would be the slope of the tangent line at that point. I understand intuitively understand that the tangent line is increasing in a linear manner so to find the slope it makes sense to take the $2$ away from $x^2$ and make it linear but what is the intuition of bringing that $2$ to the front as $2x$? Sorry if this is basic I always just applied the rules and never really thought about this until now !

Answer 1

Intuitively, the derivative of $x^n$ should be approximately $(x+1)^n-x^n$. When you multiply out $(x+1)^n$, you get one $x^n$ term (which cancels the $-x^n$) and $n$ $x^{n-1}$ terms. This is related to the binomial theorem.

The idea of estimating the slope of the tangent line at a point by taking difference quotients near the point leads to the formal definition of the derivative, and when we apply that definition, the same thing with the binomial theorem happens (and the limit business destroys all the other terms).

Answer 2

Imagine a 3D box with dimensions $10\times 10\times x$. As $x$ grows (at 1 unit per second, say), the box grows in one direction. In $h$ seconds, we add a volume $10\times 10\times h$ to one side, and that's it. The rate of change of the volume is $\frac{10\times 10\times h}h=100$.

Now what if the box's dimensions are $x\times x \times x$ instead? As we increase $x$, the box grows in three directions at once. When we increase $x$ by $h$, we're adding a square layer of size $x\times x\times h$ to each of three sides (plus some smaller edge and corner bits, which are less significant the smaller $h$ is), so the box's volume increases by $3x^2h$. The rate of change of the volume at any given moment comes out to $3x^2$, where $x$ is the side length at that moment.

Similarly, in 2D, an $x\times x$ square grows by two $x\times h$ strips when we increase $x$ by $h$, because it's growing in two directions. In general, an $n$-dimensional hypercube of side length $x$ (and therefore volume $x^n$) grows in $n$ directions at once, at rate $x^{n-1}$ in each direction, so the rate of change is $nx^{n-1}$.

This is connected to the binomial theorem stuff in my other answer: the volume of an $n$-cube of side length $x+h$ is $(x+h)^n$, and algebraically expanding this product gives us $n$ copies of the term $x^{n-1}h$. The other terms in the binomial expansion correspond to the different "edge and corner bits", which become negligible as $h\to0$ because they have a factor of $h^2$.

Answer 3

Easiest to restrict the focus to exponents that are positive integers. For $~n \in \Bbb{Z^+},~$ and for any $~x,h \in \Bbb{R},~$ you have that

$$(x + h)^n = \sum_{k=0}^n \binom{n}{k}x^{n-k}h^k$$

$$= \binom{n}{0} x^n h^0 + \binom{n}{1} x^{n-1} h^1 + \binom{n}{2} x^{n-2} h^2 + \cdots + \binom{n}{n} x^0 h^n. \tag1 $$

Now consider the expression

$$\frac{(x+h)^n - x^n}{h}.$$

Using (1) above, this equals

$$\frac{1}{h} \times \left[ ~ \binom{n}{1} x^{n-1} h^1 + \binom{n}{2} x^{n-2} h^2 + \cdots + \binom{n}{n} x^0 h^n ~\right] $$

$$= \binom{n}{1} x^{n-1} h^0 + \binom{n}{2} x^{n-2} h^1 + \cdots + \binom{n}{n} x^0 h^{n-1}. \tag2 $$

By definition:

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

$$= \lim_{h\to 0} \left[ ~ \binom{n}{1} x^{n-1} h^0 + \binom{n}{2} x^{n-2} h^1 + \cdots + \binom{n}{n} x^0 h^{n-1} ~\right]. \tag3 $$

In (3) above, every term except the first one has a factor of $~h^k ~: ~k \in \Bbb{Z^+}.$

Therefore, (3) above evaluates to $~\displaystyle \binom{n}{1} x^{n-1} h^0 = \binom{n}{1} x^{n-1}.$