Deriving the derivative rule for monomials

Question

I'm a big fan of deriving things from scratch so I wanted to try this one, where $f(x) = cx^n$

$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

$$f'(x) = \lim_{h \to 0} \frac{c \cdot (x+h)^n-cx^n}{h}$$

$$f'(x) = c \cdot \lim_{h \to 0} \frac{\left(\sum_{k=0}^{n} \binom{n}{k}x^kh^{n-k}\right)-x^n}{h}$$

$$f'(x) = c \cdot \lim_{h \to 0} \frac{\left(\sum_{k=0}^{n} \binom{n}{k}x^kh^{n-k}\right)- \binom{n}{n} x^nh^{n-n}}{h}$$

$$f'(x) = c \cdot \lim_{h \to 0} \frac{\left(\sum_{k=0}^{n-1} \binom{n}{k}x^kh^{n-k}\right)}{h}$$

$$f'(x) = c \cdot \lim_{h \to 0} \sum_{k=0}^{n-1} \binom{n}{k}x^kh^{n-k-1}$$

Let $m = n-1$. Then:

$$f'(x) = c \cdot \lim_{h \to 0} \sum_{k=0}^{m} \binom{m+1}{k}x^kh^{m-k}$$

$$f'(x) = c \cdot (m + 1) \cdot \lim_{h \to 0} \sum_{k=0}^{m} \binom{m}{k}x^kh^{m-k}$$

$$f'(x) = c \cdot (m + 1) \cdot \lim_{h \to 0} (x + h)^m$$

$$f'(x) = c \cdot (m + 1) \cdot x^m$$

$$f'(x) = cnx^{n-1}$$

I was really surprised that this worked for some reason! Is this a correct proof? Is there a simpler proof?

Answer 1

Yes it is correct, it is the standard way for this proof by binomial theorem.

Note that from here

$$f'(x) = \lim_{h \to 0} \frac{c \cdot (x+h)^n-cx^n}{h}$$

you can simply note that

• $(x+h)^n=x^n+nx^{n-1}h+$ smaller terms (i.e. $h^2, h^3,...)$

and conclude suddenly.

Answer 2

Maybe it will be interesting that you can derive this relation easily by utilizing mathematical induction.Prove the relation for n=1,2 suppose it for n=k and prove it for n=k+1. So it will be true for any n.Also this relation is true for any real value of n but can not be proved in these ways and you should use Ln properties