Proof of Polynomial Differentiation
A proof of polynomial differentiation that even a dog can understand.
I found that people who are not from science and engineering majors seem to be very afraid of this, so I post it.
$$ \begin{align*} \frac{d}{dx} x^n & = \lim_{h \to 0} \frac{(x + h)^n - x^n}{h} \\ & = \lim_{h \to 0} \frac{\left[ \sum_{k = 0}^{n} \binom{n}{k} x^{n - k} h^k \right] - x^n}{h} \\ & = \lim_{h \to 0} \frac{\left[ x^n + n x^{n - 1} h + \sum_{k = 2}^{n} \binom{n}{k} x^{n - k} h^k \right] - x^n}{h} \\ & = \lim_{h \to 0} \frac{n x^{n - 1} h + \sum_{k = 2}^{n} \binom{n}{k} x^{n - k} h^k}{h} \\ & = \lim_{h \to 0} n x^{n - 1} + \sum_{k = 2}^{n} \binom{n}{k} x^{n - k} h^{k - 1} \\ & = n x^{n - 1} \end{align*} $$It seems less scary if it is replaced with a dog πΆ.
$$ \begin{align*} \frac{d}{dπΆ} πΆ^n & = \lim_{h \to 0} \frac{(πΆ + h)^n - πΆ^n}{h} \\ & = \lim_{h \to 0} \frac{\left[ \sum_{k = 0}^{n} \binom{n}{k} πΆ^{n - k} h^k \right] - πΆ^n}{h} \\ & = \lim_{h \to 0} \frac{\left[ πΆ^n + n πΆ^{n - 1} h + \sum_{k = 2}^{n} \binom{n}{k} πΆ^{n - k} h^k \right] - πΆ^n}{h} \\ & = \lim_{h \to 0} \frac{n πΆ^{n - 1} h + \sum_{k = 2}^{n} \binom{n}{k} πΆ^{n - k} h^k}{h} \\ & = \lim_{h \to 0} n πΆ^{n - 1} + \sum_{k = 2}^{n} \binom{n}{k} πΆ^{n - k} h^{k - 1} \\ & = n πΆ^{n - 1} \end{align*} $$A cat π± would be even better.
$$ \begin{align*} \frac{d}{dπΆ} πΆ^n & = \lim_{π± \to 0} \frac{(πΆ + π±)^n - πΆ^n}{π±} \\ & = \lim_{π± \to 0} \frac{\left[ \sum_{k = 0}^{n} \binom{n}{k} πΆ^{n - k} π±^k \right] - πΆ^n}{π±} \\ & = \lim_{π± \to 0} \frac{\left[ πΆ^n + n πΆ^{n - 1} π± + \sum_{k = 2}^{n} \binom{n}{k} πΆ^{n - k} π±^k \right] - πΆ^n}{π±} \\ & = \lim_{π± \to 0} \frac{n πΆ^{n - 1} π± + \sum_{k = 2}^{n} \binom{n}{k} πΆ^{n - k} π±^k}{π±} \\ & = \lim_{π± \to 0} n πΆ^{n - 1} + \sum_{k = 2}^{n} \binom{n}{k} πΆ^{n - k} π±^{k - 1} \\ & = n πΆ^{n - 1} \end{align*} $$