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education/math/MATH1210 (calc 1)/Derivatives.md
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@ -56,4 +56,9 @@ $$ f'(x) = \lim_{h \to 0} \dfrac{(x + h)^n - x^n}{h} $$
Note that the coefficient follows the associated level of Pascal's Triangle (`1 3 3 1`), and $x$'s power decrements, while $h$'s power increments. The coefficients of each pair will always add up to $n$. Eg, $3 + 0$, $2 + 1$, $1 + 2$, and so on. The **second** term in the polynomial created will have a coefficient of $n$. Note that the coefficient follows the associated level of Pascal's Triangle (`1 3 3 1`), and $x$'s power decrements, while $h$'s power increments. The coefficients of each pair will always add up to $n$. Eg, $3 + 0$, $2 + 1$, $1 + 2$, and so on. The **second** term in the polynomial created will have a coefficient of $n$.
$$ \dfrac{(x + h)^n - x^n}{h} = \lim_{h \to 0} \dfrac{x^n + nx^{n-1}h + p_{n3}x^{n-2}h^2 + \cdots h^n}{} $$ $$ \dfrac{(x + h)^n - x^n}{h} = \lim_{h \to 0} \dfrac{(x^n + nx^{n-1}h + P_{n3}x^{n-2}h^2 + \cdots + h^n)-x^n}{h} $$ $P$ denotes some coefficient found using Pascal's triangle.
$x^n$ cancels out, and then $h$ can be factored out of the binomial series.
This leaves us with:
$$ \lim_{h \to 0} nx^{n-1} + P_{n3} x^{} $$