Intuition & Definition

How can instant rate of change be defined at a point? Call our function of choice y: Slope of y between x_1 and x_2:


\frac {y_1 - y_2} {x_1 - x_2}


= \frac {y(x_1) - y(x_2)} {x_1 - x_2}

However, x_1 \neq x_2 due to division by 0.

Definitions

Avoid division by 0 via using a limit such that x_1 \to x_2:


\frac {dy} {dx} = \lim_{x_1 \to x_2} \frac {y(x_1) - y(x_2)} {x_1 - x_2}

Changing variables:


\frac {dy} {dx} = \lim_{a \to x} \frac {y(a) - y(x)} {a - x}

Define a = \lim _{\Delta x \to 0}(x + \Delta x):


\frac {dy} {dx} = \lim_{\Delta x \to 0} \frac {y(x + \Delta x) - y(x)} {\Delta x}

Derivative Rules

Constant Rule

When y = a and a is constant:


\frac {dy} {dx}


= \lim_{\Delta x \to 0} \frac {a - a} {\Delta x}


= \lim_{\Delta x \to 0} \frac {0} {\Delta x}

= 0

Sum and Difference Rule


\frac {df} {dx} + \frac {dg} {dx}


= \lim_{\Delta x \to 0} \frac {f(x + \Delta x) - f(x)} {\Delta x} + \frac {g(x + \Delta x) - g(x)} {\Delta x}


= \lim_{\Delta x \to 0} \frac {[f(x + \Delta x) + g(x + \Delta x)] - [f(x) + g(x)]}{\Delta x}


= \frac d {dx} (f + g)

Power Rule

Note: This proof of power rule only extends to n \in \mathbb{N}. Power rule can be extended to n \in \mathbb{Z} through the use of the derivative of \ln, but this article does not cover such a proof as of now.


\frac {d} {dx} x^n


= \lim_{\Delta x \to 0} \frac {(x + \Delta x)^n - x^n} {\Delta x}

Use a binomial expansion:


= \lim_{\Delta x \to 0} \frac {\sum_{i = 0}^n {n \choose i}x^i{\Delta x}^{n - i} - x^n} {\Delta x}

Take out last term in sum:


= \lim_{\Delta x \to 0} \frac {x^n + \sum_{i = 0}^{n - 1} {n \choose i}x^i{\Delta x}^{n - i} - x^n} {\Delta x}


= \lim_{\Delta x \to 0} \frac {\sum_{i = 0}^{n - 1} {n \choose i}x^i{\Delta x}^{n - i}} {\Delta x}


= \lim_{\Delta x \to 0} \sum_{i = 0}^{n - 1} {n \choose i}x^i{\Delta x}^{n - i - 1}

Bring limit inside sum:


= \sum_{i = 0}^{n - 1} \left[{n \choose i}x^i \lim_{\Delta x \to 0} {\Delta x}^{n - i - 1}\right]

For i < n - 1, \lim {\Delta x \to 0} \Delta x^{n - i - 1} = 0, so only the case where i = n - 1 matters:


= {n \choose {n - 1}} x^{n - 1}


= nx^{n - 1}

Therefore:
\frac d {dx} x^n = nx^{n - 1}

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