Derivatives

Calculus Math

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, $x_1\ne x_2$ due to division by $0$.

Definitions

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

$$\frac{dy}{dx}=\underset{x_1\to x_2}{\lim }\frac{y(x_1)-y(x_2)}{x_1-x_2}$$

Changing variables:

$$\frac{dy}{dx}=\underset{a\to x}{\lim }\frac{y(a)-y(x)}{a-x}$$

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

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

Derivative Rules

Constant Rule

When $y=a$ and $a$ is constant:

$$\frac{dy}{dx}$$ $$=\underset{\Delta x\to 0}{\lim }\frac{a-a}{\Delta x}$$ $$=\underset{\Delta x\to 0}{\lim }\frac{0}{\Delta x}$$ $$=0$$

Sum and Difference Rule

$$\frac{df}{dx}+\frac{dg}{dx}$$ $$=\underset{\Delta x\to 0}{\lim }\frac{f(x+\Delta x)-f(x)}{\Delta x}+\frac{g(x+\Delta x)-g(x)}{\Delta x}$$ $$=\underset{\Delta x\to 0}{\lim }\frac{[f(x+\Delta x)+g(x+\Delta x)]-[f(x)+g(x)]}{\Delta x}$$ $$=\frac{d}{dx}(f+g)$$

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.

$$\frac{d}{dx}{x}^n$$ $$=\underset{\Delta x\to 0}{\lim }\frac{(x+\Delta x{)}^n-x^n}{\Delta x}$$

Use a binomial expansion:

$$=\underset{\Delta x\to 0}{\lim }\frac{{\sum }_{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)x^i{\Delta x}^{n-i}-x^n}{\Delta x}$$

Take out last term in sum:

$$=\underset{\Delta x\to 0}{\lim }\frac{x^n+{\sum }_{i=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{i}\right)x^i{\Delta x}^{n-i}-x^n}{\Delta x}$$ $$=\underset{\Delta x\to 0}{\lim }\frac{{\sum }_{i=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{i}\right)x^i{\Delta x}^{n-i}}{\Delta x}$$ $$=\underset{\Delta x\to 0}{\lim }\sum _{i=0}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{i}\right)x^i{\Delta x}^{n-i-1}$$

Bring limit inside sum:

$$=\sum _{i=0}^{n-1}\left[\left(\genfrac{}{}{0ex}{}{n}{i}\right)x^i\underset{\Delta x\to 0}{\lim }{\Delta x}^{n-i-1}\right]$$

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

$$=\left(\genfrac{}{}{0ex}{}{n}{n-1}\right)x^{n-1}$$ $$=n{x}^{n-1}$$

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