The derivative of a function $f(x)$ is another function, which we will call $f'(x)$ that tells you the slope of $f(x)$. For example, the constant function $f(x)=c$ has slope $f'(x)=0$, since a constant function is flat. What is the derivative of a line $f(x)=mx+b$? The derivative is the slope right, so we must have $f'(x)=m$. What about more complicated functions?

The derivative of a function is defined as: \[ f'(x) \equiv \lim_{ \epsilon \rightarrow 0}\frac{f(x+\epsilon)-f(x)}{\epsilon}. \] You can think of $\epsilon$ as a really small number. I mean really small. The above formula is nothing more than the rise-over-run rule for calculating the slope of a line, \[ \frac{ rise } { run } = \frac{ \Delta y } { \Delta x } = \frac{y_f - y_i}{x_f - x_i} = \frac{f(x+\epsilon)\ - \ f(x)}{x + \epsilon \ -\ x}, \] but by taking $\epsilon$ to be really small, we will get the slope at the point $x$.

Derivatives occur so often in math that people have come up with many different notations for them. Don't be fooled by that. All of them mean the same thing $Df(x) = f'(x)=\frac{df}{dx}=\dot{f}=\nabla f$.

Knowing how to take derivatives is very useful in life. Given some phenomenon described by $f(x)$ you can say how it changes over time. Many times we don't actually care about the value of $f'(x)$, just its sign. If the derivative is positive $f'(x) > 0$, then the function is increasing. If $f'(x) < 0$ then the function is decreasing.

When the function is flat at a certain $x$ then $f'(x)=0$. The points where $f'(x)=0$ (the roots of $f'(x)$) are very important for finding the maximum and minimum values of $f(x)$. Recall how we calculated the maximum height $h$ that projectile reaches by first finding the time $t_{top}$ when its velocity in the $y$ direction was zero $y^\prime(t_{top})=v(t_{top})=0$ and then substituting this time in $y(t)$ to obtain $h=\max\{ y(t) \} =y(t_{top})$.

Now let's take a derivative of $f(x)=2x^2 + 3$ to see how that complicated-looking formula works: \[ f'(x)=\lim_{\epsilon \rightarrow 0} \frac{f(x+\epsilon)-f(x)}{\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{2(x+\epsilon)^2+3 \ \ - \ \ 2x^2 + 3}{\epsilon}. \] Let's simplify the right-hand side a bit \[ \frac{2x^2+ 4x\epsilon +\epsilon^2 - 2x^2}{\epsilon} = \frac{4x\epsilon +\epsilon^2}{\epsilon}= \frac{4x\epsilon}{\epsilon} + \frac{\epsilon^2}{\epsilon}. \] Now when we take the limit, the second term disappears: \[ f'(x) = \lim_{\epsilon \rightarrow 0} \left( \frac{4x\epsilon}{\epsilon} + \frac{\epsilon^2}{\epsilon} \right) = 4x + 0. \] Congratulations, you have just taken your first derivative! The calculations were not that complicated, but it was pretty long and tedious. The good news is that you only need to calculate the derivative from first principles only once. Once you find a derivative formula for a particular function, you can use the formula every time you see a function of that form.

\[ f(x) = x^n \qquad \Rightarrow \qquad f'(x) = n x^{n-1}. \]

Use the above formula to find the derivatives of the following three functions: \[ f(x) = x^{10}, \quad g(x) = \sqrt{x^3}, \qquad h(x) = \frac{1}{x^3}. \] In the first case, we use the formula directly to find the derivative $f'(x)=10x^9$. In the second case, we first use the fact that square root is equivalent to an exponent of $\frac{1}{2}$ to rewrite the function as $g(x)=x^{\frac{3}{2} }$, then using the formula we find that $g'(x)=\frac{3}{2}x^{\frac{1}{2} } =\frac{3}{2}\sqrt{x}$. We can also rewrite the third function as $h(x)=x^{-3}$ and then compute the derivative $h'(x)=-3x^{-4}=\frac{-3}{x^4}$ using the formula.

In the next section we will develop derivative formulas for other functions.