Let $$y$$ be a function of $$x$$. A common way to express “the derivate of $$y$$ with respect to $$x$$” is the notation

$\frac{dy}{dy}$

Using this notation, we can write that if $$y = f(x)$$, then the derivative of $$y$$ with respect to $$x$$ is $$\frac{dy}{dx} = f^{\prime}(x)$$.

There are a handful of rules that you can use when evaluating derivatives.

The Power Rule states that

$\frac{d}{dx}x^k = kx^{k - 1}$

The derivative of a constant function is 0.

$\frac{d}{dx}c = 0$

The derivative of a constant times a function is the constant times the derivative of the function.

$\frac{d}{dx}[cf(x)] = c\frac{d}{dx}f(x)$

The derivative of a sum is the sum of the derivatives.

$\frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)$

The derivative of a difference is the difference of the derivatives.

$\frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}f(x) - \frac{d}{dx}g(x)$

All these rules, and more, are already encoded in your symbolic mathematical program of choice, Mathematica, Sympy, Sage, and so on. There’s nothing to memorize.

In general, you can use the derivative to find the slope of a tangent line.

Lastly, some functions are always increasing or always decreasing. The equation of the derivative of these function will tell you what behavior you can expect from the function.

Let’s look at some textbook problems.

Let’s say that the volume $$V$$ of a spherical tumor can be approximated by

$V(r) = \frac{4}{3}\pi r^3$

where $$r$$ is the radius of the tumor, in centimeters. (1) Find the rate of change of the volume with respect to the radius. (2) Find the rate of change of the volume at $$r = 1.2 cm$$.

from math import pi
import numpy as np
from sympy import *
import matplotlib.pyplot as plt

r, v, x = symbols("r, v, x")

v = 4/3 * pi * r**3

fprime = diff(v, r)
fprime
## 4.0*pi*r**2
fprime.evalf(subs = {r: 1.2})
## 18.0955736846772

When the radius is $$1.2 cm$$, the volume is changing at the rate of $$18 cm^3$$.

Let’s take the case of an always increasing function $$f(x) = x^3 + 2x$$.

def f(x):
return x ** 3 + 2 *x

x_values = np.linspace(-2, 2)
y_values = [f(x) for x in x_values]
plt.plot(x_values, y_values)

The derivative is

f = x ** 3 + 2 * x
fprime = diff(f, x)
fprime
## 3*x**2 + 2

Thus, for any $$x$$-value, $$x^2$$ will be nonnegative.