Calc 1 - April 22, 2024: Derivatives and Antiderivatives

Welcome to Professor Baker's Math Class! Today's lesson focuses on understanding derivatives and antiderivatives, which are fundamental concepts in Calculus. Let's dive in and explore how these concepts work. Remember, practice is key, so work through the examples and don't hesitate to ask questions!

Derivatives

We'll start with a quick review of derivatives. Recall that the derivative of a function, denoted as $f'(x)$, represents the instantaneous rate of change of the function.

For example, let's find the derivative of the function:

$$f(x) = 5x^3 + e^x + \ln|x| + 6$$

Applying the power rule, the derivative of $5x^3$ is $15x^2$. The derivative of $e^x$ is simply $e^x$, and the derivative of $\ln|x|$ is $\frac{1}{x}$. The derivative of a constant (like 6) is 0. Therefore, the derivative of the function is:

$$f'(x) = 15x^2 + e^x + \frac{1}{x}$$

Antiderivatives (Integrals)

Now, let's discuss antiderivatives, also known as integrals. An antiderivative of a function $f(x)$ is a function $F(x)$ such that $F'(x) = f(x)$. We denote the antiderivative using the integral symbol: $\int f(x) dx = F(x) + C$, where $C$ is the constant of integration.

Let's find the antiderivative of $f(x) = 5x^3 + e^x + \ln|x|$. We have:

  • The antiderivative of $5x^3$ is $\frac{5}{4}x^4$.
  • The antiderivative of $e^x$ is $e^x$.
  • The antiderivative of $\frac{1}{x}$ is $\ln|x|$.

Thus, the antiderivative of $f(x)$ is:

$$F(x) = \frac{5}{4}x^4 + e^x + \ln|x| + C$$

Important Note: Don't forget the constant of integration, $C$! This represents the family of functions that have the same derivative.

Antidifferentiation Formulas

Here's a table summarizing some common antiderivative formulas:

  • $\int cf(x) dx = cF(x)$
  • $\int [f(x) + g(x)] dx = F(x) + G(x)$
  • $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (where $n \neq -1$)
  • $\int \frac{1}{x} dx = \ln|x| + C$
  • $\int e^x dx = e^x + C$
  • $\int \cos x dx = \sin x + C$
  • $\int \sin x dx = -\cos x + C$

Example Problems

Let's work through a couple of examples to solidify our understanding.

Example 1: Find $g(x)$ if $g'(x) = 4\sin x + \frac{2x^5 - \sqrt{x}}{x}$.

First, simplify $g'(x)$: $g'(x) = 4 \sin x + 2x^4 - x^{-1/2}$. Then, integrate:

$$g(x) = -4\cos x + \frac{2}{5}x^5 - 2\sqrt{x} + C$$

Example 2: A particle moves in a straight line with acceleration $a(t) = 6t + 4$. Its initial velocity is $v(0) = -6$ cm/s, and its initial displacement is $s(0) = 9$ cm. Find its position function $s(t)$.

  1. First, find the velocity function $v(t)$ by integrating $a(t)$: $v(t) = \int (6t + 4) dt = 3t^2 + 4t + C$. Use the initial condition $v(0) = -6$ to find $C$: $v(0) = 3(0)^2 + 4(0) + C = -6$, so $C = -6$. Thus, $v(t) = 3t^2 + 4t - 6$.
  2. Next, find the position function $s(t)$ by integrating $v(t)$: $s(t) = \int (3t^2 + 4t - 6) dt = t^3 + 2t^2 - 6t + D$. Use the initial condition $s(0) = 9$ to find $D$: $s(0) = (0)^3 + 2(0)^2 - 6(0) + D = 9$, so $D = 9$. Thus, $s(t) = t^3 + 2t^2 - 6t + 9$.

Keep practicing these concepts, and you'll master them in no time. Good luck, and see you in the next class!