Section 3-3: Derivatives of Trigonometric Functions

Welcome to Section 3-3! In this section, we'll be diving into the derivatives of trigonometric functions. While the video for this lesson is unavailable, we've provided detailed notes to guide you through the key concepts and examples.

Derivatives of Sine and Cosine

Let's start with the fundamental derivatives of sine and cosine. These are essential building blocks for finding derivatives of more complex trigonometric expressions.

  • The derivative of $f(x) = sin(x)$ is $f'(x) = cos(x)$.
  • The derivative of $f(x) = cos(x)$ is $f'(x) = -sin(x)$.

Figure 1 visually represents the relationship between $sin(x)$ and its derivative $cos(x)$:

Sine and Cosine Graphs

Let's walk through how we find the derivative of $sin(x)$ from first principles using the limit definition of a derivative:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

For $f(x) = sin(x)$, this becomes:

$$f'(x) = \lim_{h \to 0} \frac{sin(x+h) - sin(x)}{h}$$

Using the trigonometric identity $sin(x+h) = sin(x)cos(h) + cos(x)sin(h)$, we get:

$$f'(x) = \lim_{h \to 0} \frac{sin(x)cos(h) + cos(x)sin(h) - sin(x)}{h}$$

Rearranging and splitting the limit:

$$f'(x) = \lim_{h \to 0} \frac{sin(x)(cos(h) - 1)}{h} + \lim_{h \to 0} \frac{cos(x)sin(h)}{h}$$

Now, using the following limits:

  • $\lim_{h \to 0} \frac{sin(h)}{h} = 1$
  • $\lim_{h \to 0} \frac{cos(h) - 1}{h} = 0$

We arrive at:

$$f'(x) = sin(x) * 0 + cos(x) * 1 = cos(x)$$

Thus, the derivative of $sin(x)$ is indeed $cos(x)$.

Example: Applying the Product Rule

Let's differentiate $y = x^2 sin(x)$. We'll use the product rule, which states that if $y = f(x)g(x)$, then $y' = f'(x)g(x) + f(x)g'(x)$.

Here, let $f(x) = x^2$ and $g(x) = sin(x)$. Then, $f'(x) = 2x$ and $g'(x) = cos(x)$. Applying the product rule:

$$y' = (2x)sin(x) + (cos(x))(x^2) = 2xsin(x) + x^2cos(x)$$

Example: Finding Higher Order Derivatives

Let's determine the 27th derivative of $cos(x)$. Notice the pattern in the derivatives of $cos(x)$:

  • $y = cos(x)$
  • $y' = -sin(x)$
  • $y'' = -cos(x)$
  • $y''' = sin(x)$
  • $y^{(4)} = cos(x)$ (The pattern repeats every 4 derivatives)

To find the 27th derivative, divide 27 by 4: $27 \div 4 = 6$ with a remainder of 3. This means the 27th derivative will be the same as the 3rd derivative, which is $y^{(27)} = sin(x)$.

Derivative of Tangent

To find the derivative of $tan(x) = \frac{sin(x)}{cos(x)}$, we can use the quotient rule. If $y = \frac{f(x)}{g(x)}$, then $y' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$.

Let $f(x) = sin(x)$ and $g(x) = cos(x)$. Then $f'(x) = cos(x)$ and $g'(x) = -sin(x)$. Applying the quotient rule:

$$y' = \frac{(cos(x))(cos(x)) - (sin(x))(-sin(x))}{cos^2(x)} = \frac{cos^2(x) + sin^2(x)}{cos^2(x)} = \frac{1}{cos^2(x)} = sec^2(x)$$

Therefore, the derivative of $tan(x)$ is $sec^2(x)$.

Summary of Trigonometric Derivatives

  • $\frac{d}{dx}(sin(x)) = cos(x)$
  • $\frac{d}{dx}(cos(x)) = -sin(x)$
  • $\frac{d}{dx}(tan(x)) = sec^2(x)$
  • $\frac{d}{dx}(csc(x)) = -csc(x)cot(x)$
  • $\frac{d}{dx}(sec(x)) = sec(x)tan(x)$
  • $\frac{d}{dx}(cot(x)) = -csc^2(x)$

Keep practicing, and you'll master these derivatives in no time!