Calc 1 Section 3-3: Derivatives of Trigonometric Functions

Welcome to Section 3.3! In this section, we'll dive into the derivatives of trigonometric functions. This is a crucial step in your Calculus 1 journey, building upon your understanding of limits and derivatives.

Derivative of Sine

Let's start with the derivative of $f(x) = \sin(x)$. Using the limit definition of a derivative, we have:

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

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

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

Since $\lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0$ and $\lim_{h \to 0} \frac{\sin(h)}{h} = 1$, we have:

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

Therefore, the derivative of $\sin(x)$ is $\cos(x)$. We can write this as:

$$\frac{d}{dx}(\sin x) = \cos x$$

Derivatives of Other Trigonometric Functions

Using similar methods, or by employing the quotient rule, we can find the derivatives of other trigonometric functions:

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

Example: Derivative of $y = \sin(\theta)\cos(\theta)$

Let's find the derivative of $y = \sin(\theta)\cos(\theta)$. We'll use the product rule:

If $f(\theta) = \sin(\theta)$ and $g(\theta) = \cos(\theta)$, then $f'(\theta) = \cos(\theta)$ and $g'(\theta) = -\sin(\theta)$.

Applying the product rule, $y' = f'(\theta)g(\theta) + f(\theta)g'(\theta)$, we get:

$$y' = (\cos \theta)(\cos \theta) + (\sin \theta)(-\sin \theta) = \cos^2 \theta - \sin^2 \theta$$

Recall the trigonometric identity $cos(2\theta) = \cos^2 \theta - \sin^2 \theta$. Therefore, $y' = \cos(2\theta)$.

Quotient Rule with Trig Functions

Let's find the derivative of $f(x) = \frac{\sec x}{1 + \tan x}$. Here we need the quotient rule.

Let $n(x) = \sec x$, so $n'(x) = \sec x \tan x$. Let $d(x) = 1 + \tan x$, so $d'(x) = \sec^2 x$.

Then:

$$f'(x) = \frac{(\sec x \tan x)(1 + \tan x) - (\sec^2 x)(\sec x)}{(1 + \tan x)^2}$$
$$f'(x) = \frac{\sec x(\tan x - 1)}{(1 + \tan x)^2}$$

Higher-Order Derivatives

We can also find higher-order derivatives of trigonometric functions. For example, if $f(x) = \sin x$, then:

  • $f'(x) = \cos x$
  • $f''(x) = -\sin x$
  • $f'''(x) = -\cos x$
  • $f^{(4)}(x) = \sin x$

Notice the pattern! The derivatives cycle through $\sin x$, $\cos x$, $-\sin x$, and $-\cos x$.

Applications

Understanding derivatives of trigonometric functions is essential for solving problems in physics and engineering, especially those involving oscillatory motion (like the spring example shown). Keep practicing, and you'll master these concepts in no time! Remember, Professor Baker is here to support you every step of the way.