Derivatives of Trigonometric Functions: Fresh Take

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Derivatives of Trigonometric Functions: Fresh Take

Calculate the derivatives of sine and cosine functions, including second derivatives and beyond
Determine the derivatives for basic trig functions like tangent, cotangent, secant, and cosecant

Derivatives of the Sine and Cosine Functions

The Main Idea

Derivative of Sine: $\frac{d}{dx}(\sin x) = \cos x$
Derivative of Cosine: $\frac{d}{dx}(\cos x) = -\sin x$
Graphical Interpretation:
- Where sine has a maximum or minimum, cosine (its derivative) is zero
- Where cosine has a maximum or minimum, negative sine (its derivative) is zero

Find the derivative of $f(x)= \sin x \cos x.$

$f^{\prime}(x)=\cos^2 x-\sin^2 x$

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “3.5 Derivatives of Trigonometric Functions (edited)” here (opens in new window).

Find the derivative of $f(x)=\dfrac{x}{\cos x}$ .

$\dfrac{\cos x+x \sin x}{\cos^2 x}$

Watch the following video to see the worked solution to this example

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “3.5 Derivatives of Trigonometric Functions (edited)” here (opens in new window).

Find the derivative of $f(x) = 5x^3 \sin x$ .

Using the product rule:

$\begin{array}{rcl} f'(x) &=& \frac{d}{dx}(5x^3) \cdot \sin x + 5x^3 \cdot \frac{d}{dx}(\sin x) \\ &=& 15x^2 \cdot \sin x + 5x^3 \cdot \cos x \\ &=& 15x^2 \sin x + 5x^3 \cos x \end{array}$

A particle moves along a coordinate axis with position $s(t) = 2\sin t - t$ for $0 \leq t \leq 2\pi$ . At what times is the particle at rest?

Find velocity function:

$v(t) = s'(t) = 2\cos t - 1$

Set velocity to zero:

$2\cos t - 1 = 0$

Solve:

$\cos t = \frac{1}{2}$

In the given interval:

$t = \frac{\pi}{3}$ and $t = \frac{5\pi}{3}$

The particle is at rest at $t = \frac{\pi}{3}$ and $t = \frac{5\pi}{3}$ .

A particle moves along a coordinate axis. Its position at time $t$ is given by $s(t)=\sqrt{3}t+2 \cos t$ for $0\le t\le 2\pi$ . At what times is the particle at rest?

$t=\frac{\pi}{3}, \, t=\frac{2\pi}{3}$

Derivatives of Other Trigonometric Functions

The Main Idea

Derivatives of Other Trigonometric Functions:
- $\frac{d}{dx}(\tan x) = \sec^2 x$
- $\frac{d}{dx}(\cot x) = -\csc^2 x$
- $\frac{d}{dx}(\sec x) = \sec x \tan x$
- $\frac{d}{dx}(\csc x) = -\csc x \cot x$
Derivation Methods:
- Use quotient rule for $\tan$ and $\cot$
- Apply chain rule and trigonometric identities for $\sec$ and $\csc$

Find the derivative of $f(x)=2 \tan x-3 \cot x$ .

$f^{\prime}(x)=2 \sec^2 x+3 \csc^2 x$

Find the derivative of $f(x) = \csc x + x \tan x$ .

Use the sum rule:

$f'(x) = \frac{d}{dx}(\csc x) + \frac{d}{dx}(x \tan x)$

Apply known derivative and product rule:

$f'(x) = -\csc x \cot x + (1 \cdot \tan x + x \cdot \sec^2 x)$

Simplify:

$f'(x) = -\csc x \cot x + \tan x + x \sec^2 x$

Find the slope of the line tangent to the graph of $f(x)= \tan x$ at $x=\frac{\pi}{6}$ .

Evaluate the derivative at $x=\frac{\pi}{6}$ .

$\frac{4}{3}$

Find the equation of a line tangent to the graph of $f(x) = \cot x$ at $x = \frac{\pi}{4}$ .

Find the point: $f(\frac{\pi}{4}) = \cot \frac{\pi}{4} = 1$

Point: $(\frac{\pi}{4}, 1)$

Find the slope:

$f'(x) = -\csc^2 x$
$f'(\frac{\pi}{4}) = -\csc^2 \frac{\pi}{4} = -2$

Use point-slope form:

$y - 1 = -2(x - \frac{\pi}{4})$

Simplify to slope-intercept form:

$y = -2x + 1 + \frac{\pi}{2}$

Higher-Order Derivatives of Trig Functions

The Main Idea

Cyclic Pattern of Derivatives:
- For $\sin x$ : $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$
- For $\cos x$ : $\cos x \to -\sin x \to -\cos x \to \sin x \to \cos x$
Predicting Higher-Order Derivatives:
- Use the remainder when the derivative order is divided by $4$
- Pattern repeats every four derivatives

Problem-Solving Strategy

Find the order of the derivative ( $n$ )
Calculate $n \bmod 4$ (remainder when $n$ is divided by $4$ )
Use the remainder to determine the derivative:
- For sinx:
  - Remainder $0$ : $\sin x$
  - Remainder $1$ : $\cos x$
  - Remainder $2$ : $-\sin x$
  - Remainder $3$ : $-\cos x$
- For cosx:
  - Remainder $0$ : $\cos x$
  - Remainder $1$ : $-\sin x$
  - Remainder $2$ : $-\cos x$
  - Remainder $3$ : $\sin x$

For $y= \sin x$ , find $\dfrac{d^{59}}{dx^{59}}(\sin x)$ .

$\dfrac{d^{59}}{dx^{59}}(\sin x)=\dfrac{d^{4(14)+3}}{dx^{4(14)+3}}(\sin x)$

$−\cos x$

Find $\dfrac{d^{74}}{dx^{74}}(\sin x)$ .

$74 \div 4 = 18$ remainder $2$

For $\sin x$ , remainder $2$ corresponds to $-\sin x$

Therefore, $\dfrac{d^{74}}{dx^{74}}(\sin x) = -\sin x$

A block attached to a spring is moving vertically. Its position at time $t$ is given by $s(t)=2 \sin t$ .

Find $v\left(\frac{5\pi}{6}\right)$ and $a\left(\frac{5\pi}{6}\right)$ . Compare these values and decide whether the block is speeding up or slowing down.

$v\left(\frac{5\pi}{6}\right)=−\sqrt{3}<0$ and $a\left(\frac{5\pi}{6}\right)=-1<0$ . The block is speeding up.