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3.5 Derivatives of Trig Functions

Let \( f(x) = \sin x \). What is \( f'(x) \)?

Graph of f(x) = sin(x) with horizontal tangent lines marked at its extrema.

Sketch \( f'(x) \) using slopes of tangent lines

Graph of the derivative function f'(x), showing a cosine wave starting at (0,1).

This looks just like the graph of \( \cos x \).

So, \[ \frac{d}{dx} \sin x = \cos x \]

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Following the same process, we can find:

\[ \frac{d}{dx} \cos x = -\sin x \]

Knowing the derivatives of \( \sin x \) and \( \cos x \), we can use them and the rules of differentiation to find the deriv. of the others.

Example: \( \frac{d}{dx} (\tan x) \)

We know \( \tan x = \frac{\sin x}{\cos x} \)

\( \frac{d}{dx} (\tan x) = \frac{d}{dx} \left( \frac{\sin x}{\cos x} \right) \) now use quotient rule

\[ = \frac{(\cos x) \frac{d}{dx} (\sin x) - (\sin x) \frac{d}{dx} (\cos x)}{(\cos x)^2} \]

\[ = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} \]

\[ = \left( \frac{1}{\cos x} \right)^2 = (\sec x)^2 = \boxed{\sec^2 x} \]

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Derivatives of Basic Trigonometric Functions

Following the same process, we find the derivatives of the other basic trig functions:

\[ \frac{d}{dx} \sin x = \cos x \]\[ \frac{d}{dx} \tan x = \sec^2 x \]\[ \frac{d}{dx} \sec x = \sec x \tan x \]
\[ \frac{d}{dx} \cos x = -\sin x \]\[ \frac{d}{dx} \cot x = -\csc^2 x \]\[ \frac{d}{dx} \csc x = -\csc x \cot x \]

Observations

  • Derivative of co-function is negative.
  • They are related to their co-function, too.

e.g.

\[ \frac{d}{dx} \sin x = \cos x \]\[ \frac{d}{dx} \cos x = -\sin x \]

All \( x \) here are in RADIANS. Do NOT use degrees.

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Example: Quotient Rule with Trig Functions

\[ y = \frac{\tan x - 1}{\sec x} \]

We can differentiate it directly, using the quotient rule:

\[ y' = \frac{(\sec x) \frac{d}{dx}(\tan x - 1) - (\tan x - 1) \frac{d}{dx}(\sec x)}{(\sec x)^2} \]\[ = \frac{(\sec x)(\sec^2 x) - (\tan x - 1)(\sec x \tan x)}{(\sec x)^2} \]
\[ = \frac{(\sec x) [(\sec^2 x) - (\tan x - 1)(\tan x)]}{(\sec x)^2} \]
Factor out \( \sec x \)
\[ = \frac{(\sec^2 x - \tan^2 x) + \tan x}{\sec x} \]\[ = \frac{1 + \tan x}{\sec x} \]

Identities:

\( \sin^2 x + \cos^2 x = 1 \)

Divide by \( \cos^2 x \):

\( \tan^2 x + 1 = \sec^2 x \)

\( \sec^2 x - \tan^2 x = 1 \)

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Alternative Differentiation Strategy

Another option: rewrite \( y \) in terms of \( \sin x \) and \( \cos x \) then differentiate.

\[ y = \frac{\tan x - 1}{\sec x} = \frac{\frac{\sin x}{\cos x} - 1}{\frac{1}{\cos x}} \cdot \frac{\cos x}{\cos x} = \frac{\sin x - \cos x}{1} \]

Same as:

\[ y = \sin x - \cos x \]
\[ y' = \cos x - (-\sin x) = \cos x + \sin x \]

much easier!

Verification

Verify that \( \cos x + \sin x = \frac{1 + \tan x}{\sec x} \)

\[ \frac{1 + \tan x}{\sec x} = \frac{1 + \frac{\sin x}{\cos x}}{\frac{1}{\cos x}} \cdot \frac{\cos x}{\cos x} = \frac{\cos x + \sin x}{1} \]

So the derivatives match.

Always consider rewriting in terms of \( \sin x \) and \( \cos x \) first.

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Special Trig Limits

\[ \lim_{x \to 0} \frac{\sin x}{x} \]

If \( x = 0 \), \( \frac{\sin x}{x} \to \frac{0}{0} \) indeterminate, limit = ?

There are many ways to find the limit, table of values is easiest.

Table of values for sin(x)/x as x approaches 0
\( x \)-0.01-0.00100.0010.01
\( \frac{\sin x}{x} \)0.999980.9999998X0.99999980.99998

x: in radians

Clearly,

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

Similarly, we can find

\[ \lim_{x \to 0} \frac{\cos x - 1}{x} = 0 \]
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Limits Involving Trigonometric Functions

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

We can replace "\(x\)" with an identical thing without changing the limit as long as that thing \(\to 0\).

Examples

For example,

\[ \lim_{x \to 0} \frac{\sin(x^2)}{(x^2)} = 1 \]
\[ \lim_{x \to 0} \frac{\sin(\sqrt{x})}{(\sqrt{x})} = 1 \]
\[ \lim_{x \to 0} \frac{\sin(x^2 - x)}{(x^2 - x)} = 1 \]

But,

\[ \lim_{x \to 0} \frac{\sin(5x)}{x} \neq 1 \quad \text{since they are not identical} \]

Solving Non-Identical Limits

To find \(\lim_{x \to 0} \frac{\sin(5x)}{x}\), we do this:

\[ \lim_{x \to 0} \frac{\sin(5x)}{(5x)} \cdot \frac{5}{1} = \left( \underbrace{\lim_{x \to 0} \frac{\sin(5x)}{(5x)}}_{1} \right) \cdot \left( \underbrace{\lim_{x \to 0} \frac{5}{1}}_{5} \right) = 5 \]
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Example: Complex Trigonometric Limit

\[ \lim_{x \to 0} \frac{\sin 19x}{\tan x} \]

As \(x \to 0\), \(\frac{\sin 19x}{\tan x} \to \frac{0}{0} = ?\)

Try to bring \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) into this.

\[ \lim_{x \to 0} \frac{\sin 19x}{1} \cdot \frac{1}{\tan x} = \lim_{x \to 0} \frac{\sin 19x}{1} \cdot \frac{1}{\frac{\sin x}{\cos x}} \]
\[ \lim_{x \to 0} \frac{\sin 19x}{1} \cdot \frac{\cos x}{\sin x} = \lim_{x \to 0} \underbrace{\frac{\sin 19x}{1}}_{\text{if } 19x \text{ in denom then limit is } 1} \cdot \underbrace{\frac{1}{\sin x}}_{\text{if } x \text{ in numerator then limit is } 1} \cdot \underbrace{\cos x}_{\text{goes to } 1 \text{ as } x \to 0} \]

Recall:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{x}{\sin x} = 1 \]

Final Calculation

\[ \lim_{x \to 0} \underbrace{\frac{\sin 19x}{19x}}_{1} \cdot \underbrace{\frac{x}{\sin x}}_{1} \cdot \underbrace{\cos x}_{1} \cdot \underbrace{\frac{19x}{x}}_{19} = \boxed{19} \]