Derivatives of Trig Functions for MATH 139
Exam Relevance for MATH 139
Memorize all six trig derivatives. Frequently combined with chain rule and product/quotient rules.
This skill appears on:
Derivatives of Trigonometric Functions
The six basic trig functions have derivatives you need to memorize. The good news: once you know sine and cosine, you can derive the rest using the quotient rule!
The Big Two: Sine and Cosine
These are the foundation—memorize them:
$$\frac{d}{dx}[\sin x] = \cos x$$
$$\frac{d}{dx}[\cos x] = -\sin x$$
🔑 Memory trick:
- Sine → Cosine (positive)
- Cosine → negative Sine (watch the minus sign!)
In this graph, notice how the derivative (cosine) is positive exactly when sine is increasing, and negative when sine is decreasing. At the peaks and valleys of sine, cosine crosses zero.
The Other Four: Derived from Sine and Cosine
You can derive these using the quotient rule, but it's faster to memorize them:
$$\frac{d}{dx}[\tan x] = \sec^2 x$$
$$\frac{d}{dx}[\cot x] = -\csc^2 x$$
$$\frac{d}{dx}[\sec x] = \sec x \tan x$$
$$\frac{d}{dx}[\csc x] = -\csc x \cot x$$
🔑 Pattern recognition:
- The "co-" functions (cosine, cotangent, cosecant) all have negative derivatives
- Tangent and cotangent give squared functions
- Secant and cosecant give products of two trig functions
Complete Reference Table
| Function | Derivative |
|---|---|
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\cot x$ | $-\csc^2 x$ |
| $\sec x$ | $\sec x \tan x$ |
| $\csc x$ | $-\csc x \cot x$ |
Problem: Find $\frac{d}{dx}[3\sin x - 2\cos x]$
Using linearity of the derivative: $$\frac{d}{dx}[3\sin x - 2\cos x] = 3\frac{d}{dx}[\sin x] - 2\frac{d}{dx}[\cos x]$$ $$= 3\cos x - 2(-\sin x)$$ $$= \boxed{3\cos x + 2\sin x}$$
Problem: Find $\frac{d}{dx}[x^2 \sin x]$
Step 1: Identify this as a product: $f(x) = x^2$ and $g(x) = \sin x$.
Step 2: Apply the product rule: $(fg)' = f'g + fg'$
$$\frac{d}{dx}[x^2 \sin x] = \frac{d}{dx}[x^2] \cdot \sin x + x^2 \cdot \frac{d}{dx}[\sin x]$$ $$= 2x \cdot \sin x + x^2 \cdot \cos x$$ $$= \boxed{2x\sin x + x^2\cos x}$$
Problem: Find $\frac{d}{dx}\left[\frac{\sin x}{x}\right]$
Step 1: Apply the quotient rule: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$
Let $f(x) = \sin x$ and $g(x) = x$.
Step 2: Calculate: $$\frac{d}{dx}\left[\frac{\sin x}{x}\right] = \frac{\cos x \cdot x - \sin x \cdot 1}{x^2}$$ $$= \boxed{\frac{x\cos x - \sin x}{x^2}}$$
Problem: Prove that $\frac{d}{dx}[\tan x] = \sec^2 x$
Step 1: Write tangent in terms of sine and cosine: $$\tan x = \frac{\sin x}{\cos x}$$
Step 2: Apply the quotient rule: $$\frac{d}{dx}[\tan x] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}$$ $$= \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$$
Step 3: Use the Pythagorean identity $\cos^2 x + \sin^2 x = 1$: $$= \frac{1}{\cos^2 x} = \boxed{\sec^2 x}$$
Problem: Find all $x$ in $[0, 2\pi]$ where the tangent line to $f(x) = \sin x + \cos x$ is horizontal.
Step 1: A horizontal tangent means $f'(x) = 0$.
$$f'(x) = \cos x - \sin x$$
Step 2: Solve $\cos x - \sin x = 0$: $$\cos x = \sin x$$ $$1 = \frac{\sin x}{\cos x} = \tan x$$ $$x = \arctan(1)$$
Step 3: Find all solutions in $[0, 2\pi]$: $$x = \frac{\pi}{4} \quad \text{and} \quad x = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$$
$$\boxed{x = \frac{\pi}{4}, \frac{5\pi}{4}}$$
Problem: Find $\frac{d}{dx}[\sin(3x)]$
Step 1: This requires the chain rule (covered in detail in another skill).
If $u = 3x$, then: $$\frac{d}{dx}[\sin(3x)] = \cos(3x) \cdot \frac{d}{dx}[3x] = \cos(3x) \cdot 3$$
$$= \boxed{3\cos(3x)}$$
Note: When the argument isn't just $x$, you must multiply by the derivative of the inside function (chain rule).
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting the negative sign on cosine's derivative
Wrong: $\frac{d}{dx}[\cos x] = \sin x$
Why it's wrong: The derivative of cosine is negative sine.
Correct: $\frac{d}{dx}[\cos x] = -\sin x$
❌ Mistake: Mixing up $\sec^2 x$ and $\tan^2 x$
Wrong: $\frac{d}{dx}[\tan x] = \tan^2 x$
Why it's wrong: The derivative of tangent involves secant squared, not tangent squared.
Correct: $\frac{d}{dx}[\tan x] = \sec^2 x$
❌ Mistake: Forgetting the chain rule when the argument isn't just $x$
Wrong: $\frac{d}{dx}[\sin(5x)] = \cos(5x)$
Why it's wrong: You must multiply by the derivative of the inside function.
Correct: $\frac{d}{dx}[\sin(5x)] = \cos(5x) \cdot 5 = 5\cos(5x)$
Quick Memory Aid
All "co-" functions have negative derivatives:
- cosine → negative sine
- cotangent → negative cosecant squared
- cosecant → negative cosecant cotangent
The others are positive:
- sine → cosine
- tangent → secant squared
- secant → secant tangent
Derivative of Sine
The derivative of sine is cosine. This is one of the two fundamental trig derivatives to memorize.
Variables:
- $x$:
- the angle (in radians)
Derivative of Cosine
The derivative of cosine is NEGATIVE sine. Don't forget the minus sign!
Variables:
- $x$:
- the angle (in radians)
Derivative of Tangent
The derivative of tangent is secant squared. Can be derived using quotient rule on sin/cos.
Variables:
- $x$:
- the angle (in radians)
Derivative of Cotangent
The derivative of cotangent is NEGATIVE cosecant squared. All 'co-' functions have negative derivatives.
Variables:
- $x$:
- the angle (in radians)
Derivative of Secant
The derivative of secant is secant times tangent.
Variables:
- $x$:
- the angle (in radians)
Derivative of Cosecant
The derivative of cosecant is NEGATIVE cosecant times cotangent. Another 'co-' function with a negative derivative.
Variables:
- $x$:
- the angle (in radians)
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