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

$\frac{d}{dx}[\tan x] = \sec^2 x$

Secondary Functions

$\frac{d}{dx}[\csc x] = -\csc x \cot x$

$\frac{d}{dx}[\sec x] = \sec x \tan x$

$\frac{d}{dx}[\cot x] = -\csc^2 x$

💡 Pattern: Notice that the derivatives of "co-functions" (cosine, cosecant, cotangent) have negative signs!

Memory Tricks

Trick 1: The Sign Pattern

• CO-functions (cos, csc, cot) → derivatives are NEGATIVE
• Non-co-functions (sin, sec, tan) → derivatives are positive

Trick 2: Pairs

• $\sin x$ and $\cos x$ trade back and forth (with a sign change for cosine)
• $\sec x$ and $\tan x$ appear together: $\sec x \tan x$
• $\csc x$ and $\cot x$ appear together: $-\csc x \cot x$

Trick 3: Squares

• Derivative of $\tan x$ is $\sec^2 x$ (square!)
• Derivative of $\cot x$ is $-\csc^2 x$ (square with negative!)

Using the Chain Rule with Trig Functions

When the inside is NOT just $x$, use the Chain Rule!

General Formula

$\frac{d}{dx}[\sin(u)] = \cos(u) \cdot u'$

$\frac{d}{dx}[\cos(u)] = -\sin(u) \cdot u'$

$\frac{d}{dx}[\tan(u)] = \sec^2(u) \cdot u'$

Examples

1. $\frac{d}{dx}[\sin(3x)] = \cos(3x) \cdot 3 = 3\cos(3x)$

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

3. $\frac{d}{dx}[\tan(5x + 1)] = \sec^2(5x + 1) \cdot 5 = 5\sec^2(5x + 1)$

Powers of Trig Functions

Notation Warning

$\sin^2 x$ means $(\sin x)^2$, NOT $\sin(\sin x)$

Similarly: $\cos^3 x = (\cos x)^3$

Taking Derivatives

Use the Chain Rule with the power on the outside!

Example: $\frac{d}{dx}[\sin^2 x]$

Think of this as $(\sin x)^2$:

• Outside: square function, so $2(\sin x)$
• Inside: sine function, so $\cos x$
• Answer: $2\sin x \cos x$

Example: $\frac{d}{dx}[\cos^3 x]$

Think of this as $(\cos x)^3$:

• Outside: cube function, so $3(\cos x)^2$
• Inside: cosine function, so $-\sin x$
• Answer: $3\cos^2 x \cdot (-\sin x) = -3\cos^2 x \sin x$

Common Applications

Application 1: Motion Problems

If $s(t) = 5\sin(2t)$ represents position, find velocity at $t = \frac{\pi}{4}$:

$v(t) = s'(t) = 5\cos(2t) \cdot 2 = 10\cos(2t)$

$v\left(\frac{\pi}{4}\right) = 10\cos\left(\frac{\pi}{2}\right) = 10(0) = 0$

Application 2: Rate of Change

The height of a Ferris wheel car: $h(t) = 30 + 25\sin\left(\frac{\pi t}{10}\right)$ feet

Rate of change of height:

$h'(t) = 25\cos\left(\frac{\pi t}{10}\right) \cdot \frac{\pi}{10} = \frac{5\pi}{2}\cos\left(\frac{\pi t}{10}\right)$

⚠️ Common Mistakes

Mistake 1: Forgetting the Negative

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

Mistake 2: Wrong Square

❌ $\frac{d}{dx}[\tan x] = \tan^2 x$ ✅ $\frac{d}{dx}[\tan x] = \sec^2 x$

Mistake 3: Forgetting Chain Rule

❌ $\frac{d}{dx}[\sin(2x)] = \cos(2x)$ ✅ $\frac{d}{dx}[\sin(2x)] = 2\cos(2x)$

Mistake 4: Confusing Notation

Remember: $\sin^2 x = (\sin x)^2$, not $\sin(\sin x)$

Derivatives in Radians

IMPORTANT: All trig derivatives assume angles are measured in radians, not degrees!

If working in degrees, you need conversion factors. But on the AP Calculus exam, always use radians.

📝 Practice Tips

1. Memorize the six basic trig derivatives - you'll use them constantly
2. Remember the co-function negative sign rule
3. Always use the Chain Rule when the inside is not just $x$
4. Rewrite powers of trig functions: $\sin^2 x = (\sin x)^2$
5. Check your work by looking at the signs

Step 2: Derivative of the 4th power

$\frac{dy}{dx} = 4[\sin(2x)]^3 \cdot \frac{d}{dx}[\sin(2x)]$

Step 3: Derivative of sine (with Chain Rule)

$\frac{d}{dx}[\sin(2x)] = \cos(2x) \cdot 2 = 2\cos(2x)$

Step 4: Combine everything

$\frac{dy}{dx} = 4[\sin(2x)]^3 \cdot 2\cos(2x)$

$= 8\sin^3(2x)\cos(2x)$

Answer: $\displaystyle\frac{dy}{dx} = 8\sin^3(2x)\cos(2x)$

This gives us the slope of the tangent line:

$m = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$

Step 3: Find the y-coordinate

$y = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$

So the point is $\left(\frac{\pi}{3}, \frac{1}{2}\right)$

Step 4: Use point-slope form

$y - y_1 = m(x - x_1)$

$y - \frac{1}{2} = -\frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right)$

$y = -\frac{\sqrt{3}}{2}x + \frac{\sqrt{3}\pi}{6} + \frac{1}{2}$

Answer: $y = -\frac{\sqrt{3}}{2}x + \frac{\sqrt{3}\pi}{6} + \frac{1}{2}$

(Or in point-slope form): $y - \frac{1}{2} = -\frac{\sqrt{3}}{2}\left(x - \frac{\pi}{3}\right)$