Trigonometric Derivatives: Complete Guide

Master all trigonometric derivatives with our comprehensive guide. Learn the derivatives of sin, cos, tan, sec, csc, and cot with proofs, patterns, memory tricks, and 20+ solved examples.

📊 Six Trigonometric Functions Overview

There are six trigonometric functions, and knowing their derivatives is essential for calculus. The good news? There are clear patterns that make them easier to remember!

sin(x)
cos(x)
cos(x)
-sin(x)
tan(x)
sec²(x)
csc(x)
-csc(x)cot(x)
sec(x)
sec(x)tan(x)
cot(x)
-csc²(x)
💡 Key Pattern

Co-functions have negative derivatives! The three co-functions are cos, cot, and csc (they all start with "co"). Their derivatives all have negative signs. This single rule helps you remember half of the signs!

Quick Reference Table

Function Derivative Memory Aid
d/dx[sin(x)] cos(x) Positive, switches to cosine
d/dx[cos(x)] -sin(x) Negative (co-function!)
d/dx[tan(x)] sec²(x) Squared secant
d/dx[cot(x)] -csc²(x) Negative (co-function!), squared cosecant
d/dx[sec(x)] sec(x)tan(x) Function times tangent
d/dx[csc(x)] -csc(x)cot(x) Negative (co-function!), function times cotangent

🔵 The Basic Three: sin, cos, tan

These are the most commonly used trig functions. Master these first, and the others follow naturally.

1. Derivative of sin(x)

📐 d/dx[sin(x)] = cos(x)

Why it works: As x increases, sin(x) increases at a rate determined by cos(x). When cos(x) is large (near x = 0), sin(x) is changing rapidly. When cos(x) is zero (at x = π/2), sin(x) reaches its peak and momentarily stops changing.

Examples: d/dx[sin(x)] = cos(x) d/dx[sin(2x)] = cos(2x) · 2 = 2cos(2x) [chain rule] d/dx[3sin(x)] = 3cos(x) [constant multiple] d/dx[sin(x²)] = cos(x²) · 2x = 2x·cos(x²) [chain rule]

2. Derivative of cos(x)

📐 d/dx[cos(x)] = -sin(x)

Why the negative? As x increases from 0, cos(x) decreases (it starts at 1 and heads toward 0). A decreasing function has a negative derivative. The rate of this decrease is measured by sin(x).

Examples: d/dx[cos(x)] = -sin(x) d/dx[cos(3x)] = -sin(3x) · 3 = -3sin(3x) [chain rule] d/dx[5cos(x)] = -5sin(x) [constant multiple] d/dx[cos(x³)] = -sin(x³) · 3x² = -3x²sin(x³) [chain rule]
⚠️ Common Mistake

Don't forget the negative sign! d/dx[cos(x)] = -sin(x), NOT sin(x). This is the #1 error students make with trig derivatives.

3. Derivative of tan(x)

📐 d/dx[tan(x)] = sec²(x)

Alternative form: Since sec(x) = 1/cos(x), we can also write this as d/dx[tan(x)] = 1/cos²(x)

Examples: d/dx[tan(x)] = sec²(x) d/dx[tan(4x)] = sec²(4x) · 4 = 4sec²(4x) [chain rule] d/dx[2tan(x)] = 2sec²(x) [constant multiple] d/dx[tan(x²)] = sec²(x²) · 2x = 2x·sec²(x²) [chain rule]
📝 Proof Using Quotient Rule

Since tan(x) = sin(x)/cos(x):

d/dx[tan(x)] = d/dx[sin(x)/cos(x)] Using quotient rule: [cos(x)·cos(x) - sin(x)·(-sin(x))] / cos²(x) = [cos²(x) + sin²(x)] / cos²(x) = 1 / cos²(x) [since cos²(x) + sin²(x) = 1] = sec²(x)

🔄 The Sin-Cos Cycle

The derivatives of sin and cos repeat every 4 steps:

sin(x)
cos(x)
-sin(x)
-cos(x)
sin(x)

This means: d²/dx²[sin(x)] = -sin(x), d³/dx³[sin(x)] = -cos(x), d⁴/dx⁴[sin(x)] = sin(x)

🔴 The Reciprocal Three: csc, sec, cot

These functions are reciprocals of sin, cos, and tan. Their derivatives follow patterns that connect back to the basic three.

4. Derivative of csc(x)

📐 d/dx[csc(x)] = -csc(x)cot(x)

Pattern: The derivative includes the function itself (csc) multiplied by the co-function (cot), with a negative sign.

Examples: d/dx[csc(x)] = -csc(x)cot(x) d/dx[csc(2x)] = -csc(2x)cot(2x) · 2 = -2csc(2x)cot(2x) d/dx[4csc(x)] = -4csc(x)cot(x) d/dx[csc(x²)] = -csc(x²)cot(x²) · 2x = -2x·csc(x²)cot(x²)

5. Derivative of sec(x)

📐 d/dx[sec(x)] = sec(x)tan(x)

Pattern: The derivative includes the function itself (sec) multiplied by the related function (tan), with a positive sign.

Examples: d/dx[sec(x)] = sec(x)tan(x) d/dx[sec(3x)] = sec(3x)tan(3x) · 3 = 3sec(3x)tan(3x) d/dx[2sec(x)] = 2sec(x)tan(x) d/dx[sec(πx)] = sec(πx)tan(πx) · π = π·sec(πx)tan(πx)

6. Derivative of cot(x)

📐 d/dx[cot(x)] = -csc²(x)

Pattern: Similar to tan(x) → sec²(x), but cot(x) → -csc²(x) (negative because it's a co-function).

Examples: d/dx[cot(x)] = -csc²(x) d/dx[cot(5x)] = -csc²(5x) · 5 = -5csc²(5x) d/dx[3cot(x)] = -3csc²(x) d/dx[cot(x/2)] = -csc²(x/2) · (1/2) = -(1/2)csc²(x/2)
✨ Beautiful Symmetry

Notice the pairs:

  • tan(x) → sec²(x) and cot(x) → -csc²(x)
  • sec(x) → sec(x)tan(x) and csc(x) → -csc(x)cot(x)
  • The derivatives of sec and csc include the function times another function
  • The derivatives of tan and cot are squared functions

🧠 Patterns & Memory Aids

Pattern 1: Co-Function Rule

🎯 The Co-Function Pattern

All co-functions (cos, cot, csc) have NEGATIVE derivatives!

Regular functions (sin, tan, sec) have positive derivatives (or include the function positively).

This single rule gets you all the signs correct!

Pattern 2: Squared vs. Product

📊 Two Types of Derivatives

Squared Derivatives:

  • d/dx[tan(x)] = sec²(x)
  • d/dx[cot(x)] = -csc²(x)

Product Derivatives:

  • d/dx[sec(x)] = sec(x)tan(x)
  • d/dx[csc(x)] = -csc(x)cot(x)

Pattern 3: Function Relationships

Function Pair Derivative Pattern
tan & sec tan → sec² and sec → sec·tan
cot & csc cot → -csc² and csc → -csc·cot
sin & cos sin → cos and cos → -sin

Memory Mnemonics

🎵 The "Co-Negative" Song

"Co-functions are negative, that's the key to success!"

Cos, Cot, Csc → all negative derivatives

🔢 The "Square or Multiply" Rule

Tangent/Cotangent → SQUARE the reciprocal

tan → sec², cot → -csc²

Secant/Cosecant → MULTIPLY by complement

sec → sec·tan, csc → -csc·cot

📐 The "Sin-Cos Flip"

Sin becomes Cos (positive)

Cos becomes -Sin (negative - it's a co-function!)

🔬 Why These Derivatives Work

Understanding WHY these formulas work helps you remember them and builds deeper mathematical intuition.

📝 Proof: d/dx[sin(x)] = cos(x)
Using the limit definition: d/dx[sin(x)] = lim[h→0] [sin(x+h) - sin(x)] / h Using the sum formula: sin(A+B) = sin(A)cos(B) + cos(A)sin(B) = lim[h→0] [sin(x)cos(h) + cos(x)sin(h) - sin(x)] / h = lim[h→0] [sin(x)(cos(h) - 1) + cos(x)sin(h)] / h = lim[h→0] [sin(x)(cos(h) - 1)/h + cos(x)sin(h)/h] Known limits: lim[h→0] sin(h)/h = 1 and lim[h→0] (cos(h)-1)/h = 0 = sin(x)·0 + cos(x)·1 = cos(x) ✓
📝 Proof: d/dx[sec(x)] = sec(x)tan(x)
Since sec(x) = 1/cos(x), use the quotient rule: d/dx[1/cos(x)] = [0·cos(x) - 1·(-sin(x))] / cos²(x) = sin(x) / cos²(x) = (sin(x)/cos(x)) · (1/cos(x)) = tan(x) · sec(x) = sec(x)tan(x) ✓
💡 Insight

All six trig derivatives can be proven using: (1) the limit definition, (2) trig identities, and (3) quotient/chain rules. Once you prove sin and cos, the other four follow from their definitions as ratios!

📝 Solved Examples

Example 1: Basic Derivatives
Find the derivatives: a) f(x) = 5sin(x) + 3cos(x) f'(x) = 5cos(x) + 3(-sin(x)) f'(x) = 5cos(x) - 3sin(x) b) g(x) = tan(x) - 2cot(x) g'(x) = sec²(x) - 2(-csc²(x)) g'(x) = sec²(x) + 2csc²(x) c) h(x) = 4sec(x) + csc(x) h'(x) = 4sec(x)tan(x) + (-csc(x)cot(x)) h'(x) = 4sec(x)tan(x) - csc(x)cot(x)
Example 2: Product Rule with Trig

Find: d/dx[x²sin(x)]

Using product rule: d/dx[uv] = u'v + uv' u = x², u' = 2x v = sin(x), v' = cos(x) d/dx[x²sin(x)] = 2x·sin(x) + x²·cos(x) = 2x·sin(x) + x²cos(x)

Answer: 2x·sin(x) + x²cos(x)

Example 3: Quotient Rule with Trig

Find: d/dx[sin(x)/x]

Using quotient rule: d/dx[u/v] = (u'v - uv')/v² u = sin(x), u' = cos(x) v = x, v' = 1 d/dx[sin(x)/x] = [cos(x)·x - sin(x)·1] / x² = [x·cos(x) - sin(x)] / x² = (x·cos(x) - sin(x)) / x²

Answer: (x·cos(x) - sin(x)) / x²

🔗 Trigonometric Derivatives with Chain Rule

When trig functions have something other than just x inside, you MUST use the chain rule!

⚠️ Critical Rule

d/dx[sin(u)] = cos(u)·du/dx

Where u is any function of x. The same pattern applies to all trig functions!

Example 4: Chain Rule Applications
a) d/dx[sin(3x)] u = 3x, du/dx = 3 d/dx[sin(3x)] = cos(3x)·3 = 3cos(3x) b) d/dx[cos(x²)] u = x², du/dx = 2x d/dx[cos(x²)] = -sin(x²)·2x = -2x·sin(x²) c) d/dx[tan(5x)] u = 5x, du/dx = 5 d/dx[tan(5x)] = sec²(5x)·5 = 5sec²(5x) d) d/dx[sec(√x)] u = √x = x^(1/2), du/dx = 1/(2√x) d/dx[sec(√x)] = sec(√x)tan(√x)·1/(2√x) = [sec(√x)tan(√x)] / (2√x) e) d/dx[sin³(x)] This is [sin(x)]³, so u = sin(x), du/dx = cos(x) d/dx[sin³(x)] = 3sin²(x)·cos(x) = 3sin²(x)cos(x) f) d/dx[cos(2x + π/4)] u = 2x + π/4, du/dx = 2 d/dx[cos(2x + π/4)] = -sin(2x + π/4)·2 = -2sin(2x + π/4)
💡 Chain Rule Pattern

For any trig function f and inner function u:

d/dx[trig(u)] = (derivative of trig at u) × (derivative of u)

Examples: sin(u) → cos(u)·u', tan(u) → sec²(u)·u', sec(u) → sec(u)tan(u)·u'

🔄 Inverse Trigonometric Derivatives

Inverse trig functions (arcsin, arccos, arctan, etc.) have their own special derivatives involving square roots and algebraic expressions.

The Six Inverse Trig Derivatives

Function Derivative Domain Restriction
d/dx[arcsin(x)] 1/√(1 - x²) -1 < x < 1
d/dx[arccos(x)] -1/√(1 - x²) -1 < x < 1
d/dx[arctan(x)] 1/(1 + x²) all x
d/dx[arccot(x)] -1/(1 + x²) all x
d/dx[arcsec(x)] 1/(|x|√(x² - 1)) |x| > 1
d/dx[arccsc(x)] -1/(|x|√(x² - 1)) |x| > 1
🔍 Pattern Recognition

Notice that co-functions have negative derivatives here too!

  • arcsin → positive, arccos → negative
  • arctan → positive, arccot → negative
  • arcsec → positive, arccsc → negative
Example 5: Inverse Trig with Chain Rule
a) d/dx[arcsin(2x)] u = 2x, du/dx = 2 d/dx[arcsin(2x)] = 1/√(1 - (2x)²) · 2 = 2/√(1 - 4x²) b) d/dx[arctan(x²)] u = x², du/dx = 2x d/dx[arctan(x²)] = 1/(1 + (x²)²) · 2x = 2x/(1 + x⁴) c) d/dx[arccos(1/x)] u = 1/x = x⁻¹, du/dx = -1/x² d/dx[arccos(1/x)] = -1/√(1 - (1/x)²) · (-1/x²) = 1/(x²√(1 - 1/x²)) = 1/(x²√((x² - 1)/x²)) = 1/(x√(x² - 1))

💡 Tips & Common Mistakes

Top 5 Common Mistakes

❌ Mistake #1: Forgetting Negative Signs

Wrong: d/dx[cos(x)] = sin(x)

Right: d/dx[cos(x)] = -sin(x)

Remember: ALL co-functions (cos, cot, csc) have negative derivatives!

❌ Mistake #2: Forgetting Chain Rule

Wrong: d/dx[sin(3x)] = cos(3x)

Right: d/dx[sin(3x)] = 3cos(3x)

Always multiply by the derivative of the inside function!

❌ Mistake #3: Confusing sec and csc

Remember: sec = 1/cos (secant goes with cosine)

csc = 1/sin (cosecant goes with sine)

sec²(x) shows up in tan derivative, csc²(x) in cot derivative

❌ Mistake #4: Wrong Power on sec²/csc²

Wrong: d/dx[tan(x)] = sec(x)

Right: d/dx[tan(x)] = sec²(x)

The derivative of tan and cot involve SQUARED functions!

❌ Mistake #5: Not Simplifying

Before simplifying: d/dx[sin(x)cos(x)] = cos(x)cos(x) + sin(x)(-sin(x))

After simplifying: cos²(x) - sin²(x) or cos(2x)

Use trig identities to simplify when possible!

Study Tips

✅ How to Master Trig Derivatives
  1. Start with sin and cos - These are the foundation
  2. Learn the co-function rule - Gets you all the signs
  3. Memorize the patterns - Squared vs. product types
  4. Practice chain rule constantly - Most problems involve it
  5. Make flashcards - One function per card
  6. Do 10 problems daily - Repetition builds automaticity
  7. Test yourself without notes - Can you write all 6 from memory?

Quick Mental Checks

Before Submitting Your Answer, Ask:

  • ✓ Is this a co-function? If yes, is my sign negative?
  • ✓ Did I use chain rule if there's something other than x inside?
  • ✓ For tan/cot derivatives, did I square the result?
  • ✓ For sec/csc derivatives, did I multiply by the related function?
  • ✓ Can I simplify using trig identities?

❓ Frequently Asked Questions

Q1: What is the derivative of sin(x)?

Answer: The derivative of sin(x) is cos(x). This can be proven using the limit definition of the derivative and trigonometric sum formulas. Intuitively, cos(x) measures the rate of change of sin(x) - when cos(x) is large, sin(x) is changing rapidly; when cos(x) is zero, sin(x) reaches a maximum or minimum.

Q2: Why does cos(x) have a negative derivative?

Answer: The derivative of cos(x) is -sin(x) because cos(x) is a decreasing function for x in (0, π). Since it's decreasing, its derivative must be negative. The magnitude of the decrease is measured by sin(x). This also fits the co-function pattern - all co-functions have negative derivatives.

Q3: How do I remember all six trig derivatives?

Answer: Use these patterns: (1) Co-functions (cos, cot, csc) all have negative derivatives, (2) tan and cot derivatives are squared functions (sec² and -csc²), (3) sec and csc derivatives are products (sec·tan and -csc·cot). Master sin → cos first, then the patterns give you the rest!

Q4: When do I need to use the chain rule with trig functions?

Answer: Use the chain rule whenever there's anything other than just "x" inside the trig function. Examples: sin(2x), cos(x²), tan(3x+1) all need chain rule. If it's just sin(x), cos(x), etc., no chain rule needed - use the basic derivative formula directly.

Q5: What's the difference between sec and csc?

Answer: sec(x) = 1/cos(x) (secant is reciprocal of cosine), and csc(x) = 1/sin(x) (cosecant is reciprocal of sine). Their derivatives are: d/dx[sec(x)] = sec(x)tan(x) and d/dx[csc(x)] = -csc(x)cot(x). Notice csc has a negative derivative (co-function rule!).

Q6: Why is d/dx[tan(x)] = sec²(x) and not just sec(x)?

Answer: This comes from tan(x) = sin(x)/cos(x). Using the quotient rule: [cos²(x) + sin²(x)]/cos²(x) = 1/cos²(x) = sec²(x). The squared comes from having cos²(x) in the denominator. Similarly, d/dx[cot(x)] = -csc²(x) for the same reason.

Q7: How are inverse trig derivatives different?

Answer: Inverse trig derivatives involve algebraic expressions (often with square roots) rather than other trig functions. For example, d/dx[arcsin(x)] = 1/√(1-x²) and d/dx[arctan(x)] = 1/(1+x²). They're derived using implicit differentiation and are used in integration.

Q8: Do I need to memorize all trig identities for derivatives?

Answer: No, but knowing basic identities helps simplify answers: sin²(x) + cos²(x) = 1, tan(x) = sin(x)/cos(x), sec(x) = 1/cos(x), and the double angle formulas. These help you recognize when to simplify d/dx results and verify your answers.

🚀 Practice Trig Derivatives!

Use our free calculators to verify your trig derivatives and see step-by-step solutions!

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📌 Key Takeaways

  • Six trig derivatives: sin→cos, cos→-sin, tan→sec², cot→-csc², sec→sec·tan, csc→-csc·cot
  • Co-function rule: cos, cot, and csc ALL have negative derivatives
  • Pattern recognition: tan/cot give squared results, sec/csc give products
  • Always use chain rule when trig functions have anything other than x inside
  • Sin-cos cycle repeats: sin→cos→-sin→-cos→sin (every 4 derivatives)
  • Inverse trig derivatives involve algebraic expressions with square roots
  • Common mistake: forgetting the negative sign on cos, cot, and csc derivatives
  • Practice makes perfect: do 10 problems daily until automatic
  • Proofs help understanding: knowing WHY helps you remember formulas
  • Use calculators to check: verify your work and learn from mistakes

Next Steps in Your Learning

  1. Memorize the six basic formulas - Write them 10 times each
  2. Learn the co-function pattern - This gives you all the signs
  3. Practice with chain rule - Most real problems need it
  4. Work mixed problems - Combine trig derivatives with product/quotient rules
  5. Master inverse trig derivatives - Important for integration later
  6. Use our calculators to verify your answers and see detailed solutions