Derivative of Sin- Rules and Examples
The Derivative of Sin: What You Actually Need to Know
If you're taking calculus, you've probably encountered sin(x) more times than you'd like. The derivative of sin is one of the first "non-trivial" derivatives you'll memorize, and it's the foundation for derivatives of other trig functions you'll learn later.
Here's the rule. Memorize it now:
d/dx[sin(x)] = cos(x)
That's it. The derivative of sine is cosine. Simple, but there are complications when you apply the chain rule, product rule, or quotient rule.
The Basic Derivative Rule
The derivative of sin(x) with respect to x equals cos(x). This comes from the limit definition of a derivative combined with the angle addition formula. You don't need to derive it every time—you just need to remember it.
Some examples with just x as the variable:
- d/dx[sin(x)] = cos(x)
- d/dx[sin(3x)] = 3cos(3x) (chain rule applied)
- d/dx[sin(x²)] = 2x·cos(x²) (chain rule applied)
Chain Rule: The Most Common Complication
When you have sin(u) where u is a function of x, you multiply by the derivative of u. The formula is:
d/dx[sin(u)] = cos(u) · du/dx
Let's work through examples that actually trip people up.
Example 1: sin(5x)
u = 5x, so du/dx = 5
d/dx[sin(5x)] = cos(5x) · 5 = 5cos(5x)
Example 2: sin(x² + 3x)
u = x² + 3x, so du/dx = 2x + 3
d/dx[sin(x² + 3x)] = cos(x² + 3x) · (2x + 3)
That's the full answer. Don't simplify the cosine argument—that's already simplified.
Example 3: sin(√x)
u = √x = x^(1/2), so du/dx = (1/2)x^(-1/2) = 1/(2√x)
d/dx[sin(√x)] = cos(√x) · 1/(2√x)
The answer is cos(√x)/(2√x). You can leave it like this or write it as (1/2)x^(-1/2)cos(√x).
Product Rule and Quotient Rule Applications
Sometimes you'll have sin multiplied or divided by another function. That's when you bring in the product rule or quotient rule.
Product Rule Example: x·sin(x)
Product rule: d/dx[f·g] = f'·g + f·g'
f = x, f' = 1
g = sin(x), g' = cos(x)
d/dx[x·sin(x)] = 1·sin(x) + x·cos(x)
Answer: sin(x) + x·cos(x)
Quotient Rule Example: sin(x)/x
Quotient rule: d/dx[f/g] = (f'·g - f·g')/g²
f = sin(x), f' = cos(x)
g = x, g' = 1
d/dx[sin(x)/x] = (cos(x)·x - sin(x)·1)/x²
Answer: (x·cos(x) - sin(x))/x²
Second Derivative: d²/dx²[sin(x)]
Take the derivative twice:
First derivative: cos(x)
Second derivative: d/dx[cos(x)] = -sin(x)
Notice the pattern. The derivatives of sin cycle every four:
- sin(x) → cos(x) → -sin(x) → -cos(x) → sin(x) → ...
This cycle is useful when you're solving differential equations or working with higher-order derivatives.
Derivatives of Related Trig Functions
While you're at it, here are the other trig derivatives you'll need:
| Function | Derivative |
|---|---|
| sin(x) | cos(x) |
| cos(x) | -sin(x) |
| tan(x) | sec²(x) |
| cot(x) | -csc²(x) |
| sec(x) | sec(x)tan(x) |
| csc(x) | -csc(x)cot(x) |
The derivative of cos(x) is the one people confuse most often. It's -sin(x), not sin(x).
Getting Started: How to Find the Derivative of Any Sine Function
Follow these steps in order:
- Identify the inner function u. What is inside the sin()? That's your u.
- Find du/dx. Differentiate u with respect to x.
- Write cos(u). Take the cosine of the inner function.
- Multiply by du/dx. This is your final answer.
- Simplify if needed. Expand, combine like terms, or factor if it makes the answer cleaner.
Work through 5-10 problems using this checklist. By the third problem, you'll stop thinking about the steps and just do them automatically.
Common Mistakes to Avoid
Forgetting the chain rule. This is the #1 error. d/dx[sin(3x)] is NOT cos(3x). It's 3cos(3x).
Dropping the negative sign. The derivative of cos(x) is -sin(x). The negative matters.
Not simplifying arguments. cos(2x+1) stays as cos(2x+1). Don't try to distribute the cosine.
Over-complicating simple problems. d/dx[sin(x)] is just cos(x). Don't add unnecessary steps.
When You'll Actually Use This
Physics problems involving waves, oscillations, or circular motion. Engineering problems with alternating current. Any situation where something varies sinusoidally with time.
If you're just in a calculus class to fulfill a requirement, you still need to know this for the exam. The chain rule with trig functions appears on nearly every calculus test.