The derivative of $\text{cot}(x)$ with respect to $x$ is $-\text{csc}^2x$.
Expressed mathematically, this means:
$\frac{d}{dx}(\text{cot }x) = -\text{csc}^2x$
This can also be concisely written using prime notation as:
$(\text{cot }x)' = -\text{csc}^2x$
Understanding the Derivative of cot(x)
The derivative of a function represents its instantaneous rate of change. For trigonometric functions like $\text{cot}(x)$, these derivatives are fundamental in calculus and various scientific and engineering applications.
As stated, the derivative of $\text{cot }x$ with respect to $x$ is -1 times the square of $\text{csc }x$. Here's a breakdown:
- $\text{cot}(x)$: The cotangent function, which is the reciprocal of $\text{tan}(x)$, or $\frac{\text{cos }x}{\text{sin }x}$.
- $\text{csc}(x)$: The cosecant function, which is the reciprocal of $\text{sin}(x)$, or $\frac{1}{\text{sin }x}$.
- $-\text{csc}^2x$: This indicates that the result is negative and that the cosecant function is squared, meaning $(\text{csc }x)^2$.
This specific derivative is derived using the quotient rule on $\frac{\text{cos }x}{\text{sin }x}$ or through other fundamental trigonometric identities and derivative rules.
Key Trigonometric Derivatives Table
Understanding the derivative of $\text{cot}(x)$ is part of a broader set of standard trigonometric derivatives. Here's a quick reference for commonly used derivatives:
| Function ($f(x)$) | Derivative ($f'(x)$ or $\frac{d}{dx}f(x)$) |
|---|---|
| $\text{sin }x$ | $\text{cos }x$ |
| $\text{cos }x$ | $-\text{sin }x$ |
| $\text{tan }x$ | $\text{sec}^2x$ |
| $\text{cot }x$ | $-\text{csc}^2x$ |
| $\text{sec }x$ | $\text{sec }x \text{ tan }x$ |
| $\text{csc }x$ | $-\text{csc }x \text{ cot }x$ |
Applying the Derivative: Examples
The derivative formula for $\text{cot}(x)$ is straightforward to apply in various scenarios.
Example 1: Basic Application
Question: Find the derivative of $f(x) = 5\text{cot }x$.
Solution:
Using the constant multiple rule, we can pull the constant out:
$\frac{d}{dx}(5\text{cot }x) = 5 \cdot \frac{d}{dx}(\text{cot }x)$
Substitute the known derivative of $\text{cot }x$:
$= 5(-\text{csc}^2x)$
$= -5\text{csc}^2x$
Example 2: Using the Chain Rule
Question: Find the derivative of $g(x) = \text{cot}(3x^2 + 1)$.
Solution:
Here, we need to apply the chain rule, which states $\frac{d}{dx}f(u) = f'(u) \cdot \frac{du}{dx}$.
Let $u = 3x^2 + 1$.
Then $\frac{du}{dx} = \frac{d}{dx}(3x^2 + 1) = 6x$.
Now, substitute into the chain rule formula:
$\frac{d}{dx}(\text{cot}(u)) = -\text{csc}^2(u) \cdot \frac{du}{dx}$
Substitute back $u$ and $\frac{du}{dx}$:
$= -\text{csc}^2(3x^2 + 1) \cdot (6x)$
$= -6x\text{csc}^2(3x^2 + 1)$
Why is this Derivative Important?
The derivative of $\text{cot}(x)$ is crucial for:
- Solving Related Rates Problems: Calculating how the rate of change of one quantity affects another.
- Optimization: Finding maximum or minimum values of functions involving trigonometric terms.
- Physics and Engineering: Describing oscillatory motion, wave phenomena, and electrical circuits.
- Graphing: Determining the slopes of tangent lines to cotangent graphs and identifying intervals of increase or decrease.
Further Resources
For more in-depth understanding of derivatives and their applications, you can explore resources like Khan Academy's Calculus section.
