Cot Derivative
- Introduction
- Features of `cot` Function
- Formula for Derivative of `cot x`
- Proof of Derivative of `cot x` by First Principle
- Proof by Derivative of `cot x` by Quotient Rule
- Proof of Derivative of `cot x` by First Principle
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
In a right-angled triangle, `cot x`, also called the `\text{cotangent}` `x`, is the ratio of the adjacent side of `x` to the opposite side of `x`. More specifically, `cot(x)` is defined as the reciprocal of the tangent function, or `cot(x) = 1/tan(x)`. Alternatively, the cot function can also be expressed in terms of the cosine and sine functions as `cot(x) = cos(x)/sin(x)` and is used in the differentiation of `cot (x)`. The cot derivative is the rate of change of the cotangent function with respect to the angle `x`.
Features of `cot` function
Here is the definition and properties of the cotangent function, \( \cot(x) \). It's a valuable addition to understanding this trigonometric function.
`1`. Definition: \( \cot(x) \) represents the ratio of the adjacent side of `x` to the opposite side of `x` inf a right triangle. It is defined as the reciprocal of the tangent function `(cot(x) = \frac{1}{\tan(x)})` or alternatively in terms of the `\text{cosine}` and `\text{sine}` functions `cot(x) = \frac{\cos(x)}{\sin(x)}`.
`2`. Periodicity: The `\text{cotangent}` function is periodic with a period of \( \pi \).
`3`. Singularities: It has singularities at the zeros of the `\text{sine}` function, corresponding to the vertical asymptotes of the graph. These occur where \( \sin(x) = 0 \).
`4`. Range: The range of \( \cot(x) \) is all real numbers except `0`, as it cannot attain `0` due to the fact that both \( \sin(x) \) and \( \cos(x) \) cannot be zero simultaneously.
Formula for Derivative of `cot x`
The derivative of the cotangent function \( \cot(x) \) can be found using calculus. Let's denote it as `\frac{d}{dx}[\cot(x)]`. The derivative of the cotangent function `cot(x)` is `-\text{cosec}^2(x)`, where `\text{cosec}``(x)` is the cosecant function. This indicates that the negative of the square of the cosecant function evaluated at `x` equals the rate of change of the tangent function with respect to `x`. It may be written as follows in mathematical notation:
`\frac{d}{dx} \cot(x) = -\csc^2(x)`
Proving Derivative of `cot x` Using First Principle
To prove the derivative of \( \cot(x) \) using the first principle, we start with the definition of the cotangent function:
`\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}`
Now, let's find the derivative of \( \cot(x) \) using the definition of the derivative:
`\cot'(x) = \lim_{h \to 0} \frac{\cot(x + h) - \cot(x)}{h}`
Substitute the expression for \( \cot(x) \):
`= \lim_{h \to 0} \frac{\frac{\cos(x + h)}{\sin(x + h)} - \frac{\cos(x)}{\sin(x)}}{h}`
`= \lim_{h \to 0} \frac{\cos(x + h)\sin(x) - \cos(x)\sin(x + h)}{h\sin(x)\sin(x + h)}`
Using trigonometric identity \( \sin A \cos B - \cos A \sin B = \sin(A-B) \), we can write
`\cot'(x) = \lim_{h \to 0} \sin(x-(x + h))/{h\sin(x)\sin(x + h)`
`= \lim_{h \to 0} \frac{\sin(- h)}{h\sin(x)\sin(x + h)}`
We have \(\sin(-h) = -\sin(h) \), therefore
`\cot'(x) = -\lim_{h \to 0} \frac{\sin(h)}{h}\lim_{h \to 0}\frac{1}{\sin(x)\sin(x + h)}`
Since `\lim_{h \to 0} \frac{\sin(h)}{h} = 1`, applying these limits, we get:
`\cot'(x) = -1\cdot\frac{1}{\sin(x)\sin(x + 0)} = \frac{-1}{\sin^2(x)}`
Hence
\( \cot'(x) = -\csc^2(x) \)
So, the derivative of \( \cot(x) \) using the first principle is \( \cot'(x) = -\csc^2(x) \).
Proving Derivative of `cot x` Using Chain Rule
Below steps demonstrate how to use the chain rule to prove the derivative of \( \cot(x) \).
`1`. Reciprocal property: We start by expressing \( y = \cot(x) \) as `y = (1/\tan(x)) = (\tan(x))^{-1}`, recognizing that \( \cot(x) \) and \( \tan(x) \) are reciprocals of each other.
`2`. Apply power rule and chain rule: Since there's a power `(-1)` involved, we apply the power rule. Then, using the chain rule, we differentiate \( \tan(x) \) with respect to \( x \), which yields \( \sec^2(x) \).
`3`. Substitute derivative of \( \tan(x) \): Substitute the derivative of \( \tan(x) \) into the expression obtained from the power rule.
`4`. Simplify using trigonometric identities: Recognize that `\cot(x) = \frac{\cos(x)}{\sin(x)}` and `\sec(x) = \frac{1}{\cos(x)}`. Substitute these into the expression and simplify.
`5`. Final Result: Obtain the final result, \( y' = -\csc^2(x) \), which confirms the derivative of \( \cot(x) \).
It shows how to derive the derivative of \( \cot(x) \) using the chain rule and trigonometric identities.
Starting with `y' = (-1) \cdot \frac{1}{\tan^2(x)} \cdot \frac{d}{dx}(\tan(x))`, where `\frac{d}{dx}(\tan(x)) = \sec^2(x)`, we have:
`y' = -\frac{1}{\tan^2(x)} \cdot \sec^2(x)`
Since `\tan(x) = \frac{\sin(x)}{\cos(x)}`, we can rewrite `\frac{1}{\tan^2(x)}` as `\frac{\cos^2(x)}{\sin^2(x)}`.
So, \( y' \) becomes:
`y' = -\frac{\cos^2(x)}{\sin^2(x)} \cdot \sec^2(x)`
Now, `\sec(x) = \frac{1}{\cos(x)}`, so `\sec^2(x) = \frac{1}{\cos^2(x)}`.
Substituting this, we get:
`y' = -\frac{\cos^2(x)}{\sin^2(x)} \cdot \frac{1}{\cos^2(x)}`
`= -\frac{1}{\sin^2(x)}`
Since, `\frac{1}{\sin(x)} = \csc(x)`, `\frac{1}{\sin^2(x)} = \csc^2(x)`.
Therefore, \( y' = -\csc^2(x) \) Proved
Therefore, the \( \cot(x) \) derivative with respect to \( x \) using the chain rule is \( -\csc^2(x) \).
Proving Derivative of `cot x` Using Quotient Rule
To find the derivative of \( \cot(x) \) using the quotient rule, we first express \( \cot(x) \) as reciprocal of \( \tan(x) \). That is `\cot(x) = \frac{1}{\tan(x)}`.
The quotient rule states that if `y = \frac{u}{v}`, then the derivative of \( y \) with respect to \( x \) is given by:
`y' = \frac{u'v - uv'}{v^2}`
where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) with respect to \( x \), respectively.
In the case of `\cot(x) = \frac{1}{\tan(x)}`, let \( u = 1 \) and \( v = \tan(x) \).
Then, \( u' = 0 \) (since the derivative of a constant is zero) and \( v' = \sec^2(x) \) (the derivative of \( \tan(x) \)).
Now, applying the quotient rule:
`\frac{d}{dx}[\cot(x)] = \frac{(0)(\tan(x)) - (1)(\sec^2(x))}{(\tan(x))^2}`
`= -\frac{\sec^2(x)}{\tan^2(x)}`
`= -\frac{\sec^2(x)}{\sec^2(x) - 1}`
Using the trigonometric identity \( \tan^2(x) = \sec^2(x) - 1 \), we can simplify further:
`= -\frac{\sec^2(x)}{\sec^2(x) - 1}`
`= -\frac{1}{1 - \cos^2(x)}`
`= -\frac{1}{\sin^2(x)}`
\( = -\csc^2(x) \) Proved
Therefore, the \( \cot(x) \) derivative with respect to \( x \) using the quotient rule is \( -\csc^2(x) \). which is consistent with the result obtained using the chain rule or the first principle.
Solved Examples
Example `1`: Find the derivative of \( y = \cot(6x) \).
Solution:
Using the chain rule and derivative formula for \( \cot(x) \), which is \( -\csc^2(x) \), the derivative of \( y = \cot(6x) \) is \( -6\csc^2(x) \).
Example `2`: Differentiate \( y = \cot^2(x) + 3x \).
Solution:
To find the derivative of \( y = \cot^2(x) + 3x \), we differentiate each term separately:
`\frac{d}{dx}(\cot^2(x)) = -2\cot(x) \csc^2(x)`
`\frac{d}{dx}(3x) = 3`
So, the derivative of \( y = \cot^2(x) + 3x \) is \( -2\cot(x) \csc^2(x) + 3 \).
Example `3`: Find `\frac{dy}{dx}` for `y = \frac{\cot(x)}{x^2}`.
Solution:
Using the quotient rule, we have:
`\frac{d}{dx}\left(\frac{\cot(x)}{x^2}\right) = \frac{x^2 \cdot \frac{d}{dx}(\cot(x)) - \cot(x) \cdot 2x}{(x^2)^2}`
`= \frac{x^2 \cdot (-\csc^2(x)) - 2x \cot(x)}{x^4}`
`= -\frac{x^2 \csc^2(x) + 2x \cot(x)}{x^4}`
`= -\frac{x \csc^2(x) + 2 \cot(x)}{x^3}`
Therefore, the derivative of `y = \frac{\cot(x)}{x^2}` is `-\frac{x \csc^2(x) + 2 \cot(x)}{x^3}`.
Example `4`: Find the derivative of `y = \frac{\cot(x)}{\sin(x)}`.
Solution:
Using the quotient rule, we have:
`\frac{d}{dx}\left(\frac{\cot(x)}{\sin(x)}\right) = \frac{\sin(x) \cdot \frac{d}{dx}(\cot(x)) - \cot(x) \cdot \frac{d}{dx}(\sin(x))}{(\sin(x))^2}`
`= \frac{\sin(x) \cdot (-\csc^2(x)) - \cot(x) \cdot \cos(x)}{\sin^2(x)}`
`= -\frac{\sin(x)}{\sin^2(x)} \cdot \csc^2(x) - \frac{\cot(x) \cdot \cos(x)}{\sin^2(x)}`
`= -\csc(x) \csc^2(x) - \csc(x) \cot^2(x)`
`= -\csc(x) (\csc^2(x) + \cot^2(x))`
Therefore, the derivative of `y = \frac{\cot(x)}{\sin(x)}` is \( -\csc(x) (\csc^2(x) + \cot^2(x)) \).
Example `5`: Find the derivative of \( y = \cot(x) \sin(x) \).
Solution:
To find the derivative of \( y = \cot(x) \sin(x) \), we use the product rule:
`\frac{d}{dx}(\cot(x) \sin(x)) = \cot(x) \frac{d}{dx}(\sin(x)) + \sin(x) \frac{d}{dx}(\cot(x))`
\( = \cot(x) \cdot \cos(x) + \sin(x) \cdot (-\csc^2(x)) \)
\( = \cot(x) \cos(x) - \sin(x) \csc^2(x) \)
`= \frac{\cos^2(x)}{\sin (x)} - \frac{1}{\sin (x)}`
`= -\frac{1}{\sin(x)}(1 - \cos^2(x))`
`= -\frac{1}{\sin(x)}\sin^2(x)`
\( = -\sin(x) \)
Therefore, the derivative of \( y = \cot(x) \sin(x) \) is \( -\sin(x) \).
Practice Problems
Q`1`. Find the derivative of \( y = \cot^2(x) \).
- \( -2\cot(x) \csc^2(x) \)
- \( 2\cot(x) \csc^2(x) \)
- \( -2\cot(x) \csc^3(x) \)
- \( -2\cot(x) \cos^2(x) \)
Answer: a
Q`2`. Find the derivative of \( y = 2x \cot(x) \).
- \( 2\cot(x) - 2x\cos^2(x) \)
- \( 2\cot(x) - 2x\csc^2(x) \)
- \( 2\cot(x) + 2x\csc^2(x) \)
- \( 2\tan(x) - 2x\csc^2(x) \)
Answer: b
Q`3`. Find the derivative of \( y = 3\cot(x) \).
- \( -2\csc^2(x) \)
- \( -3\csc^3(x) \)
- \( -3\csc^2(x) \)
- \( 3\csc^2(x) \)
Answer: c
Q`4`. Find the derivative of `y = \frac{x^2}{\cot(x)}`.
- \( 3x \csc(x) + x^2 \sin^3(x) \)
- \( 2x \csc(x) + x^2 \cos^2(x) \)
- \( 2x \tan(x) + x^2 \sec^2(x) \)
- \( 2x \csc(x) + x^2 \sin^2(x) \)
Answer: c
Q`5`. Find the derivative of \( y = e^x \cot(x) \).
- \( -e^x \csc^2(x) + e^x \cot(x) \)
- \( -e^x \cos^2(x) + e^x \cot(x) \)
- \( -e^x \csc^3(x) + e^x \cot(x) \)
- \( -e^x \csc^2(x) + e^x \tan(x) \)
Answer: a
Frequently Asked Questions
Q`1`. What is the derivative of \( \cot(x) \)?
Answer: The derivative of \( \cot(x) \) with respect to \( x \) is given by \( -\csc^2(x) \).
Q`2`. What is the relationship between \( \cot(x) \) and \( \tan(x) \)?
Answer: \( \cot(x) \) and \( \csc(x) \) are reciprocal trigonometric functions. `\cot(x) = \frac{1}{\tan(x)}`.
Q`3`. Can the derivative of \( \cot(x) \) be simplified further?
Answer: The derivative \( -\csc^2(x) \) is the simplest form of the derivative of \( \cot(x) \), and it cannot be simplified further.
Q`4`. How do you find the derivative of a function involving \( \cot(x) \)?
Answer: To find the derivative of a function involving \( \cot(x) \), apply the derivative rules such as the product rule, quotient rule, or chain rule as appropriate, and use the derivative of \( \cot(x) \) as (\( -\csc^2(x) \)) when necessary.