Derivative of `csc(x)`
- Introduction
- Exploring the Derivative of `"cosecant x"`
- Derivative of `"cosec x"` Formula
- Establishing the Proof for the Derivative of `"cosec x"`
- Derivative of `"cosec x"` Using the First Principle of Derivative
- Derivative of `"cosec x"` Using the Quotient Rule
- Derivative of `"cosec x"` Using the Chain Rule
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
The derivative of `"cosecant x"` (also written as `csc x`) represents the rate of change of the cosecant function with respect to angle `x`. `"Cosecant x"`, denoted as `csc x`, is defined as the reciprocal of the sine function, meaning it's `1` divided by `sin x`. To understand its derivative, we first recall that `"cosecant x"` relates to the sides of a right-angled triangle as the ratio of the hypotenuse to the perpendicular side. Through examples, we'll illustrate how to apply the derivative of `"cosecant x"` in practical scenarios, facilitating a comprehensive understanding of this fundamental trigonometric concept.
Exploring the Derivative of `"Cosecant x"`
To comprehend the differentiation of `"cosecant x"` with respect to angle `x`, we express it as `{d("cosec x")}/dx = "-cot x cosec x"`. This equation denotes the rate of change of `"cosecant x"` concerning the angle `x`. The derivative of `csc x` can be computed using the derivative of `"sine x"`. Various methods, including the chain rule, first principle, and quotient rule, are available to differentiate `"cosecant x"`. By utilizing existing trigonometric identities and differentiation techniques, we establish that the derivative of `csc x` is `"-cot x cosec x"`.
The derivative value of `"cosec x"` has the following formula:
- `"d(cosec x)"/dx = "-cot x cosec x"`
- `"(cosec x)"’ = "-cot x cosec x"`
Proving Derivative of `"Cosec x"` Using the First Principle
There are several ways to establish the derivative of `"cosecant x"`:
- By using the First Principle of Derivative
- By applying the Quotient Rule
- By applying the Chain Rule
To derive the derivative of \( \csc x \) using the first principle of derivatives, we begin by understanding that a derivative essentially measures the rate of change. To find the derivative of \( \csc(x) \) using the first principles (also known as the definition of the derivative), we'll apply the limit definition of the derivative:
`f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}`
Now let’s start differentiating `"cosec x"`:
Let \(y = f(x) = \csc(x)\)
Then \(f(x+h) = \csc(x+h)\)
Accordingly, from the first principle,
`\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}`
`= \lim_{h \to 0} \frac{\csc(x + h) - \csc(x)}{h}`
`= \lim_{h \to 0} \frac{1}{h} (\frac{1}{\sin(x + h)} - \frac{1}{\sin x})`
`= \lim_{h \to 0} \frac{\sin x - \sin(x + h)}{h \cdot \sin x \cdot \sin(x + h)}`
`= \lim_{h \to 0} \frac{- (\sin(x + h) - \sin x)}{h \cdot \sin x \cdot \sin(x + h)}`
`= \lim_{h \to 0} -\frac{\sin(x + h) - \sin x}{h} \times \lim_{h \to 0} \frac{1}{\sin x \cdot \sin(x + h)}`
Since `\lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h} = \cos x`, we can write
`= -\cos x \times \frac{1}{\sin^2 x}`
`= -\frac{\cos x}{\sin x} \times \frac{1}{\sin x}`
`= -\cot x \csc x`
Therefore, the differentiation of \(\csc x\) is \(-\cot x \csc x\) using the first principle of derivatives.
Proving Derivative of `"Cosec x"` Using the Quotient Rule
To prove the derivative of \( \csc x \) using the quotient rule, we'll express \(\csc x\) as \(\frac{1}{\sin x}\). Then, we apply the quotient rule, which states that if we have a function of the form \( \frac{u(x)}{v(x)} \), then its derivative is given by:
\( \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} \)
Given \( y = \csc x \), we have \( y = \frac{1}{\sin x} \).
\( y' = \frac{d}{dx}(\frac{1}{\sin x}) \)
By applying the Quotient Rule:
`y' = \frac{\frac{d}{dx}(1) \cdot \sin x - 1 \cdot \frac{d}{dx}(\sin x)}{\sin^2 x}`
`= \frac{(0 \cdot \sin x) - (1 \cdot \cos x)}{\sin^2 x}` `(\text{since} \frac{d}{dx}(1)) = 0`
`= -\frac{\cos x}{\sin^2 x}`
`= -\frac{\cos x}{\sin x} \times \frac{1}{\sin x}`
`= -\cot x \csc x`
Therefore, the differentiation of \(\csc x\) is \(-\cot x \csc x\) using the quotient rule.
Proving Derivative of `"Cosec x"` Using the Chain Rule
To prove the derivative of \( \csc x \) using the chain rule, we begin by expressing \( \csc x \) as \( y = \frac{1}{\sin x} = \sin^{-1} x \). Then, apply the chain rule, which states that if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
Given \( y = \csc x \), we have \( y = \frac{1}{\sin x} \).
\( y = \sin^{-1} x \)
By applying the Chain Rule:
\(\frac{dy}{dx} = -(\sin x)^{-2} \cdot \frac{d(\sin x)}{dx} \)
\(= -(\sin x)^{-2} \cdot \cos x\)
\(= -\frac{\cos x}{\sin^2 x} \)
\(= -\frac{\cos x}{\sin x} \times \frac{1}{\sin x}\)
\(= -\cot x \csc x \)
Therefore, the differentiation of \(\csc x\) is \(-\cot x \csc x\) using the chain rule.
Solved Examples
Example `1`. Compute \( \frac{d}{dx} (\csc(2x)) \).
Solution:
We'll use the chain rule of differentiation:
\(\frac{d}{dx} (f(g(x))) = f'(g(x)) \cdot g'(x) \)
In this case, \( f(g(x)) = \csc(2x) \) and \( g(x) = 2x \).
First, let's find the derivative of the outer function.
\( f'(g(x)) = \frac{d}{dx} (\csc(2x)) = -\csc(2x) \cot(2x) \)
Now, let's find the derivative of the inner function.
\( g'(x) = \frac{d}{dx} (2x) = 2 \)
By applying the chain rule, we get \( \frac{d}{dx} (\csc(2x)) = -2\csc(2x) \cot(2x) \).
Example `2`. Determine \( \frac{d}{dx} (\csc^2(x)) \).
Solution:
Using the chain rule of differentiation:
\( \frac{d}{dx} ((f(x))^n) = n(f(x))^{n-1} \cdot f'(x) \)
\(\csc^2(x)\) can be written as \((\csc(x))^2\)
`\frac{d}{dx} ((\csc(x))^2) = 2(\csc(x))^{2-1} \frac{d}{dx}(\csc(x))`
`= 2\csc(x) (-\cot(x) \csc(x))`
`= -2\csc^2(x) \cot(x)`
By applying the chain rule, we get \( \frac{d}{dx} (\csc^2(x)) = -2\csc^2(x) \cot(x)\).
Example `3`. Find \( \frac{d}{dx} (\frac{\csc(x)}{x^2}) \).
Solution:
Given the function \( \frac{d}{dx} (\frac{\csc(x)}{x^2}) \), we'll use the quotient rule of differentiation:
\( \frac{d}{dx} \left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)
Here, \( u(x) = \csc(x) \) and \( v(x) = x^2 \).
First, let's find the derivative of \( u(x) = \csc(x) \).
\( u'(x) = \frac{d}{dx} (\csc(x)) = -\csc(x) \cot(x) \)
Next, let's find the derivative of \( v(x) = x^2 \).
\(v'(x) = \frac{d}{dx} (x^2) = 2x \)
Now, apply the quotient rule:
`\frac{d}{dx} (\frac{\csc(x)}{x^2}) = \frac{(-\csc(x) \cot(x)) \cdot x^2 - \csc(x) \cdot 2x}{(x^2)^2}`
`= \frac{-x^2 \csc(x) \cot(x) - 2x\csc(x)}{x^4}`
`= \frac{-x\csc(x)(x \cot(x) + 2)}{x^4}`
`\text{Now, simplify this expression:}`
`= \frac{-\csc(x)(x \cot(x) + 2)}{x^3}`
By applying the quotient rule, we get \( \frac{d}{dx} (\frac{\csc(x)}{x^2}) = \frac{-\csc(x)(x \cot(x) + 2)}{x^3} \).
Example `4`. Compute \( \frac{d^2}{dx^2} (\csc(x)) \).
Solution:
We know that the first derivative of \( \csc x \) is \( -\csc x \cot x \).
To compute the second derivative of \( \csc x \), we differentiate \( -\csc x \cot x \) using the product rule.
\( \frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)\)
Here, \( u(x) = -\csc(x) \) and \( v(x) = \cot(x) \).
First, let's find the derivative of \( u(x) = -\csc(x) \).
\( u'(x) = \frac{d}{dx} (-\csc(x)) = \csc(x) \cot(x) \)
Next, let's find the derivative of \( v(x) = \cot(x) \).
\(v'(x) = \frac{d}{dx} (\cot(x)) = -\csc^2(x) \)
Now, apply the product rule:
`\frac{d}{dx} \left[-\csc(x) \cot(x)\right] = \csc(x) \cot(x) \cdot \cot(x) + (-\csc(x)) \cdot (-\csc^2(x))`
`= \csc(x) \cot^2(x) + \csc^3(x)`
`= \csc(x) (\cot^2(x) + \csc^2(x))`
So, the second derivative of \( \csc(x) \) is \( \frac{d^2}{dx^2} (\csc(x)) = \csc(x) (\cot^2(x) + \csc^2(x)) \).
Example `5`. Calculate \( \frac{d}{dx}(2\csc(3x)) \).
Solution:
We'll use the chain rule of differentiation:
\(\frac{d}{dx} (f(g(x))) = f'(g(x)) \cdot g'(x) \)
In this case, \( f(g(x)) = \csc(3x) \) and \( g(x) = 3x \).
First, let's find the derivative of the outer function.
\( f'(g(x)) = \frac{d}{dx} (\csc(3x)) = -\csc(3x) \cot(3x) \)
Now, let's find the derivative of the inner function.
\( g'(x) = \frac{d}{dx} (3x) = 3 \)
Now, apply the chain rule:
\( \frac{d}{dx} (2\csc(3x)) = 2(-\csc(3x) \cot(3x)) \cdot 3 \)
\( \frac{d}{dx} (2\csc(3x)) = -6\csc(3x) \cot(3x) \)
By applying the chain rule, we get \( \frac{d}{dx} (2\csc(3x)) = -6\csc(3x) \cot(3x) \).
Practice Problems
Q`1`. Determine the derivative of \( f(x) = x^2\csc(x) \).
- \( x(2-x\cot(x))\csc(x) \)
- \( -\frac{6\csc^2(x) (x\cot(x) - 1)}{x^3} \)
- \( x(x\cot^2(x))\csc(x) \)
- \( -\frac{6\csc^2(x) (\cot(x) + 1)}{x^3} \)
Answer: a
Q`2`. Find the derivative of \( h(x) = \csc^2(x) \cdot \sin(x) \).
- \( -sin(x)cos(x)\)
- \( cot^2(x)\csc(x) \)
- \( cot(x)\csc^2(x) \)
- \( -cot(x)\csc(x) \)
Answer: d
Q`3`. Calculate the derivative of \( f(x) = \frac{\csc(x^2)}{x} \).
- \( -\frac{2x^2 (\cot(x^2)+1)\csc(x^2)}{x^2} \)
- \( -\frac{\csc(x^2)(2x\cot(x^2) + 1)}{x^2} \)
- \( -\frac{2x\csc(x^2) \cot(x^2) - \csc(x^2)}{x} \)
- \( -\frac{\csc(x^2)(2x^2\cot(x^2) + 1)}{x^2} \)
Answer: a
Q`4`. Calculate \( \frac{d}{dx}(\csc^2(3x)) \).
- \( -6\csc^2(3x) \cot(3x) \)
- \( -6\csc^2(3x) \)
- \( -3\csc(3x) \cot^2(3x) \)
- \( -3\csc(3x) \)
Answer: a
Q`5`. Find \( \frac{d}{dx}(\csc(5x)) \).
- \( -5\csc(5x) \)
- \( -5\csc(5x) \cot(5x) \)
- \( -\frac{1}{x} \csc(5x) \cot(5x) \)
- \( -\frac{1}{x} \csc(5x) \)
Answer: b
Frequently Asked Questions
Q`1`. What is the derivative of \( \csc x \)?
Answer: The derivative of \( \csc x \) with respect to \( x \) is given by \( -\csc x \cot x \).
Q`2`. How do you find the derivative of \( \csc x \)?
Answer: To find the derivative of \( \csc x \), you can use various methods, including the chain rule, limit definition, and quotient rule.
Q`3`. Can you explain the chain rule when finding the derivative of \( \csc x \)?
Answer: When applying the chain rule to \( \csc x \), you differentiate the outer function (\( \csc x \)) and then multiply it by the derivative of the inner function (\( x \)).
Q`4`. How does the derivative of \( \csc x \) relate to the graph of \( \csc x \)?
Answer: The derivative of \( \csc x \) gives the slope of the tangent line to the graph of \( \csc x \) at any point. It helps in understanding the rate of change of \( \csc x \) function across its domain.
Q`5`. What is the relationship between the derivatives of \( \csc x \) and \( \sin x \)?
Answer: The derivative of \( \csc x \) is the same as the derivative of the reciprocal of \( \sin x \). They are related through the reciprocal trigonometric identity and the derivative of \( \sin x \).