How to find the derivative of cosecant x using the quotient rule in Calculus

📚 Understanding the Cosecant Function

The cosecant function, denoted as $\csc(x)$, is a trigonometric function defined as the reciprocal of the sine function. Mathematically, this is expressed as:

$\csc(x) = \frac{1}{\sin(x)}$

To find the derivative of $\csc(x)$ using the quotient rule, we first need to express it in terms of sine and then apply the rule.

📜 History and Background

Trigonometric functions have ancient roots, dating back to the study of astronomy and geometry by Greek mathematicians like Hipparchus and Ptolemy. The cosecant function, as a reciprocal of the sine function, became more formally defined as trigonometry developed through the work of mathematicians in India and later in Europe during the medieval and Renaissance periods. Understanding these functions was crucial for advancements in fields like navigation, surveying, and physics.

➗ Key Principles: The Quotient Rule

The quotient rule is a method used to find the derivative of a function that is expressed as the quotient of two other functions. If we have a function $f(x)$ defined as:

$f(x) = \frac{u(x)}{v(x)}$

Then, the derivative of $f(x)$, denoted as $f'(x)$, is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$

Where $u'(x)$ and $v'(x)$ are the derivatives of $u(x)$ and $v(x)$, respectively.

📝 Applying the Quotient Rule to $\csc(x)$

Let's find the derivative of $\csc(x)$ using the quotient rule. Recall that $\csc(x) = \frac{1}{\sin(x)}$. Here, we can consider $u(x) = 1$ and $v(x) = \sin(x)$.

• 🔍 Identify $u(x)$ and $v(x)$: $u(x) = 1$ and $v(x) = \sin(x)$.
• 💡 Find the derivatives: $u'(x) = 0$ and $v'(x) = \cos(x)$.
• ➗ Apply the quotient rule formula:

$\frac{d}{dx} \csc(x) = \frac{d}{dx} \left( \frac{1}{\sin(x)} \right) = \frac{(0)(\sin(x)) - (1)(\cos(x))}{(\sin(x))^2}$

Simplify the expression:

$\frac{-\cos(x)}{\sin^2(x)} = -\frac{\cos(x)}{\sin(x)} \cdot \frac{1}{\sin(x)}$

Recognize trigonometric identities:

Since $\cot(x) = \frac{\cos(x)}{\sin(x)}$ and $\csc(x) = \frac{1}{\sin(x)}$, we can rewrite the derivative as:

$\frac{d}{dx} \csc(x) = -\cot(x) \csc(x)$

➗ Examples

Let's go through some examples.

• 🌱 Example 1: Find the derivative of $f(x) = 5\csc(x)$.

  Solution: $f'(x) = -5\csc(x)\cot(x)$

• 🌳 Example 2: Find the derivative of $f(x) = \csc(3x)$.

  Solution: $f'(x) = -3\csc(3x)\cot(3x)$

• 🌲 Example 3: Find the derivative of $f(x) = x^2 \csc(x)$.

  Solution: Using the product rule and the derivative of $\csc(x)$, we get $f'(x) = 2x\csc(x) - x^2\csc(x)\cot(x)$

✅ Conclusion

Using the quotient rule to find the derivative of $\csc(x)$ involves expressing it as the reciprocal of $\sin(x)$ and applying the quotient rule formula. This results in the derivative being $-\csc(x)\cot(x)$. Understanding and applying this method simplifies the process and provides a clear path to the solution.