Problem 282 - Calculus Volume 1

Question

Find \(\frac{d y}{d x}\) for the given function. $$ y=\sqrt{\csc ^{-1} x} $$

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = \frac{-1}{2|x|\sqrt{x^2 - 1} \cdot \sqrt{\csc^{-1}x}}. \)

Step 1: Identify the Derivative to Calculate

We are tasked with finding the derivative \( \frac{d y}{d x} \) for the function \( y = \sqrt{\csc^{-1} x} \). This means we need the derivative of the inverse cosecant function inside a square root.

Step 2: Apply the Chain Rule

The function \( y = \sqrt{u} \) where \( u = \csc^{-1} x \). To differentiate, we use the chain rule: \( \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \). Here, \( y = u^{1/2} \), so \( \frac{d y}{d u} = \frac{1}{2} u^{-1/2} = \frac{1}{2\sqrt{u}} \).

Step 3: Differentiate the Inverse Cosecant Function

Next, differentiate \( u = \csc^{-1} x \). The derivative of \( \csc^{-1} x \) is \( \frac{-1}{|x|\sqrt{x^2 - 1}} \). Thus, \( \frac{d u}{d x} = \frac{-1}{|x|\sqrt{x^2 - 1}} \).

Step 4: Combine Derivatives Using the Chain Rule

Using the derivatives found, combine them: \[ \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} = \frac{1}{2\sqrt{\csc^{-1} x}} \times \frac{-1}{|x|\sqrt{x^2 - 1}}. \]

Step 5: Simplify the Expression

Substitute back \( u = \csc^{-1} x \) and express the derivative: \[ \frac{d y}{d x} = \frac{-1}{2|x|\sqrt{x^2 - 1} \cdot \sqrt{\csc^{-1}x}}. \] This is the simplified form of the derivative.

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus for differentiating composite functions, or functions within functions. Think of it as a way to peel back layers. To apply the Chain Rule, imagine your function is a nesting doll: the outer function and the inner function.
In the problem of finding the derivative of \( y = \sqrt{\csc^{-1} x} \), the square root function is the outer function, and the inverse cosecant is the inner function.
Here is a simple way to think about the steps:

• Identify the outer function's derivative: For \( y = u^{1/2} \), it is \( \frac{1}{2\sqrt{u}} \).
• Identify the derivative of the inner function: This is \( \csc^{-1} x \), which has its own specific rule.
• Multiply the derivatives: Combine them by multiplying the derivative of the outer function by the derivative of the inner function.

Understanding and practicing the Chain Rule unlocks numerous complex problems in calculus.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles when we know the ratios, such as sine or tangent. They are the reverse operations of standard trigonometric functions.
In this exercise, we focus on the inverse cosecant function, \( \csc^{-1} x \). Just like their counterparts, inverse trig functions have specific differentiation rules.
When differentiating \( \csc^{-1} x \), you'll see its unique formula, which includes components like the absolute value of \( x \) and radicals of the squared identity: \( \frac{-1}{|x|\sqrt{x^2 - 1}} \).
This is vital to differentiate inverse trigonometric functions since they appear frequently in calculus problems. Remember, these differentiation rules separate them from standard trig derivatives.

Derivative of Inverse Cosecant

Differentiating \( \csc^{-1} x \) requires knowing its specific formula. Unlike derivatives of simple polynomials or exponential functions, inverse trigonometric derivatives have intricate patterns.
The derivative of \( \csc^{-1} x \) is calculated as \( \frac{-1}{|x|\sqrt{x^2 - 1}} \). Here's a breakdown of which part affects the differentiation:

• \(-1\) comes from the reciprocal relationship in the cosecant function.
• The absolute value sign \(|x|\) ensures it accounts for all real values of \(x\).
• The \( \sqrt{x^2 - 1} \) derives from the trigonometric identity handlings in the function's transformation.

Each component of this derivative accounts for the behavior and constraints of \( \csc^{-1} x \). Understanding this helps in precise calculus operations.

Simplification of Derivatives

Once the derivative is found using the Chain Rule, and after differentiating inverse trigonometric functions, often you still face a complex expression.
Simplifying derivatives involves algebraic manipulation to arrive at a concise form. In this exercise, after combining using the Chain Rule, we arrive at an expression that combines both the outer and inner derivative components:
\[\frac{d y}{d x} = \frac{-1}{2|x|\sqrt{x^2 - 1} \cdot \sqrt{\csc^{-1}x}}\]
This expression looks complicated but simplifying derivatives often means factoring, reducing, or rationalizing expressions.
If ever stuck, revisit algebraic rules, and remember: practice sharpens simplification skills. Approaching such problems with a methodical attitude makes calculus less daunting.