Problem 55 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)\), then \(\frac{d y}{d x}=\) (A) \(0 \forall x \in R\) (B) \(0 \forall x \in(0, \infty)\) (A) \(0 \forall x \in R-\\{0\\}\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is B: \(0 \forall x \in (0, \infty)\).

Step 1: Understand the Function

We have two functions within the main function: the inverse secant \( \sec^{-1} \left( \frac{x+1}{x-1} \right) \) and the inverse sine \( \sin^{-1} \left( \frac{x-1}{x+1} \right) \). We aim to find the derivative \( \frac{dy}{dx} \).

Step 2: Express the Derivatives

Find the derivatives of the components: for \( y = \sec^{-1}(u) \), the derivative is \( \frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{du}{dx} \), and for \( y = \sin^{-1}(v) \), it's \( \frac{1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx} \). Identify \( u = \frac{x+1}{x-1} \) and \( v = \frac{x-1}{x+1} \).

Step 3: Differentiate Each Component

Calculate \( \frac{du}{dx} \). Let \( u = \frac{x+1}{x-1} \); use the quotient rule to find that \( \frac{du}{dx} = \frac{-2}{(x-1)^2} \). Likewise, calculate \( \frac{dv}{dx} \) for \( v = \frac{x-1}{x+1} \) using the quotient rule, resulting in \( \frac{dv}{dx} = \frac{2}{(x+1)^2} \).

Step 4: Simplify the Components

The expression for \( \frac{dy}{dx} \) considering \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) should lead to simplifications where each inverse function derivative cancels out contributions from its respective part due to symmetry in expressions of \( u \) and \( v \).

Step 5: Analyze the Expression

Due to the trigonometric identities and their symmetry, you will find that \( \frac{dy}{dx} = 0 \) for values of \( x \) where the functions are defined and differentiable.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions take standard trigonometric functions and reverse the process, essentially finding the angle given a certain trigonometric ratio. For example, the inverse secant function, written as \( \sec^{-1}(x) \), provides the angle whose secant is \( x \). Similarly, \( \sin^{-1}(x) \) is used to find an angle with a sine of \( x \).
These functions are particularly useful in situations where you have a ratio and need to determine an angle. This can often be the case in various branches of calculus and physics.

Key properties include their domains and ranges:

• The domain of \( \sec^{-1}(x) \) is \( x \geq 1 \) or \( x \leq -1 \), and its range is \( [0, \pi] \), excluding \( \pi/2 \).
• For \( \sin^{-1}(x) \), the domain is \( [-1, 1] \), and the range is \( [-\pi/2, \pi/2] \).

Understanding these properties is crucial for problem-solving, particularly when combining these functions, as seen in this calculus problem involving both inverse secant and sine.

Derivative Calculation

Calculating derivatives of inverse trigonometric functions requires understanding their derivatives' unique forms. Let's break down the derivative of each inverse function included in our original problem.
The derivative of \( \sec^{-1}(u) \) is \( \frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{du}{dx} \). This formula takes into account the absolute value of \( u \) and uses a function of \( u \) involving a square root.
Similarly, the derivative of \( \sin^{-1}(v) \) is \( \frac{1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx} \). Here, the expression uses a square root related to \( v^2 \).
By identifying and isolating \( u \) and \( v \) in the given expressions, you apply these derivatives to find individual components. Such precise calculations are essential before considering any further manipulation or simplification, like applying the quotient rule.

Quotient Rule

The quotient rule is a method used to differentiate functions that are the ratio of two differentiable functions. This rule is particularly helpful in this problem for differentiating \( u = \frac{x+1}{x-1} \) and \( v = \frac{x-1}{x+1} \).
The quotient rule states that if \( h(x) = \frac{f(x)}{g(x)} \), then its derivative is given by:

• \( h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \)

This rule is applied separately to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \), which are essential components for calculating the derivative \( \frac{dy}{dx} \) in the original equation.
When calculating \( \frac{du}{dx} \) and \( \frac{dv}{dx} \), it's important to follow the application of the quotient rule precisely, as any misstep in the calculation process can lead to incorrect results. The calculated derivatives result in expressions that contribute to zero when totaled due to the symmetrical nature of \( u \) and \( v \), hence simplifying the entire expression.