Problem 56 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\cos ^{-1}\left(\frac{x^{-1}-x}{x^{-1}+x}\right)\), then \(f^{\prime}(x)\) is (A) odd (B) even (C) periodic (D) None of these

Step-by-Step Solution

Answer

Even

Step 1: Simplify the Expression inside the Inverse Cosine Function

The function is given by \[ f(x) = \cos^{-1}\left(\frac{x^{-1} - x}{x^{-1} + x}\right) \]Let's simplify the expression inside the inverse cosine function:Start by calculating \( x^{-1} \) which is \( \frac{1}{x} \). Substitute this back into the expression:\[ \frac{x^{-1} - x}{x^{-1} + x} = \frac{\frac{1}{x} - x}{\frac{1}{x} + x} \]Multiply numerator and denominator by \( x \) to get rid of the fraction:\[ \frac{1 - x^2}{1 + x^2} \]Thus, the function becomes:\[ f(x) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]

Step 2: Apply the Derivative Rule for Inverse Trigonometric Functions

We need to find the derivative of \[ f(x) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]Recall the derivative of \( \cos^{-1}(u) \) is:\[ f'(x) = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \]Here, \( u = \frac{1 - x^2}{1 + x^2} \). Next, find \( \frac{du}{dx} \) using the quotient rule.

Step 3: Derivative Using Quotient Rule

The quotient rule states that for \( u = \frac{v}{w} \), \[ \frac{du}{dx} = \frac{w \cdot \frac{dv}{dx} - v \cdot \frac{dw}{dx}}{w^2} \]Here, let \( v = 1 - x^2 \) and \( w = 1 + x^2 \).\( \frac{dv}{dx} = -2x \) and \( \frac{dw}{dx} = 2x \).So,\[ \frac{du}{dx} = \frac{(1 + x^2)(-2x) - (1 - x^2)(2x)}{(1 + x^2)^2} \]Simplifying gives:\[ \frac{du}{dx} = \frac{-2x - 2x^3 - 2x + 2x^3}{(1 + x^2)^2} = \frac{-4x}{(1 + x^2)^2} \]

Step 4: Calculate f'(x) Using the Derivative Formula

Substitute \( u \) and \( \frac{du}{dx} \) back into the derivative formula:\[ f'(x) = -\frac{1}{\sqrt{1 - \left(\frac{1 - x^2}{1 + x^2}\right)^2}} \cdot \frac{-4x}{(1 + x^2)^2} \]Simplifying the expression inside the square root:\[ 1 - \left(\frac{1 - x^2}{1 + x^2}\right)^2 = \frac{4x^2}{(1 + x^2)^2} \]The square root becomes \( \frac{2x}{1 + x^2} \).Thus,\[ f'(x) = \frac{-4x}{\frac{2x}{1 + x^2}} \cdot \frac{1}{2x} = \frac{-2(1+x^2)}{(1+x^2)^2} \]\[ f'(x) = -2 \cdot \frac{1}{1+x^2} \]This simplifies finally to:\[ f'(x) = -2 \cdot \frac{1}{1+x^2} \]

Step 5: Determine the Nature of f'(x)

Notice that \( f'(x) = -2 \cdot \frac{1}{1+x^2} \).Since this derivative is a function of x only, examine if it is odd or even:- A function is odd if \( f(-x) = -f(x) \).- \( f'(-x) = -2 \cdot \frac{1}{1+(-x)^2} = -2 \cdot \frac{1}{1+x^2} = f'(x) \).Thus, \( f'(x) \) is an even function.

Key Concepts

Quotient Rule

When deriving complicated functions, especially those with fractions, the Quotient Rule becomes indispensable. If you have a function that is the ratio of two functions, say \( u = \frac{v}{w} \), then the derivative \( \frac{du}{dx} \) can be found using this rule. The formula is:

• \( \frac{du}{dx} = \frac{w \cdot \frac{dv}{dx} - v \cdot \frac{dw}{dx}}{w^2} \)

It essentially helps separate the tasks of differentiating the numerator \( v \) and the denominator \( w \).
In our problem, we apply the Quotient Rule to \( \frac{1 - x^2}{1 + x^2} \) where:

• \( v = 1 - x^2 \) and \( w = 1 + x^2 \).
• \( \frac{dv}{dx} = -2x \) while \( \frac{dw}{dx} = 2x \).

Substituting these into the rule, we get \( \frac{du}{dx} = \frac{-4x}{(1 + x^2)^2} \). This expression provides the needed derivative \( \frac{du}{dx} \) to further evaluate \( f'(x) \) using the derivative formula for inverse cosine functions.

Even Functions

The concept of even functions is important when analyzing the symmetry of functions. Specifically, a function \( f(x) \) is even if it satisfies the condition \( f(-x) = f(x) \). This symmetry implies the graph of the function is mirrored across the y-axis.
For derivatives, if \( f'(x) \) equals \( f'(-x) \), then the function is classified as even.
In the context of this problem, we determined \( f'(x) = -2 \cdot \frac{1}{1 + x^2} \). By evaluating \( f'(-x) \), which also results in the same expression:

• \( f'(-x) = -2 \cdot \frac{1}{1+(-x)^2} \)

It simplifies to the same form, confirming that \( f'(x) \) is indeed even.
Understanding this property helps in identifying function behavior and its graphical representation.

Simplification of Trigonometric Expressions

Simplifying trigonometric expressions is often a critical step in solving calculus problems. It helps in reducing complex fractions and expressions into more manageable forms.
Initially, the exercise involves simplifying the expression \( \frac{x^{-1} - x}{x^{-1} + x} \) by substituting \( x^{-1} \) with \( \frac{1}{x} \). This substitution transforms the expression into a typically unfriendly fraction.
To make it more approachable, multiply both numerator and denominator by \( x \). Consequently, it simplifies to \( \frac{1 - x^2}{1 + x^2} \).

• This step eliminates the fractions inside the fraction, making differentiation easier.
• The final form shows the trigonometric relationship ready for further calculus operations.

These simplification strategies allow you to see hidden symmetries and better understand the trigonometric relationships involved.