Problem 87 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f\) be a function such that \(f:(-1,1) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Let \(f\) satisfy the equation \(f(x)+f(y)=f\left(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right)\) If \(f(x)\) is differentiable on \((-1,1)\) and \(f^{\prime}(0)=1\), then \(f^{\prime}(x)\) is equal to (A) \(\frac{1}{\sqrt{1-x^{2}}}\) (B) \(-\frac{1}{\sqrt{1-x^{2}}}\) (C) \(\frac{1}{\sqrt{1+x^{2}}}\) (D) \(-\frac{1}{\sqrt{1+x^{2}}}\)

Step-by-Step Solution

Answer

The derivative \(f'(x)\) is \(\frac{1}{\sqrt{1-x^2}}\), which corresponds to option (A).

Step 1: Evaluate the Given Functional Equation at Specific Points

First, evaluate the functional equation by setting \( y = 0 \). We have \( f(x) + f(0) = f(x \sqrt{1}) = f(x) \). Therefore, \( f(0) = 0 \).

Step 2: Differentiate the Given Equation Implicitly with Respect to x

Differentiate both sides of the equation \( f(x) + f(y) = f\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \). This gives us the equation: \[ f'(x) + 0 = \frac{d}{dx}\left(f\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)\right) \].

Step 3: Apply the Chain Rule to the Right Side

Using the chain rule to differentiate \( f\left(u\right) \) where \( u = x\sqrt{1-y^2} + y\sqrt{1-x^2} \): \[ f'(u) \cdot \frac{d}{dx}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \] and since the partial derivative \( \frac{d}{dx} (x \sqrt{1-y^2} + y \sqrt{1-x^2}) = \sqrt{1-y^2} - \frac{xy}{\sqrt{1-x^2}} \), thus: \[ f'(x) = f'(u)(\sqrt{1-y^2}) \].

Step 4: Analyze Derivative at Specific Point and Generalize

By substituting \( x = 0 \) and \( y = 0 \), we find \( f'(0) = f'(0) \cdot 1 \), suggesting \( f'(0) = 1 \). This implies that for general \( x \), the form must match \( f'(x) = \frac{1}{\sqrt{1-x^2}} \), consistent with the nature of inverse functions (arcsin/arccos family).

Step 5: Compare Derivative with Given Options

Compare the derived formula with options: Option A \( \frac{1}{\sqrt{1-x^2}} \) matches the derived expression with arcsin-like differentiability properties.

Key Concepts

Functional Equation

A functional equation is a type of equation in which the variables are functions rather than simple numbers. In this exercise, we deal with the equation \( f(x) + f(y) = f\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \). Functional equations help in understanding the behavior and properties of functions by setting conditions they must satisfy.

They are often solved by substituting particular values into the variables, as seen in Step 1 where setting \( y = 0 \) allowed us to find that \( f(0) = 0 \). This technique can simplify the problem and provide insight into the function's properties.

• Helps in determining particular values or behaviors of the function.
• Can reveal symmetries or invariants.
• Provides a system to solve or reduce complexity of the equation.

Understanding functional equations is crucial because they often reveal the fundamental nature or rule governing the function's behavior over its domain.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. When you have a function within another function, like \( f(g(x)) \), the chain rule states that the derivative is the product of the outer function's derivative at the inner function's output and the inner function's derivative: \( f'(g(x)) \cdot g'(x) \).
In Step 3, the chain rule was applied because the given functional equation involves a composite expression: \( f\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \).

• First, identify the "outer" and "inner" functions.
• Differentiate the outer function with respect to its argument.
• Multiply this derivative by the derivative of the inner function.

This is crucial in handling complex differentiable functions, particularly when combined expressions are present, enabling the simplification and solution of differential equations.

Implicit Differentiation

Implicit differentiation is a method used when it is difficult or impossible to solve for one variable explicitly in terms of the other. In the original problem, we implicitly differentiated \( f(x) + f(y) = f\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \) with respect to \( x \).
This method allows differentiation without isolating \( y \) in terms of \( x \).

• Differentiate both sides of an equation with respect to one variable.
• Treat other variables as functions of the chosen variable.
• Solve for the desired derivative after differentiation.

By employing implicit differentiation, as seen in Step 2, it simplifies complex functional relationships, allowing us to determine \( f'(x) \) and connect with derivative rules as found in derivative of inverse trigonometric functions.

Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions, like arcsin and arccos, have specific derivative rules that are essential when differentiating them or compositions involving them. The derivative of \( \arcsin(x) \) is \( \frac{1}{\sqrt{1-x^2}} \), and this insight was crucial in determining \( f'(x) \).
These derivatives often appear in calculus problems that involve circular functions or transformations, as arcsines describe angles emerging from such relationships.

• The derivatives reflect the geometric relationships inherent in the unit circle.
• They help in solving problems that exhibit periodicity or rotational symmetry.
• Provide insights into the behavior of functions that mimic inverses of sine and cosine.

Recognizing the form \( \frac{1}{\sqrt{1-x^2}} \) in the derivative step allows us to conclude that the given function \( f \) behaves like an inverse sine function, thus leading us to the correct answer, which coincides with real-world arcs and angles.