Problem 96 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(u=\sin ^{-1} \frac{x}{y}+\tan ^{-1} \frac{y}{x}\), then \(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\) (A) 0 (B) 1 (C) \(-1\) (D) None of these

Step-by-Step Solution

Answer

The answer is A, 0.

Step 1: Rewrite the function

The given function is \( u = \sin^{-1} \left( \frac{x}{y} \right) + \tan^{-1} \left( \frac{y}{x} \right) \). Let's rewrite it to better understand the problem.

Step 2: Differentiate partially with respect to x

The partial derivative \( \frac{\partial u}{\partial x} \) becomes \( \frac{d}{dx} \sin^{-1} \left( \frac{x}{y} \right) + \frac{d}{dx} \tan^{-1} \left( \frac{y}{x} \right) \). Calculate each term separately.

Step 3: Evaluate partial \(\sin^{-1}\) term with respect to x

Using the chain rule, \( \frac{d}{dx} \sin^{-1} \left( \frac{x}{y} \right) \) gives \( \frac{1}{\sqrt{1-\left( \frac{x}{y} \right)^2}} \cdot \left( \frac{1}{y} \right) = \frac{y}{\sqrt{y^2 - x^2}} \).

Step 4: Evaluate partial \(\tan^{-1}\) term with respect to x

Using the chain rule, \( \frac{d}{dx} \tan^{-1} \left( \frac{y}{x} \right) \) gives \( \frac{-y}{x^2+y^2}\).

Step 5: Partial derivative with respect to x

Combine the results from Steps 3 and 4 to find \(\frac{\partial u}{\partial x} = \frac{y}{\sqrt{y^2 - x^2}} - \frac{y}{x^2 + y^2}\).

Step 6: Differentiate partially with respect to y

The partial derivative \( \frac{\partial u}{\partial y} \) becomes \( \frac{d}{dy} \sin^{-1} \left( \frac{x}{y} \right) + \frac{d}{dy} \tan^{-1} \left( \frac{y}{x} \right) \). Calculate each term separately.

Step 7: Evaluate partial \(\sin^{-1}\) term with respect to y

Using the chain rule, \( \frac{d}{dy} \sin^{-1} \left( \frac{x}{y} \right) \) gives \( \frac{-x}{y^2\sqrt{1-\left( \frac{x}{y} \right)^2}} = \frac{-x}{y\sqrt{y^2 - x^2}} \).

Step 8: Evaluate partial \(\tan^{-1}\) term with respect to y

Using the chain rule, \( \frac{d}{dy} \tan^{-1} \left( \frac{y}{x} \right) \) gives \( \frac{x}{x^2+y^2}\).

Step 9: Partial derivative with respect to y

Combine the results from Steps 7 and 8 to find \(\frac{\partial u}{\partial y} = \frac{-x}{y\sqrt{y^2 - x^2}} + \frac{x}{x^2 + y^2}\).

Step 10: Substitute into the original expression

Substitute the derivatives found: \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x \left( \frac{y}{\sqrt{y^2 - x^2}} - \frac{y}{x^2 + y^2} \right) + y \left( \frac{-x}{y\sqrt{y^2 - x^2}} + \frac{x}{x^2 + y^2} \right) \).

Step 11: Simplify the expression

Upon performing the multiplication and simplification, both terms containing \(\frac{y}{\sqrt{y^2 - x^2}}\) and \(\frac{x}{x^2 + y^2}\) cancel out, resulting in 0.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are those that reverse the process of trigonometric functions like sine, cosine, and tangent. In this case, we deal with the inverse sine (\(\sin^{-1}\)) and inverse tangent (\(\tan^{-1}\)) functions. Inverse trigonometric functions are important because they allow us to find an angle with a given trigonometric ratio.
They are often used in calculus for finding derivatives, as they have unique differentiation rules.
These rules help solve problems involving rates of change and optimization.

• For \(\sin^{-1}(x)\), the derivative is \(\frac{1}{\sqrt{1-x^2}}\).
• For \(\tan^{-1}(x)\), the derivative is \(\frac{1}{1+x^2}\).

When dealing with composite functions where the argument is not simply \(x\), such as \(\sin^{-1}\left(\frac{x}{y}\right)\), using techniques like the chain rule becomes necessary to properly differentiate.

Chain Rule in Calculus

The chain rule is an essential technique in calculus used for finding the derivative of composite functions. When a function is composed of two or more functions, we can apply the chain rule to differentiate it effectively.
This rule states that to differentiate a composite function, multiply the derivative of the outer function by the derivative of the inner function.
This is vital for solving problems involving inverse trigonometric functions, where the argument isn't a simple variable.

• If we have \(f(g(x))\), the derivative is \(f'(g(x)) \cdot g'(x)\).
• For \(\sin^{-1}\left(\frac{x}{y}\right)\), using the chain rule gives us the derivative \(\frac{1}{\sqrt{1-\left(\frac{x}{y}\right)^2}} \cdot \frac{d}{dx}\left(\frac{x}{y}\right)\).
• Similarly, \(\tan^{-1}\left(\frac{y}{x}\right)\) is differentiated using the rule to obtain \(\frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{y}{x}\right)\).

Learning to apply the chain rule correctly not only aids differentiation but enhances understanding of the relationship between functions.

Differentiation

Differentiation is one of the core concepts in calculus, involving the process of finding the derivative of a function. The derivative represents the rate of change of a function with respect to its variable.
This rate of change is pivotal in various scientific fields, from physics to economics, helping in understanding how things change over time or space.

• For a function \(f(x)\), its derivative, noted as \(f'(x)\) or \(\frac{df}{dx}\), shows how \(f(x)\) changes as \(x\) changes.
• Partial derivatives, as seen in our original problem, involve differentiating functions with respect to one of multiple variables, holding the others constant, crucial for multivariable calculus.
• For example, \(\frac{\partial}{\partial x}\) focuses on changes as the variable \(x\) varies, keeping other variables constant.

Mastering differentiation and its techniques, such as using the chain rule, significantly expands one's capability to tackle complex mathematical problems effectively.