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

Question

Let \(z=f(x, y)\) be a function of two variables \(x\) and \(y\). The partial derivative with respect to \(x\) of the function \(z=\) \(f(x, y)\) at \((x, y)\) is defined as \(\lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x, y)-f(x, y)}{\Delta x}\), provided the limit exists and is finite. It is denoted by \(\frac{\partial z}{\partial x}\) or \(\frac{\partial f}{\partial x}\) or \(f_{x^{-}}\)Clearly, \(\frac{\partial z}{\partial x}\) is the derivative of \(z=f(x, y)\) with respect to \(x\), regarding \(y\) as a constant. Similarly, we can define \(\frac{\partial z}{\partial y} \cdot \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)\), denoted by \(\frac{\partial^{2} z}{\partial x^{2}}\) or \(f_{x x}, \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)\), denoted by \(\frac{\partial^{2} z}{\partial y \partial x}\) or \(f_{y x}, \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)\), denoted by \(\frac{\partial^{2} z}{\partial x \partial y}\) or \(f_{x y}, \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)\), denoted by \(\frac{\partial^{2} z}{\partial y^{2}}\) or \(f_{y y}\) are called second order partial derivatives of \(z=f(x, y)\). 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: Understanding the Function u

First, recognize that the function given is \( u = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right) \). This function is a combination of inverse trigonometric functions in terms of \( x \) and \( y \).

Step 2: Calculating \( \frac{\partial u}{\partial x} \)

Find the partial derivative of \( u \) with respect to \( x \). Use the chain rule for inverse trigonometric functions:\[ \frac{\partial}{\partial x}\left(\sin^{-1}\left(\frac{x}{y}\right)\right) = \frac{1}{\sqrt{1-\left(\frac{x}{y}\right)^2}} \cdot \frac{1}{y} \]\[ \frac{\partial}{\partial x}\left(\tan^{-1}\left(\frac{y}{x}\right)\right) = -\frac{y}{x^2+y^2} \]Thus, \[ \frac{\partial u}{\partial x} = \frac{1}{y \sqrt{y^2-x^2}} - \frac{y}{x^2+y^2} \].

Step 3: Calculating \( \frac{\partial u}{\partial y} \)

Find the partial derivative of \( u \) with respect to \( y \). Again, use the chain rule:\[ \frac{\partial}{\partial y}\left(\sin^{-1}\left(\frac{x}{y}\right)\right) = -\frac{x}{y^2 \sqrt{y^2-x^2}} \]\[ \frac{\partial}{\partial y}\left(\tan^{-1}\left(\frac{y}{x}\right)\right) = \frac{x}{x^2+y^2} \]Thus,\[ \frac{\partial u}{\partial y} = -\frac{x}{y^2 \sqrt{y^2-x^2}} + \frac{x}{x^2+y^2} \].

Step 4: Finding the Expression \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \)

Substitute the expressions for the partial derivatives into the given condition:\[ x \frac{\partial u}{\partial x} = x\left(\frac{1}{y \sqrt{y^2-x^2}} - \frac{y}{x^2+y^2}\right) \]\[ y \frac{\partial u}{\partial y} = y\left(-\frac{x}{y^2 \sqrt{y^2-x^2}} + \frac{x}{x^2+y^2}\right) \]Simplify:\[ x \frac{1}{y \sqrt{y^2-x^2}} - \frac{xy}{x^2+y^2} - \frac{xy}{y \sqrt{y^2-x^2}} + \frac{xy}{x^2+y^2} = 0 \].

Step 5: Simplifying the Expression

Combine and simplify terms:The terms \( \frac{xy}{x^2+y^2} \) cancel out.The remaining part, \( \frac{x}{y \sqrt{y^2-x^2}} - \frac{xy}{y \sqrt{y^2-x^2}} = \frac{x}{y \sqrt{y^2-x^2}} - \frac{x}{y \sqrt{y^2-x^2}} = 0 \).

Step 6: Conclusion

The final result of the expression \( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \) simplifies to 0, thus leading to answer choice (A).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \sin^{-1}, \) and \( \tan^{-1}, \) provide the angle whose trigonometric function yields a given number. They are the counterparts to the standard trigonometric functions.
These functions allow us to work backwards from the ratio to the original angle. Let's see how they play a role in derivatives:

• The function \( u = \sin^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right) \) is an example showing inverse trigonometric functions combined.
• Calculating partial derivatives of such a function involves careful application of individual derivatives of these inverse functions.
• The derivative of \( \sin^{-1}\left(\frac{x}{y}\right) \) with respect to \( x \) can be derived using the formula:
  \[ \frac{1}{\sqrt{1-\left(\frac{x}{y}\right)^2}} \cdot \frac{1}{y} \]
• Similarly, the derivative of \( \tan^{-1}\left(\frac{y}{x}\right) \) with respect to \( x \) is obtained from
  \[ -\frac{y}{x^2+y^2} \]

Understanding these derivatives helps navigate complex problems involving inverse trigonometric functions.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. It plays a pivotal role when dealing with functions like inverse trigonometric ones.
The chain rule essentially states that to find the derivative of a composite function, you must take the derivative of the outer function and multiply it by the derivative of the inner function.

• Applying the chain rule for inverse trigonometric functions simplifies the derivative process, especially when nested inside other functions.
  For instance, in our problem, while finding \( \frac{\partial}{\partial x}\left(\sin^{-1}\left(\frac{x}{y}\right)\right) \), the derivative begins with the outer function, \( \sin^{-1} \), leading to
  \[ \frac{1}{\sqrt{1-\left(\frac{x}{y}\right)^2}} \]
• Then, by applying the chain rule to the internal fraction \( \frac{x}{y} \), we find an additional factor of \( \frac{1}{y} \) due to differentiating with respect to \( x \).

Mastering the chain rule is crucial to understanding how complex derivatives, like those with inverse trigonometric functions, are computed.

Second Order Partial Derivatives

Second order partial derivatives involve taking the derivative of a partial derivative. This is done iteratively and reveals additional properties of the function we are considering.
For functions \( z = f(x, y) \), there are more variations of derivatives based on the order and the variable:

• \( \frac{\partial^2 z}{\partial x^2} \) denotes taking the derivative with respect to \( x \), twice.
• \( \frac{\partial^2 z}{\partial y \partial x} \) or \( f_{yx} \), illustrates taking the derivative with respect to \( y \) first, followed by \( x \). The order of differentiation matters.

A practical aspect of these derivatives becomes apparent in optimizing functions, where one analyzes concavity, among other concepts. During problem-solving, particularly involving functions of multiple variables, understanding and calculating second order partial derivatives is essential; they provide insight into the behavior of the function in multiple dimensions.