Problem 69 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

$$ \begin{aligned} &\text { If } y=\sqrt{(a-x)(x-b)}-(a-b) \tan ^{-1} \sqrt{\frac{a-x}{x-b}}, \text { then } \\ &\frac{d y}{d x}= \end{aligned} $$(A) 1 (B) \(\sqrt{\frac{a-x}{x-b}}\) (C) \(\sqrt{(a-x)(x-b)}\) (D) \(\frac{1}{\sqrt{(a-x)(x-b)}}\)

Step-by-Step Solution

Answer

0 (the function does not change in response to x; constant.)

Step 1: Identify Derivatives Required

To find \( \frac{dy}{dx} \), we need to differentiate both terms of the function: \( \sqrt{(a-x)(x-b)} \) and \( (a-b) \tan^{-1}\sqrt{\frac{a-x}{x-b}} \).

Step 2: Differentiate First Term

For the expression \( \sqrt{(a-x)(x-b)} \), use the chain rule. Let \( u = (a-x)(x-b) \), then \( y = \sqrt{u} = u^{1/2} \). Differentiating \( u \) gives \( \frac{du}{dx} = -1(x-b) + (a-x)(-1) = -x + b - a + x = b-a \). Hence, \( \frac{dy_1}{dx} = \frac{1}{2\sqrt{u}} \cdot (b-a) \).

Step 3: Differentiate Second Term Using Chain Rule

For the second part, \( (a-b)\tan^{-1}\sqrt{\frac{a-x}{x-b}} \), let \( v = \tan^{-1}\sqrt{\frac{a-x}{x-b}} \). The derivative of \( \tan^{-1} z \) is \( \frac{1}{1+z^2} \), so \( \frac{dv}{dx} = \frac{1}{1 + \left(\sqrt{\frac{a-x}{x-b}}\right)^2} \cdot \frac{d}{dx}\left(\sqrt{\frac{a-x}{x-b}}\right) \).

Step 4: Simplify Second Term Derivative

Find \( \frac{d}{dx}\sqrt{\frac{a-x}{x-b}} \). Let \( z = \sqrt{\frac{a-x}{x-b}} \), then \( \frac{d}{dx}z = \frac{1}{2\sqrt{\frac{a-x}{x-b}}} \cdot \frac{-1(x-b) - (a-x)(-1)}{(x-b)^2} = \frac{-1(x-b) + (a-x)}{2\sqrt{(a-x)(x-b)^3}} = \frac{b-a}{2\sqrt{(a-x)(x-b)^3}} \).

Step 5: Combine Derivatives

Now, put everything together. For \( y = \sqrt{(a-x)(x-b)} - (a-b) \tan^{-1} \sqrt{\frac{a-x}{x-b}} \), we have \( \frac{dy}{dx} = \frac{b-a}{2\sqrt{(a-x)(x-b)}} - (a-b) \left( \frac{1}{1 + \frac{a-x}{x-b}} \right) \left( \frac{b-a}{2\sqrt{(a-x)(x-b)^3}} \right) \). After simplifying, you get \( 0 \).

Step 6: Conclusion of Derivatives

Combine and simplify both derivative parts to find that they negate each other. For the original function, \( \frac{dy}{dx} \) resolves to 0 after evaluating the simplifications. This shows that the function is effectively constant with respect to \( x \), meaning its rate of change equals 0.

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus, serving as the mathematical tool to measure how a function changes as its input changes. A derivative of a function, denoted as \( \frac{dy}{dx} \) or simply \( y' \), gives the slope of the tangent line to the curve at any given point. This signifies the rate of change of the dependent variable (usually \( y \)) with respect to the independent variable (usually \( x \)).

• Basic Principles: Imagine a curve representing a function on a graph. The derivative tells us how steep this curve is at any point.
• Real-world Example: If \( y \) represents distance traveled over time \( x \), the derivative \( \frac{dy}{dx} \) is the speed or velocity.

Differentiation, the process of finding a derivative, is central to calculus and offers a method for solving practical problems related to rates of change, motion, growth, and decay.

Chain Rule

Calculus often requires us to find derivatives of compositions of functions, known as composite functions. Here, the chain rule becomes essential. The chain rule formula is: if a function \( y \) depends on \( u \) and \( u \) depends on \( x \), then the derivative \( \frac{dy}{dx} \) is \( \frac{dy}{du} \cdot \frac{du}{dx} \).

• Form: Consider a function \( f(g(x)) \). The derivative is found using \( f'(g(x)) \cdot g'(x) \).
• Application: In the original exercise, the chain rule was used to differentiate both terms of a complex function efficiently.

The chain rule simplifies the process of differentiating complex expressions by breaking them down into manageable parts and applying derivatives sequentially. It's crucial for dealing with nested functions or compositions where one function is inside another.

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles when given certain trigonometric values. Their derivatives are particularly handy in calculus, especially when these functions are part of a composite function.

• Function Examples: Functions like \( \tan^{-1}(x) \) are inverse trigonometric functions, which reverse the tangent function.
• Basic Derivative Rules: The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \).

In the original step-by-step solution, the term \( \tan^{-1}(\sqrt{\frac{a-x}{x-b}}) \) necessitated the use of these derivative rules. Understanding how to work with inverse trigonometric functions is important, as they appear frequently in calculus, physics, and engineering problems. Their derivatives often lead to simpler expressions, enabling easier computation and simplification in complex equations.