Problem 107 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\sin y=x \sin (\alpha+y)\), then \(\frac{d y}{d x}\) is : \(\quad\) [2002] (A) \(\frac{\sin \alpha}{\sin ^{2}(\alpha+y)}\) (B) \(\frac{\sin ^{2}(\alpha+y)}{\sin \alpha}\) (C) \(\sin \alpha \sin ^{2}(\alpha+y)\) (D) \(\frac{\sin ^{2}(\alpha-y)}{\sin \alpha}\)

Step-by-Step Solution

Answer

Option (B) \( \frac{\sin^2(\alpha + y)}{\sin \alpha} \) is the correct answer.

Step 1: Differentiate with Respect to x

Given the equation \( \sin y = x \sin(\alpha + y) \), we need to find \( \frac{d y}{d x} \). Start by differentiating both sides with respect to \( x \). The derivative of the left side with respect to \( x \) using the chain rule is \( \cos y \cdot \frac{d y}{d x} \). For the right side, apply the product rule: \( \sin(\alpha + y) + x \cos(\alpha + y) \cdot \frac{d y}{d x} \).

Step 2: Set the Derivatives Equal

After applying differentiation, the equation becomes \[ \cos y \cdot \frac{d y}{d x} = \sin(\alpha + y) + x \cos(\alpha + y) \cdot \frac{d y}{d x}. \]

Step 3: Solve for \( \frac{d y}{d x} \)

Rearrange terms to isolate \( \frac{d y}{d x} \) on one side: \[ \cos y \cdot \frac{d y}{d x} - x \cos(\alpha + y) \cdot \frac{d y}{d x} = \sin(\alpha + y). \] Factor out \( \frac{d y}{d x} \): \[ \frac{d y}{d x} (\cos y - x \cos(\alpha + y)) = \sin(\alpha + y). \] Then, solve for \( \frac{d y}{d x} \): \[ \frac{d y}{d x} = \frac{\sin(\alpha + y)}{\cos y - x \cos(\alpha + y)}. \]

Step 4: Simplify the Expression

Substitute \( x = \frac{\sin y}{\sin(\alpha + y)} \) from the original equation to simplify further: \[ \frac{d y}{d x} = \frac{\sin(\alpha + y)}{\cos y - \frac{\sin y}{\sin(\alpha + y)} \cos(\alpha + y)}. \] Simplifying gives \[ \frac{d y}{d x} = \frac{\sin(\alpha + y) \sin(\alpha + y)}{\sin(\alpha + y) \cos y - \sin y \cos(\alpha + y)}. \] Using the identity \( \sin(\alpha) = \sin(\alpha + y) \cos y - \sin y \cos(\alpha + y) \), replace to yield \[ \frac{d y}{d x} = \frac{\sin^2(\alpha + y)}{\sin(\alpha)}. \]

Step 5: Select the Correct Answer

Compare the simplified expression with the choices to find \( \frac{d y}{d x} = \frac{\sin^2(\alpha + y)}{\sin(\alpha)} \), which corresponds to option (B).

Key Concepts

Understanding the Chain Rule

The chain rule is an essential concept in differential calculus, especially when dealing with composite functions. It allows us to differentiate functions with multiple layers by breaking them down into their simpler parts.
To understand the chain rule, think of it like peeling an onion layer by layer.Let's say we have a function \( f \) composed of two inner functions \( g \) and \( h \). The chain rule states:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

This means to differentiate \( f(g(x)) \), you need to:

• Differentiating \( f \) with respect to \( g \) (the inner function).
• Multiply that result by the derivative of \( g(x) \) itself.

Applying this rule to the problem, we differentiated \( \sin(y) \) with respect to \( y \) and then \( y \) itself with respect to \( x \). This gave us \( \cos(y) \cdot \frac{d y}{d x} \). The chain rule is especially useful when you encounter functions nested within each other, ensuring you accurately trace each step through layers.

Grasping the Product Rule

In differential calculus, the product rule is vital when differentiating products of two functions. It helps you find the derivative of expressions like \( u(x)v(x) \).The formula goes like this:

• \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

In this expression:

• \( u(x) \) and \( v(x) \) are functions of \( x \).
• \( u'(x) \) and \( v'(x) \) are their respective derivatives.

So, to differentiate the product, you differentiate each function one at a time, using the other function as a constant.In the given problem, we applied the product rule to the right-hand side: \( x \sin(\alpha+y) \). Here,

• \( x' \cdot \sin(\alpha+y) \) and \( \cos(\alpha+y)\cdot \frac{dy}{dx} \), which accounts for each product component.

This principle ensures you maintain accuracy when dealing with products, a frequent requirement in calculus problems.

Exploring Trigonometric Identities

Trigonometric identities are powerful tools in calculus that can simplify expressions and aid in solving equations. These identities express relationships between trigonometric functions such as sine, cosine, and more.For instance, the Pythagorean identity:

• \( \sin^2(x) + \cos^2(x) = 1 \)

This identity provides fundamental knowledge to simplify complex trigonometric expressions.In our problem, trigonometric identities facilitated the simplification of the derivative expression. We replaced terms using identities to match answers with options given. Specifically, using \( \sin(\alpha) = \sin(\alpha + y) \cos(y) - \sin(y) \cos(\alpha+y) \), helped bring the expression to a familiar form.These transformations not only made computation easier but also brought insight into choosing the correct option. Mastering these identities equips you with the ability to recognize patterns and simplify otherwise complicated calculus problems efficiently.