Problem 14 - A Textbook of Differential Calculus

Question

If \(y=\sin ^{-1}\left(\frac{\sin \alpha \sin x}{1-\cos \alpha \sin x}\right)\), then \(y^{\prime}(0)\) is (a) 1 (b) \(2 \tan \alpha\) (c) \(\frac{1}{2} \tan \alpha\) (d) \(\sin \alpha\)

Step-by-Step Solution

Answer

The answer is (d) \(\sin \alpha\).

Step 1: Understand the Problem

We are asked to find the derivative of \(y\) with respect to \(x\) at the point \(x = 0\), where \(y = \sin^{-1}\left(\frac{\sin \alpha \sin x}{1-\cos \alpha \sin x}\right)\). We will find \(y'\) and evaluate it at \(x = 0\).

Step 2: Simplify the Expression Inside the Inverse Sine

Let \(z = \frac{\sin \alpha \sin x}{1-\cos \alpha \sin x}\). We have \(y = \sin^{-1}(z)\). We need to find \(\frac{dz}{dx}\) and apply the chain rule later.

Step 3: Differentiate z with Respect to x

Use the quotient rule to differentiate \(z\):\[\frac{dz}{dx} = \frac{(\cos \alpha \sin x + \sin \alpha \cos x)(1-\cos \alpha \sin x) - (\sin \alpha \sin x)(\cos \alpha \cos x)}{(1-\cos \alpha \sin x)^2}\]Simplifying this expression will be necessary in later steps.

Step 4: Evaluate dz/dx at x=0

Notice at \(x=0\), \(\sin x = 0\) and \(\cos x = 1\). Substitute these values into the derivative:\[\frac{dz}{dx} \bigg|_{x=0} = \frac{\sin \alpha \cdot 1 - 0}{1^2} = \sin \alpha\]

Step 5: Differentiate y with Respect to z and Apply Chain Rule

Since \(y = \sin^{-1}(z)\), we have \(\frac{dy}{dz} = \frac{1}{\sqrt{1 - z^2}}\). By the chain rule, \(\frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx}\).

Step 6: Evaluate dy/dx at x=0

We have \(\frac{dz}{dx} \bigg|_{x=0} = \sin \alpha\) and initially \(z \bigg|_{x=0} = 0\). So,\[\frac{dy}{dx} \bigg|_{x=0} = \frac{1}{\sqrt{1 - 0^2}} \cdot \sin \alpha = \sin \alpha\]

Step 7: Find the Correct Option

The derivative \(y'(0)\) is \(\sin \alpha\), which corresponds to option \((d)\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the reverse operations of trigonometric functions such as sine, cosine, and tangent. They help us find angles when given the value of the trigonometric function.
For example, the function \( \sin^{-1}(x) \) returns an angle \( \theta \) such that \( \sin(\theta) = x \). These functions are crucial because they allow us to solve equations where the variable is trapped inside a trigonometric function.

• The domain of \( \sin^{-1}(x) \) is \([-1, 1]\), as sine values range between these limits.
• The range of \( \sin^{-1}(x) \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), allowing us to cover all necessary angles for sine.

In differential calculus, knowing how to manipulate these functions is vital, especially when they are part of more complex expressions. Understanding these domains and ranges is essential when solving inverse trigonometric problems, such as in this exercise where we calculate the derivative.

Derivative Computation

Derivative computation in calculus allows us to measure how a function changes at any given point. It tells us the rate of change or slope of a function. For a function \( y = f(x) \), the derivative \( y' = \frac{dy}{dx} \) indicates how \( y \) changes with respect to \( x \).
In this exercise, we compute the derivative of a composed function featuring \( \sin^{-1} \). This involves a careful breakdown:

• First, find the derivative of the inner function within the \( \sin^{-1} \).
• Apply the derivative of the inverse function. For \( \sin^{-1}(x) \), the derivative \( \frac{dy}{dz} = \frac{1}{\sqrt{1 - z^2}} \).
• Combine these derivatives using the chain rule (more on this later).

Overall, derivative computation is about understanding how small changes in \( x \) lead to changes in \( y \), allowing us to predict behavior of functions in calculus.

Chain Rule

The Chain Rule is a fundamental tool in calculus used to differentiate composite functions. A composite function is essentially a function within another function, like our exercise's \( \sin^{-1}\left( \frac{\sin \alpha \sin x}{1-\cos \alpha \sin x} \right) \).
When applying the chain rule, we differentiate the outer function first, and then multiply by the derivative of the inner function. This allows us to "chain" our way through multiple layers of functions to find the overall derivative.
Steps in applying the Chain Rule:

• Identify the outer and inner functions. Here, the outer is \( \sin^{-1}(z) \), and the inner is \( z = \frac{\sin \alpha \sin x}{1-\cos \alpha \sin x}\).
• Differentiate each component individually: \( \frac{dy}{dz} \) for outer and \( \frac{dz}{dx} \) for inner.
• Combine these derivatives to find \( \frac{dy}{dx} \) through \( \frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} \).

This process is essential for handling more complex situations in calculus where direct differentiation won't easily apply.