Problem 119 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(y=\sec \left(\tan ^{-1} x\right)\), then \(\frac{d y}{d x}\) at \(x=1\) is equal to (A) \(\frac{1}{2}\) (B) 1 (C) \(\sqrt{2}\) (D) \(\frac{1}{\sqrt{2}}\)

Step-by-Step Solution

Answer

The derivative \(\frac{dy}{dx}\) at \(x=1\) is \(\frac{1}{\sqrt{2}}\), so the answer is option (D).

Step 1: Recognize the Function and its Components

The given function is composed of an inverse trigonometric function and a trigonometric function. Specifically, we have \(y = \sec(\tan^{-1} x)\). Our task is to find \(\frac{dy}{dx}\) at \(x = 1\).

Step 2: Rewrite Secant in Terms of Tangent

Recall that if \(\theta = \tan^{-1}(x)\), then \(\tan(\theta) = x\). The identity \(\sec^2(\theta) = 1 + \tan^2(\theta)\) allows us to express \(\sec(\theta)\) in terms of \(x\):\[ \sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1 + x^2} \]Thus, \(y = \sqrt{1 + x^2}\).

Step 3: Differentiate the Expression with Respect to x

Now, differentiate \(y = \sqrt{1 + x^2}\) with respect to \(x\) using the chain rule.\[\frac{dy}{dx} = \frac{d}{dx} (1 + x^2)^{1/2} = \frac{1}{2}(1 + x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1 + x^2}}\]

Step 4: Evaluate the Derivative at x=1

Substitute \(x = 1\) into the derivative obtained:\[\frac{dy}{dx} \bigg|_{x=1} = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}}.\]

Step 5: Determine the Correct Option

The value of \(\frac{dy}{dx}\) at \(x = 1\) is \(\frac{1}{\sqrt{2}}\), which corresponds to option (D).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus as they help us find angles when given trigonometric ratios. In our exercise, we encounter the inverse tangent function, denoted as \( \tan^{-1}(x) \). This function provides an angle, \( \theta \), whose tangent is \( x \).

• In mathematical terms, if \( \theta = \tan^{-1}(x) \), then \( \tan(\theta) = x \).

These functions are continuously differentiable, meaning they have derivatives at all points in their domain. Understanding the behavior of inverse trig functions allows us to convert trigonometric problems into algebraic expressions. In our problem, this conversion is vital for moving the function \( \sec(\tan^{-1}(x)) \) to a more manageable form for differentiation.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how a function's output changes as its input changes. In the context of our exercise, we want to calculate the derivative of \( y = \sqrt{1 + x^2} \) with respect to \( x \). This is a fundamental skill in calculus, crucial for understanding rates of change and the behavior of curves.

• The chain rule is often employed when differentiating composite functions, which is done here when handling \( y = (1 + x^2)^{1/2} \).
• The derivative of \( x^2 \) is \( 2x \), and the derivative of \( (1 + x^2)^{1/2} \) includes multiplying by the derivative of the inside expression.

Ultimately, differentiating successfully leads us to the expression \( \frac{x}{\sqrt{1+x^2}} \), which can be evaluated at specific points such as \( x = 1 \), giving us meaningful insights about the function at those points.

Trigonometric Identities

Trigonometric identities are equations that are true for all values of the involved variables. They are extremely useful tools in simplifying expressions or solving equations. In the problem at hand, using the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \) is crucial for rewriting secant in terms of tangent, facilitating easier differentiation.

• This identity helps in converting the expression \( y = \sec(\tan^{-1}(x)) \) to \( y = \sqrt{1 + x^2} \).
• These identities allow swapping functions without altering relationships.

Understanding and applying trigonometric identities helps move from complex trigonometric function combinations to simpler algebraic forms, making calculus problems more manageable. Such transformations are pivotal when dealing with derivatives of trigonometric functions, especially when they are combined with inverse trig functions.