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

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

(D) \( \frac{1}{\sqrt{2}} \).

Step 1: Understand the Problem

We need to find the derivative of the function \( y = \sec(\tan^{-1}x) \) and evaluate it at \( x = 1 \).

Step 2: Expression for Inverse Tangent

Recall that \( \tan^{-1}(x) \) gives the angle \( \theta \) such that \( \tan(\theta) = x \). If \( \theta = \tan^{-1}(x) \), then \( \sec(\theta) = \sqrt{1 + \tan^2(\theta)} \).

Step 3: Simplify Using Trigonometric Identities

Since \( \tan(\theta) = x \), we have \( \sec(\theta) = \sqrt{1 + x^2} \). Thus, \( y = \sqrt{1 + x^2} \).

Step 4: Differentiate the Function

Differentiate \( y = \sqrt{1 + x^2} \) with respect to \( x \). The derivative is \( \frac{dy}{dx} = \frac{d}{dx}( (1 + x^2)^{1/2}) = \frac{1}{2} \times (1 + x^2)^{-1/2} \times 2x = \frac{x}{\sqrt{1 + x^2}} \).

Step 5: Evaluate the Derivative at \( x=1 \)

Substitute \( x = 1 \) into the derivative \( \frac{dy}{dx} \). \( \frac{dy}{dx} = \frac{1}{\sqrt{1+1^2}} = \frac{1}{\sqrt{2}} \).

Step 6: Choose the Correct Answer

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

Key Concepts

Trigonometric Identities

In calculus, trigonometric identities are essential tools for simplifying and solving derivative problems, especially when functions involve trigonometric expressions.
Understanding and using these identities correctly can make complex problems significantly easier to handle.
In this exercise, we encounter the identity \[\sec(\theta) = \sqrt{1 + \tan^2(\theta)}\],which is derived from the Pythagorean identity \[1 + \tan^2(\theta) = \sec^2(\theta)\].
This identity was crucial in simplifying the given function, \(y = \sec(\tan^{-1}x)\), to \(y = \sqrt{1 + x^2}\).
This simplification enabled us to apply more straightforward differentiation techniques later on.
Mastering trigonometric identities allows you to reduce seemingly complex functions into more manageable expressions, allowing for more efficient problem-solving in differentiation.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\tan^{-1}x\), play an important role in calculus problems involving trigonometric expressions.
These functions help you find angles corresponding to known trigonometric values.
For the inverse tangent function, \(\tan^{-1}x\) gives the angle \(\theta\) where \(\tan(\theta) = x\).
Knowing this, we can express trigonometric functions of \(\theta\) in terms of \(x\), such as \(\sec(\theta)\) leading us to \(\sec(\theta) = \sqrt{1 + x^2}\).
Understanding how to manipulate and differentiate inverse trigonometric functions is foundational in tackling advanced calculus problems.

• Inverse functions help transition between angles and ratio values.
• They often appear in problems requiring derivatives and integrals of trigonometric functions.

The use of inverse functions in this exercise is critical in expressing \(y\) in a differentiable form.

Differentiation Techniques

Differentiation techniques are crucial skills in calculus, enabling us to calculate the rate of change of a function.
In this problem, the simplified function \(y = \sqrt{1 + x^2}\) was differentiated with respect to \(x\).
To differentiate \((1 + x^2)^{1/2}\), we apply standard rules for differentiating powers of expressions.
The power rule combines with the chain rule to help us find the derivative efficiently.Here are the steps:

• The power rule tells us that the derivative of \(x^n\) is \(nx^{n-1}\).
• The chain rule allows us to differentiate complex, nested functions by multiplying the derivative of the outer function by the derivative of the inner function.

Applying these rules, we find:\[\frac{dy}{dx} = \frac{1}{2} \times (1 + x^2)^{-1/2} \times 2x = \frac{x}{\sqrt{1 + x^2}}\]
This derivative was then evaluated at \(x = 1\) to find \(\frac{1}{\sqrt{2}}\).
By mastering differentiation techniques, you can efficiently handle complex calculus problems, especially those involving trigonometric and inverse trigonometric expressions.