Problem 85 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(y=\sec \left(\tan ^{-1} x\right)\), then \(\frac{d y}{d x}\) at \(x=1\) is equal to: \([\mathbf{2 0 1 3}]\) (a) \(\frac{1}{\sqrt{2}} \quad\) (b) \(\frac{1}{2}\) (c) 1 (d) \(\sqrt{2}\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Function

We have the function \(y = \sec(\tan^{-1} x)\). Our goal is to find \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\), at \(x=1\).

Step 2: Express in Terms of Right Triangle

Consider \(\tan^{-1} x\) as an angle \(\theta\) such that \(\tan \theta = x\). In a right triangle with opposite side \(x\) and adjacent side \(1\), the hypotenuse is \(\sqrt{1 + x^2}\). Using this triangle, \(\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{\sqrt{1 + x^2}}{1} = \sqrt{1 + x^2}\). Thus, \(y = \sqrt{1 + x^2}\).

Step 3: Find the Derivative

Next, find the derivative \(\frac{dy}{dx}\) of \(y = \sqrt{1 + x^2}\). Using the chain rule: \[\frac{dy}{dx} = \frac{1}{2}(1 + x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1 + x^2}}\]

Step 4: Evaluate the Derivative at x=1

Now evaluate \(\frac{dy}{dx}\) at \(x = 1\): \[\frac{dy}{dx}\Big|_{x=1} = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}}\]

Step 5: Select the Correct Option

The evaluated derivative at \(x=1\) gives \(\frac{1}{\sqrt{2}}\). Comparing this with the options provided, (a) \(\frac{1}{\sqrt{2}}\) is the correct answer.

Key Concepts

Trigonometric Functions

Trigonometric functions are essential in mathematics, especially when dealing with angles and the relationships between the sides of right triangles. The function \(\sec(\theta)\) is the reciprocal of the cosine function, meaning \(\sec(\theta) = \frac{1}{\cos(\theta)}\). Trigonometric functions often appear in calculus problems where we need to find derivatives or integrals. These functions include sine (\(\sin\)), cosine (\(\cos\)), tangent (\(\tan\)), cosecant (\(\csc\)), secant (\(\sec\)), and cotangent (\(\cot\)).

• They relate the angles of a triangle to the ratios of its sides.
• Helpful in modeling periodic phenomena like waves and oscillations.

In this specific exercise, \(\sec(\tan^{-1} x) = \sec(\theta) = \sqrt{1+x^2} \). This is achieved by considering a right angle triangle, using \(\tan^{-1} x \) as an angle whose tangent is \(x\). Understanding this basic concept of trigonometric functions is crucial when transforming complex expressions into manageable calculations.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the roles of angles and ratios in the trigonometric functions. Each trigonometric function has an inverse, allowing us to retrieve the angle when the value of the trigonometric function is known. For instance, \(\tan^{-1} x\), also known as \(\arctan(x)\), gives the angle whose tangent is \(x\).

• They are vital when angles need to be calculated from known trigonometric values.
• Inverse functions often appear in calculus, particularly when taking derivatives.

When using inverse trigonometric functions, one often represents the result geometrically, as in the given exercise where \(\tan^{-1} x = \theta\), leads to \(\tan(\theta) = x\). From this relationship, we can construct a right triangle, helping us in expressing \(\sec(\theta)\) in different terms for further calculations. These functions provide a way to "undo" the trigonometric operations, making them highly useful in higher mathematics.

Chain Rule

The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. When you have a composite function such as \(f(g(x))\), the chain rule states that the derivative, \(\frac{d}{dx}[f(g(x))]\), is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Formally, it's written as:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

• Crucial when differentiating functions nested within one another.
• Simplifies the process of taking derivatives of complex expressions.

In the problem, the chain rule simplifies the derivative finding process. We transformed \(y = \sec(\tan^{-1} x)\) to \(y = \sqrt{1 + x^2}\), and differentiated it using the chain rule: first apply the power rule to \(\sqrt{1+x^2}\), and then multiply by the derivative of the inner function \(1+x^2\), which is \(2x\). This method is powerful in breaking down calculations that could otherwise become tricky.