If y = sec(tan⁻¹ x), then dy/dx at x = 1 is equal to (A) 1/√2 (B) 1/2 (C) 1 (D) √2

Q: Wait, why can't I just use the chain rule directly on $\sec(\tan^{-1} x)$?

You absolutely can, but it is a classic trap! Differentiating directly gives $\sec(\tan^{-1} x) \cdot \tan(\tan^{-1} x) \cdot \frac{1}{1+x^2}$. While mathematically correct, it is algebraically cumbersome and prone to errors. The JEE rewards those who simplify the expression first—it is faster, cleaner, and much less likely to lead to a calculation mistake.

Q: How do I know when to use the triangle method?

Whenever you see a composition of different trigonometric or inverse trigonometric functions, like $\sec(\tan^{-1} x)$ or $\sin(\cos^{-1} x)$, the triangle method is almost always the most efficient path. It effectively 'unpacks' the inverse function, turning a complex calculus problem into a straightforward algebraic one.

Q: Does the triangle method work if $x$ is negative?

Yes, it does! The algebraic identity $\sec(\tan^{-1} x) = \sqrt{1+x^2}$ holds true for all real $x$. The triangle method is a powerful visualization tool that remains consistent across the domain of the function. Trust the geometry; it is a robust way to handle these transformations.

Question

If $y = \sec(	an^{-1} x)$, then $\frac{dy}{dx}$ at $x = 1$ is equal to

Options:

(A) $\frac{1}{\sqrt{2}}$

(B) $\frac{1}{2}$

(C) $1$

(D) $\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are tackling a problem that serves as a perfect lesson in mathematical strategy. We are given the function $y = \sec(	an^{-1} x)$ and asked to find its derivative at $x = 1$.

At first glance, you might be tempted to dive headfirst into the chain rule. While that is a valid path, it is often the path of most resistance. In the high-stakes environment of the JEE Advanced, time and accuracy are your most precious resources.

### The Geometric Insight

Instead of brute-forcing the calculus, let us pause and look at the structure of the function. We have an inverse trigonometric function nested inside a trigonometric one. This is a signal—a flashing neon sign—that we should use geometry.

Imagine the inner part, $	heta = 	an^{-1} x$, as an angle in a right-angled triangle. By definition, this means $	an 	heta = x$. To make this useful, we write it as a ratio: $	an 	heta = \frac{x}{1}$.

Recall your basic trigonometry: tangent is the ratio of the perpendicular side to the base. In our mental canvas, we draw a right triangle where the perpendicular side is $x$ and the base is $1$. This is the "Spark" that simplifies everything.

### The Algebraic Transformation

With our triangle defined, we need to find the hypotenuse to complete the picture. The Pythagorean theorem is our trusty companion here:

$$	ext{Hypotenuse}^2 = 	ext{Base}^2 + 	ext{Perpendicular}^2$$

Substituting our values, we get $	ext{Hypotenuse}^2 = 1^2 + x^2$, which means the hypotenuse is $\sqrt{1 + x^2}$. Now, look at our original function: $y = \sec 	heta$.

In our triangle, secant is the ratio of the hypotenuse to the base. Therefore:

$$y = \sec 	heta = \frac{\sqrt{1 + x^2}}{1} = \sqrt{1 + x^2}$$

Just like that, the trigonometry has vanished, replaced by a simple algebraic expression: $y = \sqrt{1 + x^2}$.

### The Calculus

Now that we have $y = (1 + x^2)^{1/2}$, the differentiation becomes a breeze. We apply the power rule and the chain rule:

$$\frac{dy}{dx} = \frac{1}{2}(1 + x^2)^{\frac{1}{2} - 1} \cdot \frac{d}{dx}(1 + x^2)$$

The derivative of the inner term $(1 + x^2)$ is simply $2x$. So, our expression becomes:

$$\frac{dy}{dx} = \frac{1}{2}(1 + x^2)^{-\frac{1}{2}} \cdot 2x$$

The $2$ in the numerator and the $2$ in the denominator cancel out beautifully, leaving us with:

$$\frac{dy}{dx} = \frac{x}{\sqrt{1 + x^2}}$$

### The Final Step

We have arrived at the finish line. The problem asks for the value of the derivative at $x = 1$. Substituting this into our result:

$$\left. \frac{dy}{dx} ight|_{x=1} = \frac{1}{\sqrt{1 + (1)^2}} = \frac{1}{\sqrt{2}}$$

The final answer is **$\frac{1}{\sqrt{2}}$**. Remember this lesson: whenever you face a complex-looking function in the JEE, look for a way to simplify it before you start differentiating.

Animated Solution for Mathematics - Differentiation: If $y = sec (tan^{- 1} x)$ , then $\frac{d y}{d x}$ at $x = 1$ is equal to

Select Answer:

Visualized Solution

Analyzing the Function

Setting up the Angle $θ$

Mapping to a Right Triangle

Calculating the Hypotenuse

The Hypotenuse Expression

Evaluating $sec θ$

The Simplified Algebraic Function

Applying the Chain Rule

Executing the Differentiation

Simplifying the Derivative

Evaluating at $x = 1$

The Final Answer

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Geometric Insight

The Algebraic Transformation

The Calculus

The Final Step

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=sec(tan−1x), then dxdy​ at x=1 is equal to