Question

If \( \tan^{-1} x = y \), then \( \frac{dy}{dx} \) is equal to :

• A. \( (\sec^{-1} x)^2 \)
• B. \( \sec^2 y \)
• C. \( \frac{1}{\sqrt{1+x^2}} \)
• D. \( \cos^2 y \)

Hint:

When evaluating derivatives involving inverse trigonometric expressions, if the standard answer \( \frac{1}{1+x^2} \) is not present in the options, look to convert the variable \( x \) back into terms of \( y \) using identities like \( \sec^2 y = 1 + \tan^2 y = 1 + x^2 \). Thus, \( \frac{1}{1+x^2} = \frac{1}{\sec^2 y} = \cos^2 y \).

Answer 1

The Correct Option is D

Concept: The problem requires finding the first derivative of a given trigonometric relation with respect to \( x \). We can approach this problem either by using the standard formula for the derivative of the inverse trigonometric function \( \tan^{-1} x \) or by converting the inverse trigonometric expression into an explicit trigonometric form and employing implicit differentiation along with fundamental trigonometric identities.

Step 1: Expressing the equation explicitly in terms of \(x\).
The given equation is: \[ \tan^{-1} x = y \] By taking the tangent of both sides, we can rewrite the inverse function as a direct trigonometric equation: \[ x = \tan y \]

Step 2: Differentiating implicitly with respect to \(x\).
Now, we differentiate both sides of the equation \( x = \tan y \) with respect to \( x \). Applying the chain rule on the right-hand side, we get: \[ \frac{d}{dx}(x) = \frac{d}{dx}(\tan y) \] \[ 1 = \sec^2 y \cdot \frac{dy}{dx} \]

Step 3: Solving for \( \frac{dy}{dx} \) and matching the options.
To isolate \( \frac{dy}{dx} \), we divide both sides by \( \sec^2 y \): \[ \frac{dy}{dx} = \frac{1}{\sec^2 y} \] Using the reciprocal trigonometric identity where \( \frac{1}{\sec \theta} = \cos \theta \), we can simplify the expression: \[ \frac{dy}{dx} = \cos^2 y \] This precisely matches option (D).