Question

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

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

Hint:

Standard differentiation gives \(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\). Since \(\sqrt{1-x^2} = \sqrt{1-\sin^2 y} = \cos y\), the derivative can also be simply expressed in terms of \(y\) as \(\sec y\).

Answer 1

The Correct Option is D

Concept:
• If \(y = \sin^{-1} x\), then by the definition of inverse trigonometric functions, \(x = \sin y\), where \(y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
• The derivative can be determined by implicitly differentiating with respect to \(x\) or by differentiating \(x\) with respect to \(y\) using \(\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}\).
• Recall the basic reciprocal trigonometric relation: \(\frac{1}{\cos y} = \sec y\).

Step 1: Rewrite the inverse equation in direct trigonometric form
Given: \[ y = \sin^{-1} x \] Taking the sine on both sides: \[ x = \sin y \]

Step 2: Differentiate both sides with respect to \(y\)
Differentiating \(x\) with respect to \(y\): \[ \frac{dx}{dy} = \frac{d}{dy}(\sin y) = \cos y \]

Step 3: Determine \(\frac{dy}{dx}\)
Using the derivative rule for inverse functions: \[ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{\cos y} \] Using the reciprocal trigonometric identity: \[ \frac{dy}{dx} = \sec y \]