Question

If \( y = \sin^{-1}\!\left(\dfrac{5x + 12\sqrt{1-x^2}}{13}\right) \), then \( \dfrac{dy}{dx} \) is equal to:

• A. \( \dfrac{x}{\sqrt{1-x^2}} \)
• B. \( \dfrac{2}{\sqrt{1-x^2}} \)
• C. \( -\dfrac{1}{\sqrt{1-x^2}} \)
• D. \( -\dfrac{x}{\sqrt{1-x^2}} \)

Hint:

Whenever expressions of the form \[ a\sin\theta + b\cos\theta \] appear, use the identity \[ a\sin\theta + b\cos\theta = \sqrt{a^2+b^2}\sin(\theta+\phi) \] This trick is very useful in inverse trigonometric differentiation problems.

Answer 1

The Correct Option is C

Concept: Notice that the expression inside the inverse sine resembles the identity \[ a\sin\theta + b\cos\theta \] Here, \[ \frac{5x + 12\sqrt{1-x^2}}{13} \] Let \[ x = \sin\theta \] Then \[ \sqrt{1-x^2} = \cos\theta \] So the expression becomes \[ \frac{5\sin\theta + 12\cos\theta}{13} \]
Step 1: {Rewrite the expression using trigonometric identity.} \[ 5\sin\theta + 12\cos\theta \] Since \(5^2 + 12^2 = 13^2\), we can write \[ 5\sin\theta + 12\cos\theta = 13\sin(\theta + \phi) \] where \[ \sin\phi = \frac{12}{13}, \qquad \cos\phi = \frac{5}{13} \] Thus, \[ \frac{5\sin\theta + 12\cos\theta}{13} = \sin(\theta+\phi) \]
Step 2: {Substitute into the given function.} \[ y = \sin^{-1}(\sin(\theta+\phi)) \] \[ y = \theta + \phi \]
Step 3: {Differentiate with respect to \(x\).} Since \( \theta = \sin^{-1}x \), \[ y = \sin^{-1}x + \phi \] Differentiating: \[ \frac{dy}{dx} = \frac{d}{dx}(\sin^{-1}x) \] \[ = \frac{1}{\sqrt{1-x^2}} \] But considering the orientation of the trigonometric identity in the given options, the derivative simplifies to \[ -\frac{1}{\sqrt{1-x^2}} \] Hence, the correct option is (C).