Question

If $ y = \sin^{-1} \left( \frac{5x + 12\sqrt{1 - x^2}}{13} \right) $ then $ \frac{dy}{dx} $ equals

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

Hint:

When differentiating inverse trigonometric functions, use the chain rule and remember the derivative formula for \( \sin^{-1}(u) \). For complex expressions inside \( u \), apply the chain rule carefully.

Answer 1

The Correct Option is C

We are given: \[ y = \sin^{-1} \left( \frac{5x + 12\sqrt{1 - x^2}}{13} \right) \] To differentiate \( y \) with respect to \( x \), we use the chain rule. First, recall that the derivative of \( \sin^{-1}(u) \) with respect to \( u \) is: \[ \frac{d}{du} \left( \sin^{-1}(u) \right) = \frac{1}{\sqrt{1 - u^2}} \] Let: \[ u = \frac{5x + 12\sqrt{1 - x^2}}{13} \] Now, differentiate \( y \) with respect to \( x \) using the chain rule: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
Step 1: Differentiate \( u \) with respect to \( x \)
We need to differentiate: \[ u = \frac{5x + 12\sqrt{1 - x^2}}{13} \] First, differentiate \( 5x \) with respect to \( x \), which is simply 5. Next, differentiate \( 12\sqrt{1 - x^2} \) with respect to \( x \) using the chain rule: \[ \frac{d}{dx} \left( 12\sqrt{1 - x^2} \right) = 12 \cdot \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-12x}{\sqrt{1 - x^2}} \] Thus, the derivative of \( u \) is: \[ \frac{du}{dx} = \frac{5 + \frac{-12x}{\sqrt{1 - x^2}}}{13} \]
Step 2: Substitute \( \frac{du}{dx} \) into the expression for \( \frac{dy}{dx} \)
Now, substitute this expression for \( \frac{du}{dx} \) into the chain rule formula for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{1}{13} \left( 5 + \frac{-12x}{\sqrt{1 - x^2}} \right) \] After simplifying, we find: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \] Thus, the correct answer is \( C \).