Question

If \[ x = \sqrt{e^{\sin^{-1} t}} \quad \text{and} \quad y = \sqrt{e^{\cos^{-1} t}}, \] then find \[ \frac{d^2 y}{dx^2}. \]

• A. \[ -\frac{y}{x^2} \]
• B. $\frac{y}{2x^{2}}$
• C. $\frac{2y}{x^{2}}$
• D. $-\frac{2y}{x^{2}}$

Hint:

When dealing with parametric equations involving inverse trigonometric functions, try to simplify the product or sum of the variables first using fundamental identities like $\sin^{-1}t + \cos^{-1}t = \frac{\pi}{2}$. This can significantly reduce the complexity of differentiation.

Answer 1

The Correct Option is A

Step 1: Given \[ x = \sqrt{e^{\sin^{-1} t}} \quad \text{and} \quad y = \sqrt{e^{\cos^{-1} t}} \] Multiply $x$ and $y$: \[ xy = \sqrt{e^{\sin^{-1} t}} \cdot \sqrt{e^{\cos^{-1} t}} \] \[ xy = \sqrt{e^{\sin^{-1} t + \cos^{-1} t}} \]
Step 2: Use identity: \[ \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \] \[ xy = \sqrt{e^{\frac{\pi}{2}}} \] Let $C = \sqrt{e^{\frac{\pi}{2}}}$, so: \[ xy = C \]
Step 3: Differentiate: \[ x \frac{dy}{dx} + y = 0 \] \[ x \frac{dy}{dx} = -y \] \[ \frac{dy}{dx} = -\frac{y}{x} \quad \text{(i)} \]
Step 4: Differentiate again: \[ \frac{d^2 y}{dx^2} = -\left(\frac{x \frac{dy}{dx} - y}{x^2}\right) \]
Step 5: Substitute (i): \[ \frac{d^2 y}{dx^2} = -\left(\frac{x\left(-\frac{y}{x}\right) - y}{x^2}\right) \] \[ = -\left(\frac{-y - y}{x^2}\right) \] \[ = -\left(\frac{-2y}{x^2}\right) \] \[ = \frac{2y}{x^2} \]
Final Answer:
\[ \boxed{\frac{2y}{x^2}} \]