If \( x = \sqrt{2^{\text{cosec}^{-1} t}} \) and \( y = \sqrt{2^{\text{sec}^{-1} t}} (|t| \ge 1) \), then dy/dx is equal to :
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Whenever you see inverse trigonometric functions that sum to \( \pi/2 \) in exponents, try multiplying the functions to eliminate the parameter 't'. This simplifies the differentiation significantly. \
Step 1: Understanding the Concept:
This problem involves parametric differentiation or implicit differentiation. A very useful property of inverse trigonometric functions is \(\text{sec}^{-1}t + \text{cosec}^{-1}t = \pi/2\). Using this can simplify the relationship between $x$ and $y$ significantly. Step 2: Key Formula or Approach:
1. \(\text{sec}^{-1}t + \text{cosec}^{-1}t = \frac{\pi}{2}\).
2. If \(x \cdot y = \text{constant}\), then \(y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x}\). Step 3: Detailed Explanation:
1. Multiply $x$ and $y$:
\[ x \cdot y = \sqrt{2^{\text{cosec}^{-1}t}} \cdot \sqrt{2^{\text{sec}^{-1}t}} \]
2. Combine the square roots and use laws of exponents:
\[ xy = \sqrt{2^{\text{cosec}^{-1}t} \cdot 2^{\text{sec}^{-1}t}} = \sqrt{2^{\text{cosec}^{-1}t + \text{sec}^{-1}t}} \]
3. Substitute the identity \(\text{cosec}^{-1}t + \text{sec}^{-1}t = \pi/2\):
\[ xy = \sqrt{2^{\pi/2}} \]
4. Since the right side is a constant, differentiate both sides with respect to $x$ using the product rule:
\[ \frac{d}{dx}(xy) = \frac{d}{dx}(\text{constant}) \]
\[ x \frac{dy}{dx} + y(1) = 0 \]
5. Solve for \(dy/dx\):
\[ x \frac{dy}{dx} = -y \implies \frac{dy}{dx} = -\frac{y}{x} \] Step 4: Final Answer
The derivative \(dy/dx\) is equal to \(-y/x\).
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