If `x=sqrt(2^(cosec^(-1_t)))` and `y=sqrt(2^(sec^(-1_t)))`` (|t|ge1)`then (dy)/(dx) is equal to.

Options

A. y/x
B. `-y/x`
C. `-x/y`
D. x/y

Answer

To solve the problem, we need to find \( \frac{dy}{dx} \) given the equations for \( x \) and \( y \):

1. **Given:**
\[ x = \sqrt{2^{\csc^{-1}(t)}} \]
\[ y = \sqrt{2^{\sec^{-1}(t)}} \]

2. **Step 1: Differentiate \( x \) with respect to \( t \)**
\[ x = \sqrt{2^{\csc^{-1}(t)}} = (2^{\csc^{-1}(t)})^{1/2} = 2^{\frac{1}{2} \csc^{-1}(t)} \]
Using the chain rule:
\[ \frac{dx}{dt} = \frac{1}{2} \cdot 2^{\frac{1}{2} \csc^{-1}(t)} \cdot \ln(2) \cdot \frac{d}{dt}(\csc^{-1}(t)) \]

We know:
\[ \frac{d}{dt}(\csc^{-1}(t)) = -\frac{1}{\sqrt{t^2 - 1}} \quad \text{for } |t| \geq 1 \]

Thus:
\[ \frac{dx}{dt} = \frac{1}{2} \cdot 2^{\frac{1}{2} \csc^{-1}(t)} \cdot \ln(2) \cdot \left(-\frac{1}{\sqrt{t^2 - 1}}\right) \]
\[ \frac{dx}{dt} = -\frac{2^{\frac{1}{2} \csc^{-1}(t)} \ln(2)}{2 \sqrt{t^2 - 1}} \]

3. **Step 2: Differentiate \( y \) with respect to \( t \)**
\[ y = \sqrt{2^{\sec^{-1}(t)}} = (2^{\sec^{-1}(t)})^{1/2} = 2^{\frac{1}{2} \sec^{-1}(t)} \]
Using the chain rule:
\[ \frac{dy}{dt} = \frac{1}{2} \cdot 2^{\frac{1}{2} \sec^{-1}(t)} \cdot \ln(2) \cdot \frac{d}{dt}(\sec^{-1}(t)) \]

We know:
\[ \frac{d}{dt}(\sec^{-1}(t)) = \frac{1}{\sqrt{t^2 - 1}} \quad \text{for } |t| \geq 1 \]

Thus:
\[ \frac{dy}{dt} = \frac{1}{2} \cdot 2^{\frac{1}{2} \sec^{-1}(t)} \cdot \ln(2) \cdot \frac{1}{\sqrt{t^2 - 1}} \]
\[ \frac{dy}{dt} = \frac{2^{\frac{1}{2} \sec^{-1}(t)} \ln(2)}{2 \sqrt{t^2 - 1}} \]

4. **Step 3: Find \( \frac{dy}{dx} \)**
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the expressions we found:
\[ \frac{dy}{dx} = \frac{\frac{2^{\frac{1}{2} \sec^{-1}(t)} \ln(2)}{2 \sqrt{t^2 - 1}}}{-\frac{2^{\frac{1}{2} \csc^{-1}(t)} \ln(2)}{2 \sqrt{t^2 - 1}}} \]

The \( \ln(2) \) and \( 2 \sqrt{t^2 - 1} \) cancel out:
\[ \frac{dy}{dx} = -\frac{2^{\frac{1}{2} \sec^{-1}(t)}}{2^{\frac{1}{2} \csc^{-1}(t)}} \]

5. **Final Result:**
\[ \frac{dy}{dx} = -\frac{2^{\frac{1}{2} \sec^{-1}(t)}}{2^{\frac{1}{2} \csc^{-1}(t)}} \]