For `t in (0,1), Let \x= sqrt( 2 ^(sin^(-1)(t))) and y=sqrt(2 ^(cos^(-1)(t)))` then `1+ ((dy)/(dx))^(2)` equals :

Options

A. `(x ^(2))/(y ^(2))`
B. `(y ^(2))/(x ^(2))`
C. `(x ^(2) + y ^(2))/( y ^(2))`
D. `(x ^(2) + y ^(2))/( x ^(2))`

Answer

To solve the problem, we need to find the value of \( 1 + \left( \frac{dy}{dx} \right)^2 \) given the expressions for \( x \) and \( y \):

\[ x = \sqrt{2^{\sin^{-1}(t)}} \]
\[ y = \sqrt{2^{\cos^{-1}(t)}} \]

### Step 1: Rewrite \( x \) and \( y \)

We can rewrite \( x \) and \( y \) as follows:

\[ x = 2^{\frac{1}{2} \sin^{-1}(t)} \]
\[ y = 2^{\frac{1}{2} \cos^{-1}(t)} \]

### Step 2: Take the logarithm of \( x \) and \( y \)

Taking the logarithm of both sides:

\[ \log x = \frac{1}{2} \sin^{-1}(t) \log 2 \]
\[ \log y = \frac{1}{2} \cos^{-1}(t) \log 2 \]

### Step 3: Differentiate \( x \) and \( y \) with respect to \( t \)

Differentiating \( x \):

\[ \frac{1}{x} \frac{dx}{dt} = \frac{1}{2} \log 2 \cdot \frac{1}{\sqrt{1 - t^2}} \]

Thus,

\[ \frac{dx}{dt} = \frac{1}{2} \log 2 \cdot x \cdot \frac{1}{\sqrt{1 - t^2}} \]

Differentiating \( y \):

\[ \frac{1}{y} \frac{dy}{dt} = \frac{1}{2} \log 2 \cdot \left(-\frac{1}{\sqrt{1 - t^2}}\right) \]

Thus,

\[ \frac{dy}{dt} = -\frac{1}{2} \log 2 \cdot y \cdot \frac{1}{\sqrt{1 - t^2}} \]

### Step 4: Find \( \frac{dy}{dx} \)

Using the chain rule:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the expressions we found:

\[ \frac{dy}{dx} = \frac{-\frac{1}{2} \log 2 \cdot y \cdot \frac{1}{\sqrt{1 - t^2}}}{\frac{1}{2} \log 2 \cdot x \cdot \frac{1}{\sqrt{1 - t^2}}} \]

The \( \frac{1}{\sqrt{1 - t^2}} \) and \( \frac{1}{2} \log 2 \) cancel out:

\[ \frac{dy}{dx} = -\frac{y}{x} \]

### Step 5: Calculate \( \left( \frac{dy}{dx} \right)^2 \)

Now we square \( \frac{dy}{dx} \):

\[ \left( \frac{dy}{dx} \right)^2 = \left(-\frac{y}{x}\right)^2 = \frac{y^2}{x^2} \]

### Step 6: Find \( 1 + \left( \frac{dy}{dx} \right)^2 \)

Now we can find:

\[ 1 + \left( \frac{dy}{dx} \right)^2 = 1 + \frac{y^2}{x^2} \]

To combine these, we can write:

\[ 1 + \frac{y^2}{x^2} = \frac{x^2 + y^2}{x^2} \]

### Final Answer

Thus, the final result is:

\[ 1 + \left( \frac{dy}{dx} \right)^2 = \frac{x^2 + y^2}{x^2} \]