If `y=e^(sin^(-1)x)+e^(cos^(-1)x),0ltxlt1`, then

Options

A. `(dy)/(dx)=0`
B. `(dy)/(dx)=(pi)/(2)`
C. does not exist
D. none of these

Answer

To solve the problem, we need to find the derivative of the function \( y = e^{\sin^{-1}x} + e^{\cos^{-1}x} \) for \( 0 < x < 1 \).

### Step-by-Step Solution:

1. **Write the given function:**
\[ y = e^{\sin^{-1}x} + e^{\cos^{-1}x} \]

2. **Use the identity for inverse trigonometric functions:**
Recall that:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
Therefore, we can express \( e^{\cos^{-1}x} \) in terms of \( e^{\sin^{-1}x} \):
\[ e^{\cos^{-1}x} = e^{\frac{\pi}{2} - \sin^{-1}x} = \frac{e^{\frac{\pi}{2}}}{e^{\sin^{-1}x}} \]

3. **Substituting back into the equation:**
Substitute \( e^{\cos^{-1}x} \) into the equation for \( y \):
\[ y = e^{\sin^{-1}x} + \frac{e^{\frac{\pi}{2}}}{e^{\sin^{-1}x}} \]
Let \( z = e^{\sin^{-1}x} \), then:
\[ y = z + \frac{e^{\frac{\pi}{2}}}{z} \]

4. **Finding the derivative \( \frac{dy}{dx} \):**
Differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{dz}{dx} + \frac{d}{dx}\left(\frac{e^{\frac{\pi}{2}}}{z}\right) \]
Using the quotient rule:
\[ \frac{d}{dx}\left(\frac{e^{\frac{\pi}{2}}}{z}\right) = -\frac{e^{\frac{\pi}{2}} \frac{dz}{dx}}{z^2} \]

5. **Finding \( \frac{dz}{dx} \):**
We know:
\[ z = e^{\sin^{-1}x} \]
Thus,
\[ \frac{dz}{dx} = e^{\sin^{-1}x} \cdot \frac{d}{dx}(\sin^{-1}x) = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1-x^2}} \]

6. **Combining the derivatives:**
Substitute \( \frac{dz}{dx} \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1-x^2}} - \frac{e^{\frac{\pi}{2}} \cdot e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1-x^2}}}{(e^{\sin^{-1}x})^2} \]
Simplifying this gives:
\[ \frac{dy}{dx} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}} - \frac{e^{\frac{\pi}{2}}}{e^{\sin^{-1}x} \sqrt{1-x^2}} \]
\[ = \frac{e^{\sin^{-1}x} - e^{\frac{\pi}{2}}}{\sqrt{1-x^2}} \]

7. **Setting the derivative to zero:**
For \( \frac{dy}{dx} = 0 \):
\[ e^{\sin^{-1}x} - e^{\frac{\pi}{2}} = 0 \]
This implies:
\[ e^{\sin^{-1}x} = e^{\frac{\pi}{2}} \]
Therefore, \( \sin^{-1}x = \frac{\pi}{2} \) which is not possible for \( 0 < x < 1 \).

### Conclusion:
Thus, \( \frac{dy}{dx} = 0 \) for \( 0 < x < 1 \) means that \( y \) is constant in this interval.