The derivative of `e^(sin^(-1)x)` is :

Options

A. `(e^(sin^(-1)x))/(sqrt(1-x^(2)))`
B. `(e^(sin^(-1)x))/(sqrt(1+x^(2)))`
C. `e^(sin^(-1)x)(sqrt(1+x^(2)))`
D. None of these

Answer

To find the derivative of the function \( y = e^{\sin^{-1}(x)} \), we will use the chain rule. Here’s a step-by-step solution:

### Step 1: Define the function
Let
\[ y = e^{\sin^{-1}(x)} \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we apply the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then the derivative is given by:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]
In our case, \( f(u) = e^u \) where \( u = \sin^{-1}(x) \).

### Step 3: Differentiate the outer function
The derivative of \( f(u) = e^u \) is:
\[ f'(u) = e^u \]
Thus,
\[ \frac{dy}{du} = e^{\sin^{-1}(x)} \]

### Step 4: Differentiate the inner function
Next, we need to differentiate the inner function \( g(x) = \sin^{-1}(x) \). The derivative of \( \sin^{-1}(x) \) is:
\[ g'(x) = \frac{1}{\sqrt{1 - x^2}} \]

### Step 5: Combine the derivatives
Now, we can combine the derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^{\sin^{-1}(x)} \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Step 6: Write the final answer
Thus, the derivative of \( y = e^{\sin^{-1}(x)} \) is:
\[ \frac{dy}{dx} = \frac{e^{\sin^{-1}(x)}}{\sqrt{1 - x^2}} \]

### Summary
The derivative of \( e^{\sin^{-1}(x)} \) is:
\[ \frac{e^{\sin^{-1}(x)}}{\sqrt{1 - x^2}} \]

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