If `y=sin^(-1)((sinalphasinx))/(1-cos alphasinx)`, then y'(0), is

Options

A. 1
B. `2tanalpha`
C. `(1//2)tanalpha`
D. `sinalpha`

Answer

To find \( y'(0) \) for the given function

\[ y = \sin^{-1}\left(\frac{\sin \alpha \sin x}{1 - \cos \alpha \sin x}\right), \]

we will differentiate \( y \) with respect to \( x \) and then evaluate the derivative at \( x = 0 \).

### Step 1: Differentiate \( y \)

Using the chain rule and the quotient rule, we have:

\[ y' = \frac{d}{dx}\left(\sin^{-1}(u)\right) \quad \text{where } u = \frac{\sin \alpha \sin x}{1 - \cos \alpha \sin x}. \]

The derivative of \( \sin^{-1}(u) \) is given by

\[ \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \]

### Step 2: Find \( \frac{du}{dx} \)

Using the quotient rule:

\[ \frac{du}{dx} = \frac{(1 - \cos \alpha \sin x)(\sin \alpha \cos x) - (\sin \alpha \sin x)(-\cos \alpha \cos x)}{(1 - \cos \alpha \sin x)^2}. \]

This simplifies to:

\[ \frac{du}{dx} = \frac{\sin \alpha \cos x (1 - \cos \alpha \sin x) + \sin \alpha \sin x \cos \alpha \cos x}{(1 - \cos \alpha \sin x)^2}. \]

### Step 3: Substitute \( u \) and \( \frac{du}{dx} \) into \( y' \)

Now substituting \( u \) and \( \frac{du}{dx} \) into the expression for \( y' \):

\[ y' = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \]

### Step 4: Evaluate \( y'(0) \)

At \( x = 0 \):

1. Calculate \( u \):

\[ u = \frac{\sin \alpha \sin 0}{1 - \cos \alpha \sin 0} = 0. \]

2. Calculate \( \sqrt{1 - u^2} \):

\[ \sqrt{1 - 0^2} = 1. \]

3. Calculate \( \frac{du}{dx} \) at \( x = 0 \):

Substituting \( x = 0 \) into \( \frac{du}{dx} \):

\[ \frac{du}{dx} = \frac{\sin \alpha \cos 0 (1 - \cos \alpha \cdot 0) + \sin \alpha \sin 0 \cos \alpha \cos 0}{(1 - \cos \alpha \cdot 0)^2} = \frac{\sin \alpha \cdot 1 \cdot 1 + 0}{1^2} = \sin \alpha. \]

### Step 5: Final Calculation of \( y'(0) \)

Now substituting everything back into \( y' \):

\[ y'(0) = \frac{1}{1} \cdot \sin \alpha = \sin \alpha. \]

Thus, the final result is:

\[ y'(0) = \sin \alpha. \]