The differential `sin^(-1) x + sin^(-1) y = 1`, is

Options

A. `sqrt(1-x^(2)) dx + sqrt(1-y^(2)) dy = 0`
B. `sqrt(1-x^(2)) dy + sqrt(1-y^(2)) dx = 0`
C. `sqrt(1-x^(2)) dy + sqrt(1-y^(2)) dx = 0`
D. `sqrt(1-x^(2)) dx = sqrt(1-y^(2)) dy = 0`

Answer

To solve the problem given by the equation \( \sin^{-1} x + \sin^{-1} y = 1 \), we need to differentiate both sides with respect to \( x \). Let's go through the steps one by one.

### Step 1: Write down the equation
We start with the equation:
\[ \sin^{-1} x + \sin^{-1} y = 1 \]

### Step 2: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). Using the chain rule, we have:
\[ \frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} y) = \frac{d}{dx}(1) \]
The derivative of a constant (1) is 0. Thus, we have:
\[ \frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} y) = 0 \]

### Step 3: Apply the derivative formula
We know that:
\[ \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]
and by the chain rule:
\[ \frac{d}{dx}(\sin^{-1} y) = \frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} \]
Substituting these into our equation gives:
\[ \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = 0 \]

### Step 4: Rearranging the equation
We can rearrange this equation to isolate \(\frac{dy}{dx}\):
\[ \frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}} \]
Multiplying both sides by \(\sqrt{1 - y^2}\) gives:
\[ \frac{dy}{dx} = -\frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \]

### Step 5: Final expression
Thus, we have the final expression for the derivative:
\[ \frac{dy}{dx} = -\frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \]