Differential coefficient of `sin^(-1)((1-x)/(1+x))` w.r.t `sqrt(x)` is

Options

A. `1/(2sqrt(x))`
B. `sqrt(x)/sqrt(1-x)`
C. 1
D. `-2/(1+x)`

Answer

To find the differential coefficient of \( \sin^{-1}\left(\frac{1-x}{1+x}\right) \) with respect to \( \sqrt{x} \), we will follow these steps:

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

### Step 2: Differentiate \( y \) with respect to \( x \)
Using the chain rule, we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{1-x}{1+x}\right)^2}} \cdot \frac{d}{dx}\left(\frac{1-x}{1+x}\right) \]

### Step 3: Differentiate the inner function
To differentiate \( \frac{1-x}{1+x} \), we apply the quotient rule:
\[ \frac{d}{dx}\left(\frac{1-x}{1+x}\right) = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1 - x - 1 + x}{(1+x)^2} = \frac{-2}{(1+x)^2} \]

### Step 4: Substitute back into the derivative
Now substituting back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - \left(\frac{1-x}{1+x}\right)^2}} \cdot \left(\frac{-2}{(1+x)^2}\right) \]

### Step 5: Simplify the expression under the square root
We need to simplify \( 1 - \left(\frac{1-x}{1+x}\right)^2 \):
\[ \left(\frac{1-x}{1+x}\right)^2 = \frac{(1-x)^2}{(1+x)^2} = \frac{1 - 2x + x^2}{1 + 2x + x^2} \]
Thus,
\[ 1 - \left(\frac{1-x}{1+x}\right)^2 = 1 - \frac{1 - 2x + x^2}{1 + 2x + x^2} = \frac{(1 + 2x + x^2) - (1 - 2x + x^2)}{1 + 2x + x^2} = \frac{4x}{1 + 2x + x^2} \]

### Step 6: Substitute back into the derivative
Now substituting this back:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{\frac{4x}{1 + 2x + x^2}}} \cdot \left(\frac{-2}{(1+x)^2}\right) = \frac{-2}{(1+x)^2} \cdot \frac{\sqrt{1 + 2x + x^2}}{2\sqrt{x}} = \frac{-\sqrt{1 + 2x + x^2}}{(1+x)^2\sqrt{x}} \]

### Step 7: Change the variable to \( z = \sqrt{x} \)
Now, let \( z = \sqrt{x} \), then \( x = z^2 \) and \( \frac{dx}{dz} = 2z \). We need to find \( \frac{dy}{dz} \):
\[ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} = \frac{-\sqrt{1 + 2z^2 + z^4}}{(1 + z^2)^2 z} \cdot 2z = \frac{-2\sqrt{1 + 2z^2 + z^4}}{(1 + z^2)^2} \]

### Final Answer
Thus, the differential coefficient of \( \sin^{-1}\left(\frac{1-x}{1+x}\right) \) with respect to \( \sqrt{x} \) is:
\[ \frac{dy}{dz} = \frac{-2\sqrt{1 + 2z^2 + z^4}}{(1 + z^2)^2} \]