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

Options

A. `(1)/(1+x)`
B. `(-2)/(1+x)`
C. `(1)/(1+x^2)`
D. None

Answer

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

### Step 1: Substitute \( t = \sqrt{x} \)
Let \( t = \sqrt{x} \). Then, we have:
\[ x = t^2 \]

### Step 2: Rewrite the function in terms of \( t \)
Now, we can rewrite the function \( f(t) \):
\[ f(t) = \sin^{-1} \left( \frac{1 - t^2}{1 + t^2} \right) \]

### Step 3: Simplify the expression using trigonometric identities
Using the identity:
\[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos(2\theta) \]
we can set \( t = \tan(\theta) \). Thus:
\[ f(t) = \sin^{-1}(\cos(2\theta)) \]

### Step 4: Use the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \)
From the identity:
\[ \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \]
we can express \( f(t) \) as:
\[ f(t) = \frac{\pi}{2} - 2\theta \]

### Step 5: Substitute back \( \theta \)
Since \( \theta = \tan^{-1}(t) \), we have:
\[ f(t) = \frac{\pi}{2} - 2\tan^{-1}(t) \]

### Step 6: Differentiate \( f(t) \) with respect to \( t \)
Now, we differentiate \( f(t) \):
\[ f'(t) = 0 - 2 \cdot \frac{1}{1 + t^2} \cdot \frac{dt}{dt} = -\frac{2}{1 + t^2} \]

### Step 7: Relate \( \frac{dt}{dt} \) to \( \frac{dx}{d\sqrt{x}} \)
Since \( t = \sqrt{x} \), we have:
\[ \frac{dt}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2t} \]
Thus:
\[ \frac{dx}{dt} = 2t \]

### Step 8: Chain rule to find \( \frac{d}{d\sqrt{x}} \)
Using the chain rule:
\[ \frac{d}{d\sqrt{x}} = \frac{d}{dt} \cdot \frac{dt}{d\sqrt{x}} = f'(t) \cdot \frac{dt}{d\sqrt{x}} = -\frac{2}{1 + t^2} \cdot \frac{1}{2t} \]

### Step 9: Substitute back \( t \) in terms of \( x \)
Substituting \( t = \sqrt{x} \):
\[ \frac{d}{d\sqrt{x}} = -\frac{1}{\sqrt{x}(1 + x)} \]

### Final Result
Thus, the differentiation of \( \sin^{-1} \left( \frac{1-x}{1+x} \right) \) with respect to \( \sqrt{x} \) is:
\[ -\frac{1}{\sqrt{x}(1 + x)} \]