The derivative of `sin^(-1)((2x)/(1+x^(2)))` with respect to `cos^(-1)((1-x^(2))/(1+x^(2)))` is

Options

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

Answer

To find the derivative of \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \), we can use the relationships between the inverse trigonometric functions.

### Step-by-Step Solution:

1. **Define the Functions**:
Let \( u = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) and \( v = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \).

2. **Use the Relationship**:
It is known that:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}(x) \]
and
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x) \]
Therefore, we can express both \( u \) and \( v \) in terms of \( \tan^{-1}(x) \):
\[ u = 2\tan^{-1}(x), \quad v = 2\tan^{-1}(x) \]

3. **Differentiate \( u \) and \( v \)**:
To find \( \frac{du}{dv} \), we first need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

- Differentiate \( u \):
\[ \frac{du}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \]

- Differentiate \( v \):
\[ \frac{dv}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \]

4. **Calculate \( \frac{du}{dv} \)**:
Now, we can find \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} = \frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}} = 1 \]

5. **Final Result**:
Therefore, the derivative of \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \) is:
\[ \frac{du}{dv} = 1 \]