The derivative of `sin^(-1)((2x)/(1+x^(2)))`w.r.t`tan^(-1)((2x)/(1-x^(2)))` is

Options

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

Answer

To find the derivative of \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \), we will follow these steps:

### Step 1: Substitution
Let \( x = \tan(\theta) \). Then, we have:
- \( \tan^{-1}(x) = \theta \)

### Step 2: Rewrite the Functions
Now we can rewrite the expressions for \( u \) and \( v \):
- \( u = \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) \)
- \( v = \tan^{-1}\left(\frac{2\tan(\theta)}{1-\tan^2(\theta)}\right) \)

### Step 3: Simplify the Expressions
Using the double angle formulas:
- \( \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta) \)
- \( \frac{2\tan(\theta)}{1-\tan^2(\theta)} = \tan(2\theta) \)

Thus, we can rewrite:
- \( u = \sin^{-1}(\sin(2\theta)) = 2\theta \)
- \( v = \tan^{-1}(\tan(2\theta)) = 2\theta \)

### Step 4: Differentiate \( u \) and \( v \)
Now, we differentiate \( u \) and \( v \):
- \( \frac{du}{d\theta} = 2 \)
- \( \frac{dv}{d\theta} = 2 \)

### Step 5: Use the Chain Rule
To find \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{du/d\theta}{dv/d\theta} = \frac{2}{2} = 1 \]

### Final Answer
Thus, the derivative of \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) is:
\[ \frac{du}{dv} = 1 \]