Differentiate w.r.t. as indicated :
`sin^(-1)((2x)/(1+x^(2)))" w.r.t. "tan^(-1)x`

Answer

To differentiate \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( \tan^{-1}(x) \), we can follow these steps:

### Step 1: Define the Functions
Let:
- \( u = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)
- \( v = \tan^{-1}(x) \)

### Step 2: Use the Chain Rule
We need to find \( \frac{du}{dv} \). By the chain rule, we can express this as:
\[ \frac{du}{dv} = \frac{du}{dx} \cdot \frac{dx}{dv} \]
This means we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

### Step 3: Differentiate \( u \) with respect to \( x \)
To differentiate \( u \), we first simplify \( \frac{2x}{1+x^2} \):
\[ \frac{2x}{1+x^2} = \sin(2\theta) \quad \text{where } \theta = \tan^{-1}(x) \]
Thus, we can write:
\[ u = \sin^{-1}(\sin(2\theta)) = 2\theta = 2\tan^{-1}(x) \]
Now, differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} \]
So,
\[ \frac{du}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \]

### Step 4: Differentiate \( v \) with respect to \( x \)
Now, differentiate \( v \):
\[ \frac{dv}{dx} = \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} \]

### Step 5: Substitute into the Chain Rule
Now substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into the chain rule:
\[ \frac{du}{dv} = \frac{du}{dx} \cdot \frac{dx}{dv} = \frac{du}{dx} \cdot \frac{1}{\frac{dv}{dx}} = \frac{\frac{2}{1+x^2}}{\frac{1}{1+x^2}} = 2 \]

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