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

Answer

To differentiate \( u = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) with respect to \( v = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \), we will use the chain rule. We can express the derivative as:

\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]

### Step 1: Differentiate \( u \) with respect to \( x \)

First, we need to find \( \frac{du}{dx} \).

Using the formula for the derivative of the inverse sine function:
\[ \frac{d}{dx} \sin^{-1}(y) = \frac{1}{\sqrt{1-y^2}} \cdot \frac{dy}{dx} \]

Let \( y = \frac{2x}{1+x^2} \). We first find \( \frac{dy}{dx} \).

Using the quotient rule:
\[ \frac{dy}{dx} = \frac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2} = \frac{2 + 2x^2 - 4x^2}{(1+x^2)^2} = \frac{2 - 2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \]

Now, we need to find \( 1 - y^2 \):
\[ y^2 = \left(\frac{2x}{1+x^2}\right)^2 = \frac{4x^2}{(1+x^2)^2} \]
\[ 1 - y^2 = 1 - \frac{4x^2}{(1+x^2)^2} = \frac{(1+x^2)^2 - 4x^2}{(1+x^2)^2} = \frac{1 + 2x^2 + x^4 - 4x^2}{(1+x^2)^2} = \frac{1 - 2x^2 + x^4}{(1+x^2)^2} = \frac{(1-x^2)^2}{(1+x^2)^2} \]

Thus, we have:
\[ \frac{du}{dx} = \frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = \frac{1}{\sqrt{\frac{(1-x^2)^2}{(1+x^2)^2}}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \]
\[ = \frac{(1+x^2)}{(1-x^2)} \cdot \frac{2(1-x^2)}{(1+x^2)^2} = \frac{2}{1+x^2} \]

### Step 2: Differentiate \( v \) with respect to \( x \)

Now we find \( \frac{dv}{dx} \).

Using the formula for the derivative of the inverse tangent function:
\[ \frac{d}{dx} \tan^{-1}(y) = \frac{1}{1+y^2} \cdot \frac{dy}{dx} \]

Let \( y = \frac{2x}{1-x^2} \). We first find \( \frac{dy}{dx} \).

Using the quotient rule:
\[ \frac{dy}{dx} = \frac{(1-x^2)(2) - (2x)(-2x)}{(1-x^2)^2} = \frac{2 - 2x^2 + 4x^2}{(1-x^2)^2} = \frac{2 + 2x^2}{(1-x^2)^2} = \frac{2(1+x^2)}{(1-x^2)^2} \]

Now, we need to find \( 1 + y^2 \):
\[ y^2 = \left(\frac{2x}{1-x^2}\right)^2 = \frac{4x^2}{(1-x^2)^2} \]
\[ 1 + y^2 = 1 + \frac{4x^2}{(1-x^2)^2} = \frac{(1-x^2)^2 + 4x^2}{(1-x^2)^2} = \frac{1 - 2x^2 + x^4 + 4x^2}{(1-x^2)^2} = \frac{1 + 2x^2 + x^4}{(1-x^2)^2} = \frac{(1+x^2)^2}{(1-x^2)^2} \]

Thus, we have:
\[ \frac{dv}{dx} = \frac{1}{1+y^2} \cdot \frac{dy}{dx} = \frac{(1-x^2)^2}{(1+x^2)^2} \cdot \frac{2(1+x^2)}{(1-x^2)^2} = \frac{2}{1+x^2} \]

### Step 3: Compute \( \frac{du}{dv} \)

Now we can compute \( \frac{du}{dv} \):
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{2}{1+x^2}}{\frac{2}{1+x^2}} = 1 \]

### 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:
\[ \frac{du}{dv} = 1 \]