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

Answer

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

### Step 1: Define \( u \) and \( v \)
Let:
\[ u = \tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right) \]
\[ v = \sin\left(2 \cot^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right) \]

### Step 2: Differentiate \( u \) with respect to \( x \)
To differentiate \( u \), we will first simplify it. We can use the substitution \( x = \sin(\theta) \), which gives us:
\[ u = \tan^{-1}\left(\frac{\sin(\theta)}{1 + \cos(\theta)}\right) \]
Using the identity \( 1 + \cos(\theta) = 2\cos^2\left(\frac{\theta}{2}\right) \) and \( \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \), we can rewrite \( u \):
\[ u = \tan^{-1}\left(\frac{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}{2\cos^2\left(\frac{\theta}{2}\right)}\right) = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2} \]
Since \( \theta = \sin^{-1}(x) \), we have:
\[ u = \frac{1}{2} \sin^{-1}(x) \]

Now, differentiate \( u \):
\[ \frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{1}{2\sqrt{1 - x^2}} \]

### Step 3: Differentiate \( v \) with respect to \( x \)
Next, we need to differentiate \( v \). We start with:
\[ v = \sin\left(2 \cot^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right) \]
Let \( w = \cot^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right) \), then:
\[ v = \sin(2w) = 2 \sin(w) \cos(w) \]
Using the identity \( \cot(w) = \sqrt{\frac{1+x}{1-x}} \), we can find \( \sin(w) \) and \( \cos(w) \):
\[ \sin(w) = \frac{\sqrt{1-x}}{\sqrt{2}} \quad \text{and} \quad \cos(w) = \frac{\sqrt{1+x}}{\sqrt{2}} \]
Thus,
\[ v = 2 \cdot \frac{\sqrt{1-x}}{\sqrt{2}} \cdot \frac{\sqrt{1+x}}{\sqrt{2}} = \frac{2\sqrt{(1-x)(1+x)}}{2} = \sqrt{1-x^2} \]

Now, differentiate \( v \):
\[ \frac{dv}{dx} = \frac{d}{dx}(\sqrt{1-x^2}) = \frac{-x}{\sqrt{1-x^2}} \]

### Step 4: Use the chain rule to find \( \frac{du}{dv} \)
Using the chain rule:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{2\sqrt{1-x^2}}}{\frac{-x}{\sqrt{1-x^2}}} = \frac{1}{2} \cdot \frac{\sqrt{1-x^2}}{-x} = -\frac{1}{2x} \]

### Final Answer
Thus, the derivative of \( u \) with respect to \( v \) is:
\[ \frac{du}{dv} = -\frac{1}{2x} \]