Differentiate `tan^-1sqrt((1-x^2)/(1+x^2))` with respect to `cos^-1""x^2`

Answer

To differentiate the function \( u = \tan^{-1} \left( \sqrt{\frac{1 - x^2}{1 + x^2}} \right) \) with respect to \( v = \cos^{-1}(x^2) \), we will follow these steps:

### Step 1: Rewrite the Function
We start with:
\[ u = \tan^{-1} \left( \sqrt{\frac{1 - x^2}{1 + x^2}} \right) \]
We will express \( x^2 \) in terms of \( \theta \) where \( x^2 = \cos \theta \).

### Step 2: Substitute \( x^2 \)
Substituting \( x^2 = \cos \theta \) into \( u \):
\[ u = \tan^{-1} \left( \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \right) \]

### Step 3: Simplify the Expression
Using the identities \( 1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right) \) and \( 1 + \cos \theta = 2 \cos^2 \left( \frac{\theta}{2} \right) \):
\[ u = \tan^{-1} \left( \sqrt{\frac{2 \sin^2 \left( \frac{\theta}{2} \right)}{2 \cos^2 \left( \frac{\theta}{2} \right)}} \right) \]
This simplifies to:
\[ u = \tan^{-1} \left( \tan \left( \frac{\theta}{2} \right) \right) \]

### Step 4: Apply the Inverse Tangent Identity
Since \( \tan^{-1}(\tan x) = x \) (for \( x \) in the principal range):
\[ u = \frac{\theta}{2} \]

### Step 5: Differentiate \( u \) with Respect to \( \theta \)
Differentiating \( u \) with respect to \( \theta \):
\[ \frac{du}{d\theta} = \frac{1}{2} \]

### Step 6: Define \( v \)
Now, we define:
\[ v = \cos^{-1}(x^2) = \cos^{-1}(\cos \theta) = \theta \]

### Step 7: Differentiate \( v \) with Respect to \( \theta \)
Differentiating \( v \) with respect to \( \theta \):
\[ \frac{dv}{d\theta} = 1 \]

### Step 8: Use the Chain Rule to Find \( \frac{du}{dv} \)
Using the chain rule:
\[ \frac{du}{dv} = \frac{du/d\theta}{dv/d\theta} = \frac{\frac{1}{2}}{1} = \frac{1}{2} \]

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