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

Answer

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

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

### Step 2: Substitute \( x \) with \( \sin \theta \)
To simplify \( u \), we can substitute \( x \) with \( \sin \theta \):
\[ u = \tan^{-1}\left(\frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}\right) \]
Since \( \sqrt{1 - \sin^2 \theta} = \cos \theta \), we have:
\[ u = \tan^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) = \tan^{-1}(\tan \theta) = \theta \]

### Step 3: Differentiate \( u \) with respect to \( \theta \)
Now, we differentiate \( u \):
\[ \frac{du}{d\theta} = 1 \]

### Step 4: Substitute \( x \) with \( \cos \theta \) for \( v \)
Next, we can express \( v \) in terms of \( \theta \) by substituting \( x \) with \( \cos \theta \):
\[ v = \cos^{-1}(2\cos^2 \theta - 1) \]
Using the identity \( \cos 2\theta = 2\cos^2 \theta - 1 \), we have:
\[ v = \cos^{-1}(\cos 2\theta) = 2\theta \]

### Step 5: Differentiate \( v \) with respect to \( \theta \)
Now, we differentiate \( v \):
\[ \frac{dv}{d\theta} = 2 \]

### Step 6: Find \( \frac{du}{dv} \)
Using the chain rule, we find:
\[ \frac{du}{dv} = \frac{du/d\theta}{dv/d\theta} = \frac{1}{2} \]

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

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