the derivation of ` cos^(-1) "" (1-x^2)/(1+x^2) w.r.t cot^(-1) "" (1-3x^2)/( 3x-x^3)` is

Options

A. `1/2`
B. `2/3`
C. `1`
D. `3/2`

Answer

To find the derivative of \( \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \) with respect to \( \cot^{-1} \left( \frac{1 - 3x^2}{3x - x^3} \right) \), we will follow these steps:

### Step 1: Substitute \( x = \tan(\theta) \)

Let \( x = \tan(\theta) \). Then, we can express both functions in terms of \( \theta \).

### Step 2: Simplify \( \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \)

Using the identity for cosine of double angle:
\[ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} \]
Thus,
\[ \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = \cos^{-1}(\cos(2\theta)) = 2\theta \]

### Step 3: Express \( \theta \) in terms of \( x \)

From our substitution, we have:
\[ \theta = \tan^{-1}(x) \]
So,
\[ \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = 2 \tan^{-1}(x) \]

### Step 4: Simplify \( \cot^{-1} \left( \frac{1 - 3x^2}{3x - x^3} \right) \)

Using the identity for cotangent:
\[ \cot(\theta) = \frac{1}{\tan(\theta)} \]
We can express:
\[ \cot^{-1} \left( \frac{1 - 3\tan^2(\theta)}{3\tan(\theta) - \tan^3(\theta)} \right) = \tan^{-1}(3\tan(\theta)) \]

### Step 5: Express the second function in terms of \( \theta \)

Using the identity:
\[ \tan^{-1}(3\tan(\theta)) = 3\theta \]
Thus, we have:
\[ \cot^{-1} \left( \frac{1 - 3x^2}{3x - x^3} \right) = 3\tan^{-1}(x) \]

### Step 6: Differentiate

Now we need to differentiate \( 2\tan^{-1}(x) \) with respect to \( 3\tan^{-1}(x) \).

Let \( u = 2\tan^{-1}(x) \) and \( v = 3\tan^{-1}(x) \).

Using the chain rule:
\[ \frac{du}{dv} = \frac{\frac{du}{dx}}{\frac{dv}{dx}} \]

### Step 7: Calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)

1. Differentiate \( u = 2\tan^{-1}(x) \):
\[ \frac{du}{dx} = \frac{2}{1 + x^2} \]

2. Differentiate \( v = 3\tan^{-1}(x) \):
\[ \frac{dv}{dx} = \frac{3}{1 + x^2} \]

### Step 8: Find \( \frac{du}{dv} \)

Now substituting:
\[ \frac{du}{dv} = \frac{\frac{2}{1 + x^2}}{\frac{3}{1 + x^2}} = \frac{2}{3} \]

### Final Answer

Thus, the derivative of \( \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \) with respect to \( \cot^{-1} \left( \frac{1 - 3x^2}{3x - x^3} \right) \) is:
\[ \frac{2}{3} \]