Derivative of ` tan ^(-1) ((sqrt( 1+x^(2))-1)/( x)) w.r.cos ^(-1) sqrt((1+sqrt( 1+x^(2)))/( 2sqrt(1+x^(2))))` is

Options

A. 1
B. -1
C. 2
D. -2

Answer

To find the derivative of the function \( y = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right) \) with respect to \( z = \cos^{-1} \left( \sqrt{\frac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}} \right) \), we will follow these steps:

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

### Step 2: Find the Derivative \( \frac{dy}{dz} \)
Using the chain rule, we have:
\[ \frac{dy}{dz} = \frac{f'(x)}{g'(x)} \]

### Step 3: Find \( f'(x) \)
To find \( f'(x) \), we first simplify \( f(x) \):
1. Substitute \( x = \tan(\theta) \):
\[ f(x) = \tan^{-1} \left( \frac{\sqrt{1+\tan^2(\theta)} - 1}{\tan(\theta)} \right) = \tan^{-1} \left( \frac{\sec(\theta) - 1}{\tan(\theta)} \right) \]
2. Rewrite using sine and cosine:
\[ f(x) = \tan^{-1} \left( \frac{\frac{1}{\cos(\theta)} - 1}{\frac{\sin(\theta)}{\cos(\theta)}} \right) = \tan^{-1} \left( \frac{1 - \cos(\theta)}{\sin(\theta)} \right) \]
3. Using the half-angle identity:
\[ 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \quad \text{and} \quad \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \]
4. Thus, we have:
\[ f(x) = \tan^{-1} \left( \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} \right) = \tan^{-1} \left( \tan\left(\frac{\theta}{2}\right) \right) = \frac{\theta}{2} \]
5. Since \( \theta = \tan^{-1}(x) \):
\[ f(x) = \frac{1}{2} \tan^{-1}(x) \]
6. Differentiate:
\[ f'(x) = \frac{1}{2} \cdot \frac{1}{1+x^2} \]

### Step 4: Find \( g'(x) \)
Next, we simplify \( g(x) \):
1. Substitute \( x = \tan(\theta) \):
\[ g(x) = \cos^{-1} \left( \sqrt{\frac{1+\sec(\theta)}{2\sec(\theta)}} \right) \]
2. Simplify:
\[ g(x) = \cos^{-1} \left( \sqrt{\frac{2}{2\sec(\theta)}} \right) = \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\theta}{2} \]
3. Differentiate:
\[ g'(x) = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - \left(\frac{1}{\sqrt{2}}\right)^2}} = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{1}{2}}} = \frac{1}{2} \cdot \sqrt{2} \]

### Step 5: Calculate \( \frac{dy}{dz} \)
Now substituting back into our derivative:
\[ \frac{dy}{dz} = \frac{f'(x)}{g'(x)} = \frac{\frac{1}{2(1+x^2)}}{\frac{1}{2}\sqrt{2}} = \frac{1}{\sqrt{2}(1+x^2)} \]

### Final Result
Thus, the derivative \( \frac{dy}{dz} \) is:
\[ \frac{1}{\sqrt{2}(1+x^2)} \]