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

Options

A. `1`
B. `2`
C. `(1)/(2)`
D. `-2`

Answer

To find the derivative of \( f(x) = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \) with respect to \( g(x) = \tan^{-1}(x) \), we will use the chain rule. The steps are as follows:

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

### Step 2: Differentiate \( g(x) \)
The derivative of \( g(x) \) is:
\[ g'(x) = \frac{1}{1+x^2} \]

### Step 3: Simplify \( f(x) \)
To differentiate \( f(x) \), we first simplify \( f(x) \):
\[ f(x) = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \]
Substituting \( x = \tan(\theta) \), we have:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta) \]
Thus,
\[ f(x) = \tan^{-1}\left(\frac{\sec(\theta)-1}{\tan(\theta)}\right) \]

### Step 4: Rewrite in terms of sine and cosine
Using the identity \( \sec(\theta) = \frac{1}{\cos(\theta)} \) and \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we can express:
\[ f(x) = \tan^{-1}\left(\frac{\frac{1}{\cos(\theta)}-1}{\frac{\sin(\theta)}{\cos(\theta)}}\right) = \tan^{-1}\left(\frac{1 - \cos(\theta)}{\sin(\theta)}\right) \]

### Step 5: Apply half-angle identities
Using the half-angle identities:
\[ 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \]
We can rewrite:
\[ f(x) = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{\sin(\theta)}\right) \]
Since \( \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \):
\[ f(x) = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}\right) = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2} \]

### Step 6: Substitute back for \( x \)
Since \( \theta = \tan^{-1}(x) \):
\[ f(x) = \frac{1}{2} \tan^{-1}(x) \]

### Step 7: Differentiate \( f(x) \)
Now we differentiate \( f(x) \):
\[ f'(x) = \frac{1}{2} g'(x) = \frac{1}{2} \cdot \frac{1}{1+x^2} \]

### Step 8: Find the derivative \( \frac{f'(x)}{g'(x)} \)
Now we find:
\[ \frac{f'(x)}{g'(x)} = \frac{\frac{1}{2(1+x^2)}}{\frac{1}{1+x^2}} = \frac{1}{2} \]

### Final Answer
Thus, the derivative of \( f(x) \) with respect to \( g(x) \) is:
\[ \frac{f'(x)}{g'(x)} = \frac{1}{2} \]