Differentiate the following w.r.t. x :
`sqrt(x^(2)+1)-log(1/x+sqrt(1+1/x^(2)))`

Answer

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

### Step 1: Differentiate the first term \( \sqrt{x^2 + 1} \)

Using the chain rule, we have:
\[ \frac{d}{dx} \left( \sqrt{x^2 + 1} \right) = \frac{1}{2\sqrt{x^2 + 1}} \cdot \frac{d}{dx}(x^2 + 1) = \frac{1}{2\sqrt{x^2 + 1}} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}} \]

### Step 2: Differentiate the second term \( -\log\left(\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}}\right) \)

Let \( v = \frac{1}{x} + \sqrt{1 + \frac{1}{x^2}} \). Then,
\[ \frac{d}{dx}(-\log(v)) = -\frac{1}{v} \cdot \frac{dv}{dx} \]

Now, we need to find \( \frac{dv}{dx} \):
\[ v = \frac{1}{x} + \sqrt{1 + \frac{1}{x^2}} \]
Differentiating \( v \):
\[ \frac{dv}{dx} = -\frac{1}{x^2} + \frac{1}{2\sqrt{1 + \frac{1}{x^2}}} \cdot \frac{d}{dx}\left(1 + \frac{1}{x^2}\right) \]
The derivative of \( 1 + \frac{1}{x^2} \) is:
\[ \frac{d}{dx}\left(1 + \frac{1}{x^2}\right) = 0 - \frac{2}{x^3} = -\frac{2}{x^3} \]
Thus,
\[ \frac{dv}{dx} = -\frac{1}{x^2} - \frac{1}{2\sqrt{1 + \frac{1}{x^2}}} \cdot \frac{2}{x^3} = -\frac{1}{x^2} - \frac{1}{x^3\sqrt{1 + \frac{1}{x^2}}} \]

### Step 3: Substitute \( \frac{dv}{dx} \) back into the derivative of the logarithm

Now substituting back:
\[ \frac{d}{dx}(-\log(v)) = -\frac{1}{v} \cdot \left(-\frac{1}{x^2} - \frac{1}{x^3\sqrt{1 + \frac{1}{x^2}}}\right) = \frac{1}{v} \left(\frac{1}{x^2} + \frac{1}{x^3\sqrt{1 + \frac{1}{x^2}}}\right) \]

### Step 4: Combine the derivatives

Now, we can combine the derivatives:
\[ \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}} + \frac{1}{\left(\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}}\right)} \left(\frac{1}{x^2} + \frac{1}{x^3\sqrt{1 + \frac{1}{x^2}}}\right) \]

### Final Result

Thus, the derivative of the given function is:
\[ \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 1}} + \frac{1}{\left(\frac{1}{x} + \sqrt{1 + \frac{1}{x^2}}\right)} \left(\frac{1}{x^2} + \frac{1}{x^3\sqrt{1 + \frac{1}{x^2}}}\right) \]