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

Answer

To differentiate the function \( y = \log(\sin(\sqrt{1 + x^2})) \) with respect to \( x \), we will apply the chain rule and the properties of logarithmic differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate the logarithm
We start with the function:
\[ y = \log(\sin(\sqrt{1 + x^2})) \]
Using the chain rule, the derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \sin(\sqrt{1 + x^2}) \).

### Step 2: Differentiate the inner function
Now we need to differentiate \( u = \sin(\sqrt{1 + x^2}) \):
\[ \frac{du}{dx} = \cos(\sqrt{1 + x^2}) \cdot \frac{d}{dx}(\sqrt{1 + x^2}) \]

### Step 3: Differentiate \( \sqrt{1 + x^2} \)
Next, we differentiate \( \sqrt{1 + x^2} \):
\[ \frac{d}{dx}(\sqrt{1 + x^2}) = \frac{1}{2\sqrt{1 + x^2}} \cdot \frac{d}{dx}(1 + x^2) = \frac{1}{2\sqrt{1 + x^2}} \cdot 2x = \frac{x}{\sqrt{1 + x^2}} \]

### Step 4: Combine the derivatives
Now, substituting back into the derivative of \( u \):
\[ \frac{du}{dx} = \cos(\sqrt{1 + x^2}) \cdot \frac{x}{\sqrt{1 + x^2}} \]

### Step 5: Substitute back to find \( \frac{dy}{dx} \)
Now we substitute \( u \) and \( \frac{du}{dx} \) back into the derivative of \( y \):
\[ \frac{dy}{dx} = \frac{1}{\sin(\sqrt{1 + x^2})} \cdot \cos(\sqrt{1 + x^2}) \cdot \frac{x}{\sqrt{1 + x^2}} \]

### Step 6: Simplify the expression
Thus, we can simplify:
\[ \frac{dy}{dx} = \frac{x \cos(\sqrt{1 + x^2})}{\sin(\sqrt{1 + x^2}) \sqrt{1 + x^2}} = x \cdot \cot(\sqrt{1 + x^2}) \cdot \frac{1}{\sqrt{1 + x^2}} \]

### Final Answer
The derivative of \( y = \log(\sin(\sqrt{1 + x^2})) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{x \cot(\sqrt{1 + x^2})}{\sqrt{1 + x^2}} \]