`cos^-1((x)/sqrt(1+x^2))`

Answer

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

### Step 1: Identify the function
We start with the function:
\[ y = \cos^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we use the chain rule. The derivative of \( \cos^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \]
where \( u = \frac{x}{\sqrt{1+x^2}} \).

### Step 3: Calculate \( u \) and \( \frac{du}{dx} \)
First, we need to find \( u^2 \):
\[ u^2 = \left(\frac{x}{\sqrt{1+x^2}}\right)^2 = \frac{x^2}{1+x^2} \]
Now, we compute \( 1 - u^2 \):
\[ 1 - u^2 = 1 - \frac{x^2}{1+x^2} = \frac{1+x^2 - x^2}{1+x^2} = \frac{1}{1+x^2} \]

Next, we differentiate \( u \):
\[ u = \frac{x}{\sqrt{1+x^2}} \]
Using the quotient rule:
\[ \frac{du}{dx} = \frac{\sqrt{1+x^2} \cdot 1 - x \cdot \frac{1}{2\sqrt{1+x^2}} \cdot 2x}{1+x^2} \]
This simplifies to:
\[ \frac{du}{dx} = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} = \frac{1+x^2 - x^2}{(1+x^2)\sqrt{1+x^2}} = \frac{1}{(1+x^2)\sqrt{1+x^2}} \]

### Step 4: Substitute back into the derivative
Now we substitute \( u \) and \( \frac{du}{dx} \) back into the derivative:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]
Substituting \( 1 - u^2 = \frac{1}{1+x^2} \):
\[ \sqrt{1 - u^2} = \sqrt{\frac{1}{1+x^2}} = \frac{1}{\sqrt{1+x^2}} \]
Thus,
\[ \frac{dy}{dx} = -\frac{1}{\frac{1}{\sqrt{1+x^2}}} \cdot \frac{1}{(1+x^2)\sqrt{1+x^2}} = -\sqrt{1+x^2} \cdot \frac{1}{(1+x^2)\sqrt{1+x^2}} = -\frac{1}{1+x^2} \]

### Final Result
The derivative of the function \( y = \cos^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) \) is:
\[ \frac{dy}{dx} = -\frac{1}{1+x^2} \]