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

Answer

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

### Step 1: Identify the inner function
Let \( u = \frac{x}{x+1} \). We can rewrite our function as:
\[ y = \cos^{-1}(u) \]

### Step 2: Differentiate \( y \) with respect to \( u \)
The derivative of \( \cos^{-1}(u) \) with respect to \( u \) is given by:
\[ \frac{dy}{du} = -\frac{1}{\sqrt{1 - u^2}} \]

### Step 3: Differentiate \( u \) with respect to \( x \)
Now we need to find \( \frac{du}{dx} \). We will use the quotient rule for differentiation:
\[ u = \frac{x}{x+1} \]
Using the quotient rule \( \frac{du}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \), where \( u = x \) and \( v = x + 1 \):
- \( \frac{du}{dx} = 1 \)
- \( \frac{dv}{dx} = 1 \)

So,
\[ \frac{du}{dx} = \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{x + 1 - x}{(x + 1)^2} = \frac{1}{(x + 1)^2} \]

### Step 4: Apply the chain rule
Now we can apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{\sqrt{1 - u^2}} \cdot \frac{1}{(x + 1)^2} \]

### Step 5: Substitute back for \( u \)
Now we need to substitute \( u = \frac{x}{x+1} \) back into the derivative:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - \left(\frac{x}{x+1}\right)^2}} \cdot \frac{1}{(x + 1)^2} \]

### Step 6: Simplify the expression
Now we simplify \( 1 - \left(\frac{x}{x+1}\right)^2 \):
\[ 1 - \left(\frac{x}{x+1}\right)^2 = 1 - \frac{x^2}{(x+1)^2} = \frac{(x+1)^2 - x^2}{(x+1)^2} = \frac{x^2 + 2x + 1 - x^2}{(x+1)^2} = \frac{2x + 1}{(x+1)^2} \]

Thus, we have:
\[ \sqrt{1 - \left(\frac{x}{x+1}\right)^2} = \sqrt{\frac{2x + 1}{(x + 1)^2}} = \frac{\sqrt{2x + 1}}{x + 1} \]

### Final expression for \( \frac{dy}{dx} \)
Substituting this back, we get:
\[ \frac{dy}{dx} = -\frac{1}{\frac{\sqrt{2x + 1}}{x + 1}} \cdot \frac{1}{(x + 1)^2} = -\frac{(x + 1)}{\sqrt{2x + 1}(x + 1)^2} = -\frac{1}{(x + 1)\sqrt{2x + 1}} \]

### Conclusion
Thus, the derivative of \( y = \cos^{-1}\left(\frac{x}{x+1}\right) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{1}{(x + 1)\sqrt{2x + 1}} \]