`tan ^(-1)(cos sqrt(x))`

Answer

To differentiate the function \( y = \tan^{-1}(\cos(\sqrt{x})) \) with respect to \( x \), we will use the chain rule. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
Let:
- \( u = \cos(\sqrt{x}) \) (inner function)
- \( y = \tan^{-1}(u) \) (outer function)

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

### Step 3: Differentiate the inner function
Next, we need to differentiate \( u = \cos(\sqrt{x}) \) with respect to \( x \). We will apply the chain rule again:
- Let \( v = \sqrt{x} \) (inner function for \( u \))
- Then, \( u = \cos(v) \)

The derivative of \( u = \cos(v) \) with respect to \( v \) is:
\[ \frac{du}{dv} = -\sin(v) \]

Now, differentiate \( v = \sqrt{x} \) with respect to \( x \):
\[ \frac{dv}{dx} = \frac{1}{2\sqrt{x}} \]

### Step 4: Combine the derivatives using the chain rule
Now, we can find \( \frac{du}{dx} \) using the chain rule:
\[ \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = -\sin(v) \cdot \frac{1}{2\sqrt{x}} = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]

### Step 5: Combine everything to find \( \frac{dy}{dx} \)
Now we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{1 + u^2} \cdot \left(-\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\right) \]

Substituting back \( u = \cos(\sqrt{x}) \):
\[ \frac{dy}{dx} = \frac{1}{1 + \cos^2(\sqrt{x})} \cdot \left(-\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\right) \]

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

---