Differentiate `sinsqrt(x) + cos^(2)sqrt(x)`

Explanation

To differentiate the function \( y = \sin(\sqrt{x}) + \cos^2(\sqrt{x}) \), we will apply the chain rule and the product rule where necessary. Let's break it down step by step.

### Step 1: Identify the function
We have:
\[ y = \sin(\sqrt{x}) + \cos^2(\sqrt{x}) \]

### Step 2: Differentiate \( \sin(\sqrt{x}) \)
Using the chain rule:
\[ \frac{d}{dx}[\sin(\sqrt{x})] = \cos(\sqrt{x}) \cdot \frac{d}{dx}[\sqrt{x}] \]
Now, differentiate \( \sqrt{x} \):
\[ \frac{d}{dx}[\sqrt{x}] = \frac{1}{2\sqrt{x}} \]
Thus,
\[ \frac{d}{dx}[\sin(\sqrt{x})] = \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\cos(\sqrt{x})}{2\sqrt{x}} \]

### Step 3: Differentiate \( \cos^2(\sqrt{x}) \)
Using the chain rule and the product rule:
\[ \frac{d}{dx}[\cos^2(\sqrt{x})] = 2\cos(\sqrt{x}) \cdot \frac{d}{dx}[\cos(\sqrt{x})] \]
Now, differentiate \( \cos(\sqrt{x}) \):
\[ \frac{d}{dx}[\cos(\sqrt{x})] = -\sin(\sqrt{x}) \cdot \frac{d}{dx}[\sqrt{x}] = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]
So,
\[ \frac{d}{dx}[\cos^2(\sqrt{x})] = 2\cos(\sqrt{x}) \cdot \left(-\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}}\right) = -\frac{\sin(\sqrt{x}) \cos(\sqrt{x})}{\sqrt{x}} \]

### Step 4: Combine the derivatives
Now, we combine the derivatives from Steps 2 and 3:
\[ \frac{dy}{dx} = \frac{\cos(\sqrt{x})}{2\sqrt{x}} - \frac{\sin(\sqrt{x}) \cos(\sqrt{x})}{\sqrt{x}} \]
We can factor out \( \frac{1}{\sqrt{x}} \):
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x}} \left( \frac{\cos(\sqrt{x})}{2} - \sin(\sqrt{x}) \cos(\sqrt{x}) \right) \]

### Step 5: Simplify the expression
We can rewrite the expression inside the parentheses:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{x}} \left( \frac{\cos(\sqrt{x}) - 2\sin(\sqrt{x}) \cos(\sqrt{x})}{2} \right) \]
This can be expressed as:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \left( \cos(\sqrt{x}) - 2\sin(\sqrt{x}) \cos(\sqrt{x}) \right) \]

### Final Answer
Thus, the derivative of the function is:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \left( \cos(\sqrt{x}) - 2\sin(\sqrt{x}) \cos(\sqrt{x}) \right) \]

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