Using definition , find the derivative of `f(x)=sinsqrt(x)`.

Answer

To find the derivative of the function \( f(x) = \sin(\sqrt{x}) \) using the definition of the derivative, we will follow these steps:

### Step 1: Write the definition of the derivative
The derivative of a function \( f(x) \) at a point \( x \) is defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x) \) and \( f(x+h) \)
Given \( f(x) = \sin(\sqrt{x}) \), we need to find \( f(x+h) \):
\[ f(x+h) = \sin(\sqrt{x+h}) \]
Now, substituting into the derivative definition:
\[ f'(x) = \lim_{h \to 0} \frac{\sin(\sqrt{x+h}) - \sin(\sqrt{x})}{h} \]

### Step 3: Use the sine subtraction formula
We can use the sine subtraction formula:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
Let \( A = \sqrt{x+h} \) and \( B = \sqrt{x} \). Then:
\[ f'(x) = \lim_{h \to 0} \frac{2 \cos\left(\frac{\sqrt{x+h} + \sqrt{x}}{2}\right) \sin\left(\frac{\sqrt{x+h} - \sqrt{x}}{2}\right)}{h} \]

### Step 4: Simplify the expression
Now we need to simplify \( \frac{\sqrt{x+h} - \sqrt{x}}{h} \). We can rationalize this:
\[ \sqrt{x+h} - \sqrt{x} = \frac{(x+h) - x}{\sqrt{x+h} + \sqrt{x}} = \frac{h}{\sqrt{x+h} + \sqrt{x}} \]
Thus,
\[ \frac{\sqrt{x+h} - \sqrt{x}}{h} = \frac{1}{\sqrt{x+h} + \sqrt{x}} \]

### Step 5: Substitute back into the limit
Now substituting this back into our limit:
\[ f'(x) = \lim_{h \to 0} 2 \cos\left(\frac{\sqrt{x+h} + \sqrt{x}}{2}\right) \cdot \frac{1}{\sqrt{x+h} + \sqrt{x}} \cdot \frac{h}{h} \]
As \( h \to 0 \), \( \sqrt{x+h} \to \sqrt{x} \):
\[ f'(x) = 2 \cos\left(\sqrt{x}\right) \cdot \frac{1}{2\sqrt{x}} = \frac{\cos(\sqrt{x})}{\sqrt{x}} \]

### Final Answer
Thus, the derivative of \( f(x) = \sin(\sqrt{x}) \) is:
\[ f'(x) = \frac{\cos(\sqrt{x})}{\sqrt{x}} \]