Differentiate the following with respect to x using first principle method.
`sqrt(x)+(1)/(sqrt(x))`

Answer

To differentiate the function \( f(x) = \sqrt{x} + \frac{1}{\sqrt{x}} \) using the first principle of derivatives, we follow these steps:

### Step 1: Define the function
Let \( f(x) = \sqrt{x} + \frac{1}{\sqrt{x}} \).

### Step 2: Apply the first principle of derivatives
The derivative of \( f(x) \) using the first principle is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 3: Calculate \( f(x+h) \)
We need to find \( f(x+h) \):
\[ f(x+h) = \sqrt{x+h} + \frac{1}{\sqrt{x+h}} \]

### Step 4: Substitute into the limit
Now, substitute \( f(x+h) \) and \( f(x) \) into the limit:
\[ f'(x) = \lim_{h \to 0} \frac{\left(\sqrt{x+h} + \frac{1}{\sqrt{x+h}}\right) - \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)}{h} \]

### Step 5: Simplify the expression
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x} + \frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} \]

### Step 6: Split the limit
We can split this limit into two parts:
\[ f'(x) = \lim_{h \to 0} \left( \frac{\sqrt{x+h} - \sqrt{x}}{h} \right) + \lim_{h \to 0} \left( \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} \right) \]

### Step 7: Evaluate the first limit
For the first limit, we can use the identity \( a - b = \frac{(a^2 - b^2)}{a + b} \):
\[ \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \frac{1}{2\sqrt{x}} \]

### Step 8: Evaluate the second limit
For the second limit, we can also use the identity:
\[ \frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}} = \frac{\sqrt{x} - \sqrt{x+h}}{\sqrt{x}\sqrt{x+h}} = \frac{-h}{\sqrt{x}\sqrt{x+h}(\sqrt{x+h} + \sqrt{x})} \]
Thus:
\[ \lim_{h \to 0} \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} = \lim_{h \to 0} \frac{-1}{\sqrt{x}\sqrt{x+h}} \cdot \frac{1}{\sqrt{x+h} + \sqrt{x}} = -\frac{1}{2x} \]

### Step 9: Combine the results
Now, combine both limits:
\[ f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2x} \]

### Final Result
Thus, the derivative of \( f(x) = \sqrt{x} + \frac{1}{\sqrt{x}} \) is:
\[ f'(x) = \frac{1}{2\sqrt{x}} - \frac{1}{2x} \]