Find, from first principle, the derivative of the following w.r.t. x :
`e^(sqrtx)`

Answer

To find the derivative of the function \( f(x) = e^{\sqrt{x}} \) using the first principle, we can follow these steps:

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

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

### Step 3: Substitute \( f(x) \) into the derivative formula
Substituting \( f(x) \) into the formula, we get:
\[ f'(x) = \lim_{h \to 0} \frac{e^{\sqrt{x+h}} - e^{\sqrt{x}}}{h} \]

### Step 4: Factor out \( e^{\sqrt{x}} \)
We can factor out \( e^{\sqrt{x}} \) from the expression:
\[ f'(x) = e^{\sqrt{x}} \lim_{h \to 0} \frac{e^{\sqrt{x+h}} - e^{\sqrt{x}}}{h} \]

### Step 5: Simplify the limit
Now, we need to simplify the limit. To do this, we can use the property of exponents:
\[ e^{\sqrt{x+h}} - e^{\sqrt{x}} = e^{\sqrt{x}}(e^{\sqrt{x+h} - \sqrt{x}} - 1) \]
Thus, we have:
\[ f'(x) = e^{\sqrt{x}} \lim_{h \to 0} \frac{e^{\sqrt{x+h} - \sqrt{x}} - 1}{h} \]

### Step 6: Use the identity for exponential limits
We know that:
\[ \lim_{u \to 0} \frac{e^u - 1}{u} = 1 \]
Let \( u = \sqrt{x+h} - \sqrt{x} \). As \( h \to 0 \), \( u \to 0 \). We can find \( \frac{du}{dh} \):
\[ u = \sqrt{x+h} - \sqrt{x} \implies \frac{du}{dh} = \frac{1}{2\sqrt{x+h}} \to \frac{1}{2\sqrt{x}} \text{ as } h \to 0 \]

### Step 7: Rewrite the limit
Now we can rewrite the limit:
\[ f'(x) = e^{\sqrt{x}} \cdot \lim_{h \to 0} \frac{e^u - 1}{h} = e^{\sqrt{x}} \cdot \lim_{h \to 0} \frac{e^u - 1}{u} \cdot \frac{du}{dh} \]
Substituting the limits:
\[ f'(x) = e^{\sqrt{x}} \cdot 1 \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{2\sqrt{x}} \]

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