Find, from first principle, the derivative of the following w.r.t. x :
`e^(2x)`

Answer

To find the derivative of the function \( f(x) = e^{2x} \) using the first principle of derivatives, we follow these steps:

### Step 1: Write the formula for the derivative from first principles.
The derivative of a function \( f(x) \) using the first principle is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 2: Substitute the function into the formula.
For our function \( f(x) = e^{2x} \), we need to calculate \( f(x + h) \):
\[ f(x + h) = e^{2(x + h)} = e^{2x + 2h} \]
Now, substituting this into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{e^{2x + 2h} - e^{2x}}{h} \]

### Step 3: Factor out common terms.
We can factor out \( e^{2x} \) from the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{e^{2x}(e^{2h} - 1)}{h} \]

### Step 4: Simplify the limit.
Now we can take \( e^{2x} \) outside the limit:
\[ f'(x) = e^{2x} \lim_{h \to 0} \frac{e^{2h} - 1}{h} \]

### Step 5: Evaluate the limit.
To evaluate the limit \( \lim_{h \to 0} \frac{e^{2h} - 1}{h} \), we can use the fact that:
\[ \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \]
Thus, substituting \( 2h \) for \( x \):
\[ \lim_{h \to 0} \frac{e^{2h} - 1}{h} = 2 \cdot \lim_{x \to 0} \frac{e^x - 1}{x} = 2 \]

### Step 6: Final result.
Now substituting this back into our expression for \( f'(x) \):
\[ f'(x) = e^{2x} \cdot 2 = 2e^{2x} \]

### Conclusion:
The derivative of \( e^{2x} \) with respect to \( x \) is:
\[ f'(x) = 2e^{2x} \]