Differentiate `f(x)=e^(2-3x)` by using 1st principle.

Answer

To differentiate the function \( f(x) = e^{2 - 3x} \) using the first principle of differentiation, we follow these steps:

### Step 1: Write the definition of the derivative using the first principle.
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 \( f(x) \) into the formula.
Substituting \( f(x) = e^{2 - 3x} \) into the formula, we have:
\[ f'(x) = \lim_{h \to 0} \frac{e^{2 - 3(x + h)} - e^{2 - 3x}}{h} \]

### Step 3: Simplify the expression inside the limit.
We can simplify the expression:
\[ f'(x) = \lim_{h \to 0} \frac{e^{2 - 3x - 3h} - e^{2 - 3x}}{h} \]
This can be rewritten as:
\[ f'(x) = \lim_{h \to 0} \frac{e^{2 - 3x} \cdot e^{-3h} - e^{2 - 3x}}{h} \]
Factoring out \( e^{2 - 3x} \):
\[ f'(x) = e^{2 - 3x} \cdot \lim_{h \to 0} \frac{e^{-3h} - 1}{h} \]

### Step 4: Evaluate the limit.
The limit \( \lim_{h \to 0} \frac{e^{-3h} - 1}{h} \) can be evaluated using L'Hôpital's Rule since it is in the indeterminate form \( \frac{0}{0} \):
1. Differentiate the numerator: The derivative of \( e^{-3h} \) is \( -3e^{-3h} \).
2. Differentiate the denominator: The derivative of \( h \) is \( 1 \).

Thus, applying L'Hôpital's Rule:
\[ \lim_{h \to 0} \frac{e^{-3h} - 1}{h} = \lim_{h \to 0} \frac{-3e^{-3h}}{1} = -3e^{0} = -3 \]

### Step 5: Substitute back into the derivative expression.
Now substituting back into our expression for \( f'(x) \):
\[ f'(x) = e^{2 - 3x} \cdot (-3) = -3e^{2 - 3x} \]

### Final Answer:
Thus, the derivative of \( f(x) = e^{2 - 3x} \) is:
\[ f'(x) = -3e^{2 - 3x} \]