Find the derivative of `e^(x^(2)-5x)` using first principle of derivative

Answer

To find the derivative of the function \( f(x) = e^{x^2 - 5x} \) using the first principle of derivatives, we will follow these steps:

### Step 1: Write the definition of the derivative
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 definition
Substituting \( f(x) = e^{x^2 - 5x} \) into the definition:
\[ f'(x) = \lim_{h \to 0} \frac{e^{(x+h)^2 - 5(x+h)} - e^{x^2 - 5x}}{h} \]

### Step 3: Simplify \( f(x+h) \)
Now, we simplify \( f(x+h) \):
\[ f(x+h) = e^{(x+h)^2 - 5(x+h)} = e^{x^2 + 2xh + h^2 - 5x - 5h} \]
This can be rewritten as:
\[ f(x+h) = e^{x^2 - 5x} \cdot e^{2xh + h^2 - 5h} \]

### Step 4: Substitute back into the limit
Now substituting back into the limit:
\[ f'(x) = \lim_{h \to 0} \frac{e^{x^2 - 5x} \cdot e^{2xh + h^2 - 5h} - e^{x^2 - 5x}}{h} \]
Factoring out \( e^{x^2 - 5x} \):
\[ f'(x) = e^{x^2 - 5x} \cdot \lim_{h \to 0} \frac{e^{2xh + h^2 - 5h} - 1}{h} \]

### Step 5: Evaluate the limit
Now we need to evaluate the limit:
\[ \lim_{h \to 0} \frac{e^{2xh + h^2 - 5h} - 1}{h} \]
Using the fact that \( \lim_{u \to 0} \frac{e^u - 1}{u} = 1 \), we can rewrite the limit:
\[ = \lim_{h \to 0} \frac{e^{2xh + h^2 - 5h} - 1}{2xh + h^2 - 5h} \cdot (2xh + h^2 - 5h) \]
As \( h \to 0 \), \( 2xh + h^2 - 5h \to 0 \). Thus, the limit becomes:
\[ = (2x - 5) \cdot 1 = 2x - 5 \]

### Step 6: Combine results
Putting it all together, we have:
\[ f'(x) = e^{x^2 - 5x} \cdot (2x - 5) \]

### Final Answer
Thus, the derivative of \( f(x) = e^{x^2 - 5x} \) is:
\[ f'(x) = e^{x^2 - 5x} (2x - 5) \]

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