Find the derivative of `x^(2) + 5x + 3` using first principle of derivative.

Answer

To find the derivative of the function \( f(x) = x^2 + 5x + 3 \) using the first principle of derivatives, we will follow these steps:

### Step 1: Write down the first principle of derivatives
The first principle of derivatives states that:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 2: Identify \( f(x) \)
Here, we have:
\[ f(x) = x^2 + 5x + 3 \]

### Step 3: Calculate \( f(x + h) \)
Now, we need to calculate \( f(x + h) \):
\[ f(x + h) = (x + h)^2 + 5(x + h) + 3 \]
Expanding this, we get:
\[ f(x + h) = (x^2 + 2xh + h^2) + (5x + 5h) + 3 \]
Combining like terms:
\[ f(x + h) = x^2 + 2xh + h^2 + 5x + 5h + 3 \]

### Step 4: Substitute into the derivative formula
Now, we substitute \( f(x + h) \) and \( f(x) \) into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 + 5x + 5h + 3) - (x^2 + 5x + 3)}{h} \]

### Step 5: Simplify the expression
Now, simplify the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 + 5x + 5h + 3) - x^2 - 5x - 3}{h} \]
This simplifies to:
\[ f'(x) = \lim_{h \to 0} \frac{2xh + h^2 + 5h}{h} \]

### Step 6: Factor out \( h \)
We can factor \( h \) out of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{h(2x + h + 5)}{h} \]
Cancelling \( h \) (as long as \( h \neq 0 \)):
\[ f'(x) = \lim_{h \to 0} (2x + h + 5) \]

### Step 7: Evaluate the limit
Now, we take the limit as \( h \) approaches 0:
\[ f'(x) = 2x + 0 + 5 = 2x + 5 \]

### Final Answer
Thus, the derivative of \( f(x) = x^2 + 5x + 3 \) is:
\[ f'(x) = 2x + 5 \]

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