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: State the First Principle of Derivative
The first principle of derivative states that:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Calculate \( f(x+h) \)
We need to find \( f(x+h) \):
\[ f(x+h) = (x+h)^2 + 5(x+h) + 3 \]
Expanding this:
\[ = x^2 + 2xh + h^2 + 5x + 5h + 3 \]

### Step 3: Substitute into the Derivative Formula
Now, 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 4: Simplify the Expression
Now simplify the expression inside the limit:
\[ = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 + 5x + 5h + 3) - x^2 - 5x - 3}{h} \]
This simplifies to:
\[ = \lim_{h \to 0} \frac{2xh + h^2 + 5h}{h} \]

### Step 5: Factor Out \( h \)
Now, factor out \( h \) from the numerator:
\[ = \lim_{h \to 0} \frac{h(2x + h + 5)}{h} \]
Since \( h \neq 0 \) (as we are taking the limit), we can cancel \( h \):
\[ = \lim_{h \to 0} (2x + h + 5) \]

### Step 6: Evaluate the Limit
Now, evaluate the limit as \( h \) approaches 0:
\[ = 2x + 0 + 5 = 2x + 5 \]

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

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