Find the derivative of `(3x - 2) (x + 1)` using first principle of derivative.

Answer

To find the derivative of the function \( f(x) = (3x - 2)(x + 1) \) using the first principle of derivatives, we will follow these steps:

### Step 1: Define the function
We start by defining the function:
\[ f(x) = (3x - 2)(x + 1) \]

### Step 2: Expand the function
Next, we expand the function:
\[ f(x) = 3x^2 + 3x - 2x - 2 = 3x^2 + x - 2 \]

### Step 3: Write the expression for \( f(x + h) \)
Now, we need to find \( f(x + h) \):
\[ f(x + h) = 3(x + h)^2 + (x + h) - 2 \]
Expanding this:
\[ f(x + h) = 3(x^2 + 2xh + h^2) + x + h - 2 = 3x^2 + 6xh + 3h^2 + x + h - 2 \]
Combining like terms:
\[ f(x + h) = 3x^2 + (6h + 1)x + (3h^2 + h - 2) \]

### Step 4: Use the first principle of derivatives
The first principle of derivatives states:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]
Substituting \( f(x + h) \) and \( f(x) \):
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{(3x^2 + (6h + 1)x + (3h^2 + h - 2)) - (3x^2 + x - 2)}{h} \]

### Step 5: Simplify the expression
Now, simplifying the expression:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{(6h + 1)x + (3h^2 + h)}{h} \]
Breaking it down:
\[ = \lim_{h \to 0} \left( \frac{(6h + 1)x}{h} + \frac{(3h^2 + h)}{h} \right) \]
This simplifies to:
\[ = \lim_{h \to 0} \left( 6x + \frac{1}{h} + 3h + 1 \right) \]

### Step 6: Evaluate the limit
As \( h \) approaches 0, the term \( \frac{1}{h} \) will not contribute, and we are left with:
\[ \frac{dy}{dx} = 6x + 1 \]

### Final Result
Thus, the derivative of the function \( f(x) = (3x - 2)(x + 1) \) is:
\[ \frac{dy}{dx} = 6x + 1 \]

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