Differentiate from first principles:
11. `(x+1)(2x-3)`

Answer

To differentiate the function \( f(x) = (x + 1)(2x - 3) \) from first principles, we will follow these steps:

### Step 1: Write the definition of the derivative
The derivative of a function \( f(x) \) from first principles is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 2: Calculate \( f(x + h) \)
First, we need to find \( f(x + h) \):
\[ f(x + h) = (x + h + 1)(2(x + h) - 3) \]
Simplifying this:
\[ = (x + h + 1)(2x + 2h - 3) \]
Now, we can expand this expression:
\[ = (x + h + 1)(2x + 2h - 3) = (x + h)(2x + 2h - 3) + 1(2x + 2h - 3) \]
\[ = (2x^2 + 2xh - 3x + 2hx + 2h^2 - 3h) + (2x + 2h - 3) \]
Combining like terms:
\[ = 2x^2 + 4xh + 2h^2 - 3x + 2x + 2h - 3 \]
\[ = 2x^2 + 4xh + 2h^2 - x + 2h - 3 \]

### Step 3: Calculate \( f(x + h) - f(x) \)
Now we need to calculate \( f(x + h) - f(x) \):
\[ f(x) = (x + 1)(2x - 3) = 2x^2 - 3x + 2x - 3 = 2x^2 - x - 3 \]
Thus,
\[ f(x + h) - f(x) = (2x^2 + 4xh + 2h^2 - x + 2h - 3) - (2x^2 - x - 3) \]
This simplifies to:
\[ = 4xh + 2h^2 + 2h \]

### Step 4: Substitute into the limit definition
Now we substitute this into the limit:
\[ f'(x) = \lim_{h \to 0} \frac{4xh + 2h^2 + 2h}{h} \]
We can simplify this by dividing each term by \( h \):
\[ = \lim_{h \to 0} (4x + 2h + 2) \]

### Step 5: Evaluate the limit
Now we take the limit as \( h \) approaches 0:
\[ f'(x) = 4x + 2(0) + 2 = 4x + 2 \]

### Final Answer
Thus, the derivative of the function \( f(x) = (x + 1)(2x - 3) \) is:
\[ f'(x) = 4x + 2 \]