Differentiate from first principles:
2. `(x-1)^(2)`

Answer

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

### Step 1: Write down the definition of the derivative using first principles.
The derivative \( f'(x) \) can be defined as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Calculate \( f(x+h) \).
We need to find \( f(x+h) \):
\[ f(x+h) = (x+h-1)^2 \]
Expanding this using the formula \( (a+b)^2 = a^2 + 2ab + b^2 \):
\[ f(x+h) = (x-1+h)^2 = (x-1)^2 + 2(x-1)h + h^2 \]

### Step 3: Calculate \( f(x) \).
Next, we calculate \( f(x) \):
\[ f(x) = (x-1)^2 \]

### Step 4: Substitute \( f(x+h) \) and \( f(x) \) into the derivative formula.
Now we substitute these into our derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{(x-1+h)^2 - (x-1)^2}{h} \]

### Step 5: Simplify the expression.
Now we simplify the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{(x-1)^2 + 2(x-1)h + h^2 - (x-1)^2}{h} \]
The \( (x-1)^2 \) terms cancel out:
\[ f'(x) = \lim_{h \to 0} \frac{2(x-1)h + h^2}{h} \]
Now we can factor \( h \) out of the numerator:
\[ f'(x) = \lim_{h \to 0} \frac{h(2(x-1) + h)}{h} \]
Cancelling \( h \) gives:
\[ f'(x) = \lim_{h \to 0} (2(x-1) + h) \]

### Step 6: Evaluate the limit as \( h \) approaches 0.
Now we take the limit:
\[ f'(x) = 2(x-1) + 0 = 2(x-1) \]

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