If `f(x)=(x-1)^(2)`, find `f^(')` from first principles.

Answer

To find the derivative \( f'(x) \) of the function \( f(x) = (x - 1)^2 \) from first principles, we will use the definition of the derivative. The derivative from first principles is given by:

\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 1: Calculate \( f(x + h) \)

We start by calculating \( f(x + h) \):

\[ f(x + h) = (x + h - 1)^2 \]

### Step 2: Expand \( f(x + h) \)

Now, we expand \( f(x + h) \):

\[ f(x + h) = (x + h - 1)^2 = (x - 1 + h)^2 = (x - 1)^2 + 2(x - 1)h + h^2 \]

### Step 3: Substitute into the derivative formula

Next, we substitute \( f(x + h) \) and \( f(x) \) into the derivative formula:

\[ f'(x) = \lim_{h \to 0} \frac{(x - 1)^2 + 2(x - 1)h + h^2 - (x - 1)^2}{h} \]

### Step 4: Simplify the expression

Now, we simplify the expression:

\[ f'(x) = \lim_{h \to 0} \frac{2(x - 1)h + h^2}{h} \]

### Step 5: Factor out \( h \)

We can factor \( h \) out of the numerator:

\[ f'(x) = \lim_{h \to 0} \frac{h(2(x - 1) + h)}{h} \]

### Step 6: Cancel \( h \)

Since \( h \) is in both the numerator and denominator, we can cancel \( h \) (as long as \( h \neq 0 \)):

\[ f'(x) = \lim_{h \to 0} (2(x - 1) + h) \]

### Step 7: Take the limit as \( h \to 0 \)

Now, we take the limit as \( h \) approaches 0:

\[ f'(x) = 2(x - 1) + 0 = 2(x - 1) \]

### Final Answer

Thus, the derivative \( f'(x) \) is:

\[ f'(x) = 2(x - 1) \]

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