If `f(x)=x^(2)-19x+18`, then find `f^(')(2), f^(')(1)" and "f^(')(10)`, using limit process only once.

Answer

To find the derivatives \( f'(2) \), \( f'(1) \), and \( f'(10) \) for the function \( f(x) = x^2 - 19x + 18 \) using the limit process, we will follow these steps:

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

### Step 2: Substitute \( f(x) \) into the limit definition.
First, we calculate \( f(x + h) \):
\[ f(x + h) = (x + h)^2 - 19(x + h) + 18 \]
Expanding this:
\[ = x^2 + 2xh + h^2 - 19x - 19h + 18 \]
Now, substituting back into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 - 19x - 19h + 18) - (x^2 - 19x + 18)}{h} \]

### Step 3: Simplify the expression.
The \( x^2 \), \( -19x \), and \( +18 \) terms cancel out:
\[ = \lim_{h \to 0} \frac{2xh + h^2 - 19h}{h} \]
Factoring out \( h \) from the numerator:
\[ = \lim_{h \to 0} \frac{h(2x + h - 19)}{h} \]
Cancelling \( h \) (assuming \( h \neq 0 \)):
\[ = \lim_{h \to 0} (2x + h - 19) \]

### Step 4: Apply the limit.
Now, taking the limit as \( h \) approaches 0:
\[ f'(x) = 2x - 19 \]

### Step 5: Calculate \( f'(2) \), \( f'(1) \), and \( f'(10) \).
Now we can find the required derivatives:
1. For \( f'(2) \):
\[ f'(2) = 2(2) - 19 = 4 - 19 = -15 \]

2. For \( f'(1) \):
\[ f'(1) = 2(1) - 19 = 2 - 19 = -17 \]

3. For \( f'(10) \):
\[ f'(10) = 2(10) - 19 = 20 - 19 = 1 \]

### Final Answers:
- \( f'(2) = -15 \)
- \( f'(1) = -17 \)
- \( f'(10) = 1 \)

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