Calculating a Derivative by Definition

Step-by-Step: Calculating a Derivative by Definition

Let’s use the four-step process for finding the derivative of a simple quadratic function, as this approach is both fundamental and widely used in calculus¹². For this example, let’s use a general quadratic function, but if you want a specific example from the PDF, we’ll use the quadratic function $ f(x) = x^2 $ (which is used in the context of quadratic functions³).

However, the PDF also discusses the slope of a quadratic function $ f(x) = ax^2 + bx + c $, mentioning that the slope at any point is $ 2ax + b $. Let’s see how this is derived using the definition.

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Step 1: Write $ f(x + h) $ and $ f(x) $

Suppose $ f(x) = ax^2 + bx + c $ (you can set $ a=1, b=0, c=0 $ for $ f(x) = x^2 $, but we’ll keep it general).

$$ f(x + h) = a(x + h)^2 + b(x + h) + c = a(x^2 + 2xh + h^2) + b(x + h) + c $$$$ = ax^2 + 2axh + ah^2 + bx + bh + c $$$$ f(x) = ax^2 + bx + c $$

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Step 2: Compute $ f(x + h) - f(x) $

Subtract $ f(x) $ from $ f(x + h) $:

$$ f(x + h) - f(x) = (ax^2 + 2axh + ah^2 + bx + bh + c) - (ax^2 + bx + c) $$$$ = 2axh + ah^2 + bh $$$$ = h(2ax + ah + b) $$

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Step 3: Divide by $ h $

$$ \frac{f(x + h) - f(x)}{h} = \frac{h(2ax + ah + b)}{h} = 2ax + ah + b $$

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Step 4: Take the limit as $ h \to 0 $

$$ f'(x) = \lim_{h \to 0} (2ax + ah + b) = 2ax + b $$

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Summary Table

Step	Action	Expression/Result
1	Write $ f(x + h) $, $ f(x) $	$ f(x+h) = a(x+h)^2 + ··· $
2	Subtract $ f(x) $ from $ f(x+h) $	$ h(2ax + ah + b) $
3	Divide by $ h $	$ 2ax + ah + b $
4	Take the limit $ h \to 0 $	$ 2ax + b $

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Example from the PDF

The PDF states (in Section 3.2):

“For the quadratic function described as $ f(x) = ax^2 + bx + c $ where $ a \neq 0 $, the slope of $ f $ at any given point $ (x, f(x)) $ is $ (2ax + b) $.”

This matches exactly the result we derived above using the definition of the derivative³.

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Quick Example: $ f(x) = x^2 $

Let’s do the calculation for $ f(x) = x^2 $:

1. Step 1: $ f(x + h) = (x + h)^2 = x^2 + 2xh + h^2 $ $ f(x) = x^2 $
2. Step 2: $ f(x + h) - f(x) = 2xh + h^2 $
3. Step 3: $ \frac{f(x + h) - f(x)}{h} = \frac{2xh + h^2}{h} = 2x + h $
4. Step 4: $ \lim_{h \to 0} (2x + h) = 2x $

So, the derivative of $ f(x) = x^2 $ is $ f’(x) = 2x $, which matches the general result above (with $ a=1, b=0 $)³¹.

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Related Question

Q: What is the slope of the quadratic function $ f(x) = 3x^2 + 4x + 1 $ at $ x = 2 $?

A: Using the general result $ f’(x) = 2ax + b $, $ a = 3, b = 4 $, so $ f’(x) = 6x + 4 $. At $ x = 2 $: $ f’(2) = 6 \times 2 + 4 = 16 $.

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Summary

• The derivative is the slope of the tangent to the curve at a point.
• Use the four-step process:

1. Write $ f(x + h) $ and $ f(x) $.
2. Compute $ f(x + h) - f(x) $.
3. Divide by $ h $.
4. Take the limit as $ h \to 0 $.

• For quadratic functions, the derivative is $ 2ax + b $.

This method works for any differentiable function, and the PDF confirms this for quadratic functions³¹.