Differentiation from First Principles

Before you learn shortcuts like the power rule, you need to know where derivatives come from. Differentiation from first principles uses the limit definition directly, and it is a regular question in FSc Part 2, A Level and first-year university exams. Once you understand the method, every derivative rule makes sense.

The Idea: From Average to Instant Rate of Change

The gradient of a straight line is easy: rise over run. But a curve's steepness changes from point to point. To find the gradient at a single point, we:

1. Take a point \( P \) at \( x \) and a nearby point \( Q \) at \( x + h \) on the curve.
2. Find the gradient of the secant line \( PQ \).
3. Let \( Q \) slide towards \( P \) by letting \( h \to 0 \). The secant becomes the tangent.

The gradient of the secant is

$$ \frac{f(x + h) - f(x)}{(x + h) - x} = \frac{f(x + h) - f(x)}{h} $$

This is called the difference quotient.

The Definition of the Derivative

The derivative of \( f(x) \) is

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

provided this limit exists. Some textbooks use \( \delta x \) instead of \( h \), writing

$$ \frac{dy}{dx} = \lim_{\delta x \to 0} \frac{\delta y}{\delta x} $$

Both mean the same thing.

The Four-Step Method

1. Write \( f(x + h) \) by replacing every \( x \) with \( (x + h) \).
2. Subtract \( f(x) \) and simplify the numerator.
3. Divide by \( h \): factor \( h \) out of the numerator and cancel it.
4. Take the limit as \( h \to 0 \).

You cannot substitute \( h = 0 \) before step 3, because that gives \( \frac{0}{0} \). The whole point of the algebra is to cancel the \( h \) in the denominator.

Worked Example 1: A Quadratic

Question: Differentiate \( f(x) = 3x^2 - 5x \) from first principles.

Solution:

• Step 1: Find \( f(x + h) \).

$$ f(x + h) = 3(x + h)^2 - 5(x + h) = 3x^2 + 6xh + 3h^2 - 5x - 5h $$

• Step 2: Subtract \( f(x) \).

$$ f(x + h) - f(x) = (3x^2 + 6xh + 3h^2 - 5x - 5h) - (3x^2 - 5x) = 6xh + 3h^2 - 5h $$

• Step 3: Divide by \( h \).

$$ \frac{6xh + 3h^2 - 5h}{h} = 6x + 3h - 5 $$

• Step 4: Take the limit.

$$ f'(x) = \lim_{h \to 0} (6x + 3h - 5) = 6x - 5 $$

This matches the power rule: \( \frac{d}{dx}(3x^2) = 6x \) and \( \frac{d}{dx}(-5x) = -5 \).

Worked Example 2: A Reciprocal Function

Question: Differentiate \( f(x) = \frac{1}{x} \) from first principles.

Solution:

• Step 1: \( f(x + h) = \frac{1}{x + h} \).
• Step 2: Subtract using a common denominator.

$$ \frac{1}{x + h} - \frac{1}{x} = \frac{x - (x + h)}{x(x + h)} = \frac{-h}{x(x + h)} $$

• Step 3: Divide by \( h \).

$$ \frac{-h}{x(x + h)} \div h = \frac{-1}{x(x + h)} $$

• Step 4: Take the limit.

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

This agrees with the power rule for \( x^{-1} \): \( -1 \cdot x^{-2} \).

Worked Example 3: A Square Root

Question: Differentiate \( f(x) = \sqrt{x} \) from first principles.

Solution:

• Step 1: Write the difference quotient.

$$ \frac{\sqrt{x + h} - \sqrt{x}}{h} $$

• Step 2: Multiply by the conjugate.

$$ \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} = \frac{(x + h) - x}{h\left(\sqrt{x + h} + \sqrt{x}\right)} $$

• Step 3: Simplify and cancel \( h \).

$$ = \frac{h}{h\left(\sqrt{x + h} + \sqrt{x}\right)} = \frac{1}{\sqrt{x + h} + \sqrt{x}} $$

• Step 4: Take the limit.

$$ f'(x) = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}} $$

Again this matches the power rule: \( \frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2} \).

Finding the Gradient at a Specific Point

Sometimes you are asked for the gradient at a particular point, such as \( x = 2 \) for \( f(x) = x^2 \). You can use

$$ f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} = \lim_{h \to 0} \frac{(4 + 4h + h^2) - 4}{h} = \lim_{h \to 0} (4 + h) = 4 $$

So the tangent to \( y = x^2 \) at \( x = 2 \) has gradient 4.

Common Mistakes

• Expanding \( (x + h)^2 \) as \( x^2 + h^2 \). The correct expansion is \( x^2 + 2xh + h^2 \).
• Forgetting to subtract all of \( f(x) \). Use brackets to keep the signs right.
• Setting \( h = 0 \) too early, giving \( \frac{0}{0} \).
• Dropping the "lim" notation before the final step. Write \( \lim_{h \to 0} \) on every line until you substitute.
• Using the power rule when the question says "from first principles". It will receive no marks.

Tips

• For fractions, combine into a single fraction before dividing by \( h \).
• For square roots, multiply by the conjugate.
• Learn the expansions of \( (x + h)^2 \) and \( (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \).
• Check your final answer with the standard rules; they should always agree.

Key Takeaways

• The derivative is \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \).
• It turns the gradient of a secant into the gradient of a tangent.
• Follow four steps: write \( f(x + h) \), subtract \( f(x) \), divide by \( h \), take the limit.
• Use common denominators for fractions and conjugates for square roots.
• First principles explain why shortcut rules like the power rule work.

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