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1. Differentiation from First Principles0%

Differentiation · Topic 1 of 10

1. Differentiation from First Principles

Video coming soon3 worked examples

Theory

Every differentiation rule you use rests on a single definition. To find the gradient of the tangent to y=f(x)y = f(x) at the point A (x, f(x))A\,(x,\,f(x)), take a second point B (x+h, f(x+h))B\,(x+h,\,f(x+h)) on the curve and calculate the gradient of the chord ABAB:

mAB=f(x+h)−f(x)(x+h)−x=f(x+h)−f(x)hm_{AB} = \frac{f(x+h) - f(x)}{(x+h) - x} = \frac{f(x+h) - f(x)}{h}

As hh gets smaller, BB slides towards AA and the chord becomes indistinguishable from the tangent. The value the gradient approaches is the derivative:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Differentiating from first principles means using this definition directly — without any of the standard rules or known derivatives.

The Golden Rule: expand f(x+h)f(x+h) completely, then subtract f(x)f(x). Every surviving term must contain a factor of hh — if one doesn't, you have made an algebra slip. Cancel that hh against the denominator, and only then let h→0h \to 0.

⚠️ Common Examiner Traps

  • Setting h=0h = 0 too early: you cannot substitute h=0h=0 while hh is still in the denominator — that gives 00\frac{0}{0}. Cancel first, take the limit second.
  • Dropping the limit notation: write lim⁡h→0\lim_{h \to 0} in front of every line until you actually take the limit. Marks are awarded for it.
  • Expanding (x+h)3(x+h)^3 carelessly: it is x3+3x2h+3xh2+h3x^3 + 3x^2h + 3xh^2 + h^3, not x3+h3x^3 + h^3.
  • Using the rules instead: if a question says “from first principles”, quoting nxn−1nx^{n-1} earns nothing, even with the right answer.

Worked examples

Example 1

Differentiate f(x)=2x2−5xf(x) = 2x^2 - 5x from first principles.

Step 1: Write down f(x+h)f(x+h) and expand it fully:

f(x+h)=2(x+h)2−5(x+h)=2x2+4xh+2h2−5x−5hf(x+h) = 2(x+h)^2 - 5(x+h) = 2x^2 + 4xh + 2h^2 - 5x - 5h

Step 2: Subtract f(x)=2x2−5xf(x) = 2x^2 - 5x. The terms without an hh cancel:

f(x+h)−f(x)=4xh+2h2−5hf(x+h) - f(x) = 4xh + 2h^2 - 5h

Step 3: Divide by hh, cancelling the common factor:

f(x+h)−f(x)h=h(4x+2h−5)h=4x+2h−5\frac{f(x+h)-f(x)}{h} = \frac{h(4x + 2h - 5)}{h} = 4x + 2h - 5

Step 4: Now the limit can be taken safely:

f′(x)=lim⁡h→0 (4x+2h−5)=4x−5f'(x) = \lim_{h \to 0}\,(4x + 2h - 5) = 4x - 5

Example 2

Differentiate f(x)=x3−2xf(x) = x^3 - 2x from first principles.

Step 1: Expand f(x+h)f(x+h), taking care with the cube:

f(x+h)=(x+h)3−2(x+h)=x3+3x2h+3xh2+h3−2x−2hf(x+h) = (x+h)^3 - 2(x+h) = x^3 + 3x^2h + 3xh^2 + h^3 - 2x - 2h

Step 2: Subtract f(x)=x3−2xf(x) = x^3 - 2x:

f(x+h)−f(x)=3x2h+3xh2+h3−2hf(x+h) - f(x) = 3x^2h + 3xh^2 + h^3 - 2h

Step 3: Every term has a factor of hh, so divide through:

f(x+h)−f(x)h=3x2+3xh+h2−2\frac{f(x+h)-f(x)}{h} = 3x^2 + 3xh + h^2 - 2

Step 4: Let h→0h \to 0; the two terms still containing hh vanish:

f′(x)=lim⁡h→0 (3x2+3xh+h2−2)=3x2−2f'(x) = \lim_{h \to 0}\,(3x^2 + 3xh + h^2 - 2) = 3x^2 - 2

Example 3

Differentiate f(x)=1xf(x) = \dfrac{1}{x} from first principles.

Step 1: Here f(x+h)=1x+hf(x+h) = \dfrac{1}{x+h}. Subtracting gives a difference of two fractions, so combine them over a common denominator:

f(x+h)−f(x)=1x+h−1x=x−(x+h)x(x+h)=−hx(x+h)f(x+h) - f(x) = \frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}

Step 2: Dividing by hh means multiplying by 1h\dfrac{1}{h}, which cancels the hh in the numerator:

f(x+h)−f(x)h=−hx(x+h)×1h=−1x(x+h)\frac{f(x+h)-f(x)}{h} = \frac{-h}{x(x+h)} \times \frac{1}{h} = \frac{-1}{x(x+h)}

Step 3: There is no longer an hh in the denominator's way, so let h→0h \to 0:

f′(x)=lim⁡h→0−1x(x+h)=−1x⋅x=−1x2f'(x) = \lim_{h \to 0} \frac{-1}{x(x+h)} = \frac{-1}{x \cdot x} = -\frac{1}{x^2}

This agrees with the power rule: ddx(x−1)=−x−2\frac{d}{dx}(x^{-1}) = -x^{-2}.