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Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Derivative from First Principles

The derivative is the slope of the tangent line, and first principles means finding it straight from the limit definition, before any shortcut rules. Tests ask for it because it shows you know where the rules come from.

How It Works

Start with the slope of a secant line between xx and a nearby point x+hx + h. Then let hh shrink to zero, so the secant becomes the tangent:

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

For the slope at one number aa, use the same idea with aa in place of xx:

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

Substituting h=0h = 0 straight away always gives 00\frac{0}{0}. The whole method is algebra that makes the hh in the denominator cancel. Work in four steps:

1. Write f(x+h)f(x + h). Replace every xx in the function with (x+h)(x + h), brackets included. Most errors start here.

2. Subtract f(x)f(x) and simplify the numerator. Every term without an hh should cancel. If one survives, check step 1.

3. Cancel the hh. For a polynomial, factor hh out of the numerator. For a fraction, combine over a common denominator first. For a square root, multiply top and bottom by the conjugate.

4. Let h→0h \to 0. Now substitution works.

Keep writing lim⁡h→0\lim_{h \to 0} on every line until that last step. Markers look for it, and dropping it changes what the line says.

Once you have f′(a)f'(a), the tangent line at x=ax = a is y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x - a). That is the usual second part of a first principles question.

A useful check: when you finish, compare your answer with the power rule. If they disagree, the algebra slipped somewhere. The test still wants the limit work, though, because the rule is what you are proving.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    Use first principles to find f′(x)f'(x) for f(x)=x2+3xf(x) = x^2 + 3x.
    Show solution
    1. f(x+h)=(x+h)2+3(x+h)=x2+2xh+h2+3x+3hf(x + h) = (x + h)^2 + 3(x + h) = x^2 + 2xh + h^2 + 3x + 3h.
    2. f(x+h)−f(x)=2xh+h2+3h=h(2x+h+3)f(x + h) - f(x) = 2xh + h^2 + 3h = h(2x + h + 3).
    3. f′(x)=lim⁡h→0h(2x+h+3)h=lim⁡h→0(2x+h+3)=2x+3f'(x) = \lim_{h \to 0}\frac{h(2x + h + 3)}{h} = \lim_{h \to 0}(2x + h + 3) = 2x + 3.

    Answer: f′(x)=2x+3f'(x) = 2x + 3

  2. Easy

    Use first principles to find f′(3)f'(3) for f(x)=4x2−xf(x) = 4x^2 - x.
    Show solution
    1. f(3)=36−3=33f(3) = 36 - 3 = 33.
    2. f(3+h)=4(9+6h+h2)−(3+h)=33+23h+4h2f(3 + h) = 4(9 + 6h + h^2) - (3 + h) = 33 + 23h + 4h^2.
    3. f′(3)=lim⁡h→023h+4h2h=lim⁡h→0(23+4h)=23f'(3) = \lim_{h \to 0}\frac{23h + 4h^2}{h} = \lim_{h \to 0}(23 + 4h) = 23.

    Answer: f′(3)=23f'(3) = 23

  3. Medium

    Use first principles to find f′(x)f'(x) for f(x)=x3f(x) = x^3.
    Show solution
    1. Expand: (x+h)3=x3+3x2h+3xh2+h3(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3.
    2. f(x+h)−f(x)=3x2h+3xh2+h3=h(3x2+3xh+h2)f(x + h) - f(x) = 3x^2h + 3xh^2 + h^3 = h(3x^2 + 3xh + h^2).
    3. f′(x)=lim⁡h→0(3x2+3xh+h2)=3x2f'(x) = \lim_{h \to 0}(3x^2 + 3xh + h^2) = 3x^2.

    Answer: f′(x)=3x2f'(x) = 3x^2

  4. Medium

    Use first principles to find f′(x)f'(x) for f(x)=1x−2f(x) = \frac{1}{x - 2}.
    Show solution
    1. f(x+h)−f(x)=1x+h−2−1x−2f(x + h) - f(x) = \frac{1}{x + h - 2} - \frac{1}{x - 2}.
    2. Common denominator: (x−2)−(x+h−2)(x+h−2)(x−2)=−h(x+h−2)(x−2)\frac{(x - 2) - (x + h - 2)}{(x + h - 2)(x - 2)} = \frac{-h}{(x + h - 2)(x - 2)}.
    3. Divide by hh: f′(x)=lim⁡h→0−1(x+h−2)(x−2)=−1(x−2)2f'(x) = \lim_{h \to 0}\frac{-1}{(x + h - 2)(x - 2)} = \frac{-1}{(x - 2)^2}.

    Answer: f′(x)=−1(x−2)2f'(x) = -\frac{1}{(x - 2)^2}

  5. Test level

    Use first principles to find f′(x)f'(x) for f(x)=2x+1f(x) = \sqrt{2x + 1}, then find the equation of the tangent line at x=4x = 4.
    Show solution
    1. f′(x)=lim⁡h→02x+2h+1−2x+1hf'(x) = \lim_{h \to 0}\frac{\sqrt{2x + 2h + 1} - \sqrt{2x + 1}}{h}. There is a root, so multiply by the conjugate.
    2. The numerator becomes (2x+2h+1)−(2x+1)=2h(2x + 2h + 1) - (2x + 1) = 2h, so the expression is 2hh(2x+2h+1+2x+1)\frac{2h}{h(\sqrt{2x + 2h + 1} + \sqrt{2x + 1})}.
    3. Cancel hh and let h→0h \to 0: f′(x)=222x+1=12x+1f'(x) = \frac{2}{2\sqrt{2x + 1}} = \frac{1}{\sqrt{2x + 1}}.
    4. At x=4x = 4: f(4)=9=3f(4) = \sqrt{9} = 3 and f′(4)=13f'(4) = \frac{1}{3}.
    5. Tangent: y−3=13(x−4)y - 3 = \frac{1}{3}(x - 4), so y=13x+53y = \frac{1}{3}x + \frac{5}{3}.

    Answer: f′(x)=12x+1f'(x) = \frac{1}{\sqrt{2x + 1}}; tangent y=13x+53y = \frac{1}{3}x + \frac{5}{3}

Learn It on Martian Lab

Martian Lab teaches this in the Derivatives lesson of its MCV4U Derivatives unit. Watch it, then take the Limits practice tests to see if it stuck.

Watch the Derivatives lesson free

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More MCV4U topics: Chain Rule · Optimization Problems · Related Rates · Curve Sketching Steps · Dot Product and Projections · Cross Product, Area and Torque · Equations of Planes · Intersection of Three Planes · Power, Product and Quotient Rules · Velocity and Acceleration · Increasing, Decreasing and Critical Points · Concavity and the Second Derivative Test · Derivatives of e^x and a^x · Derivatives of Sin, Cos and Tan · Adding and Subtracting Vectors · Cartesian Vectors in 2D and 3D · Force and Velocity Vector Problems · Equations of Lines in 3D · Intersection of Lines and Planes · Distance from a Point to a Line or Plane

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