Mathematical Methods · Units 3 & 4

Differentiation from First Principles

Understand differentiation from first principles the easy way, with plain English intuition, a clear visual, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Learn

Picture zooming in on a curvy line with your phone camera. Keep zooming into one single point and something surprising happens. The curve stops looking curvy and starts to look perfectly straight. Differentiation from first principles is just the careful way of measuring how steep that tiny straight piece is. Get this one idea and the rest of calculus falls into place, because every other rule is built on top of it.

P Q
The dashed line through P and Q is the secant. As Q slides toward P, the secant swings around until it matches the red tangent at P. That tangent's steepness is the derivative.

What you are really measuring

A straight line is easy because it has the same steepness everywhere. A curve is sneaky, because its steepness keeps changing, so the question “how steep is it” has no answer until you say where. So here is the move. Pick your point. Drop a second point a tiny distance away. Measure the steepness of the straight line joining the two. Then slide that second point closer and closer until the gap almost vanishes.

The straight line joining the two points has steepness

f(x+h)−f(x)h\frac{f(x+h) - f(x)}{h}

where hh is the tiny gap between them.

The one line of maths that says it all

Here is the whole idea written as a single line. It looks fancy, but read it slowly and it just says “shrink the gap to zero”.

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

The lim⁡h→0\lim_{h \to 0} part is shorthand for “let the gap hh get so small it basically disappears”. Everything else is just the steepness of the line between your two points.

How to actually do it

The idea is the hard part. The method is a short recipe you repeat every time.

  1. Write f(x+h)f(x+h) by putting x+hx+h everywhere you see xx.
  2. Work out f(x+h)−f(x)f(x+h) - f(x) and tidy the top.
  3. Divide every term by hh. The top always has an hh waiting to cancel.
  4. Let h→0h \to 0 by setting the leftover hh terms to zero.

The trap that catches most students is step 3. Before hh reaches zero it is still a real number, so the hh values cancel cleanly. Cancel first, then take the limit. Do it the other way around and you are dividing by zero.

See this recipe in action in the Worked Examples tab, then test yourself in Try It.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

Write the first principles definition of the derivative.
What does the fraction f(x+h)−f(x)h\dfrac{f(x+h) - f(x)}{h} represent, before the limit?
Why must you cancel the hh before letting h→0h \to 0?
Differentiate f(x)=1xf(x) = \dfrac{1}{x} from first principles.
When is a VCE student required to use first principles rather than a shortcut rule?
Recall · The Chain Rule
To differentiate a composite like (3x+1)5(3x+1)^5, what does the chain rule tell you to do?
Recall · Functions and Graphs
What is the gradient of a straight line f(x)=mx+cf(x) = mx + c, and how does first principles confirm it?

Worked examples

Worked Example 1Matching a polynomial to a curve, from a real exam

Let g:[0,5π2]→R, g(x)=ax3+bx2+cx+dg:\left[0,\dfrac{5\pi}{2}\right]\to R,\ g(x)=ax^{3}+bx^{2}+cx+d be a polynomial, where a,b,c,d∈Ra,b,c,d\in R. Suppose g(0)=f(0)g(0)=f(0) and g′(0)=f′(0)g'(0)=f'(0), where f(x)=sin⁡(x)+1f(x)=\sin(x)+1. Show that c=1c=1 and d=1d=1.

  1. 1

    Match the values at x=0x=0. Here g(0)=dg(0)=d and f(0)=sin⁡0+1=1f(0)=\sin 0 + 1 = 1.

    d=1d = 1
  2. 2

    Match the derivatives at x=0x=0. Differentiating gives g′(x)=3ax2+2bx+cg'(x)=3ax^2+2bx+c, so g′(0)=cg'(0)=c, while f′(x)=cos⁡xf'(x)=\cos x gives f′(0)=cos⁡0=1f'(0)=\cos 0 = 1.

    g′(x)=3ax2+2bx+c,g′(0)=c=1g'(x)=3ax^2+2bx+c,\quad g'(0)=c=1
  3. 3

    As this is a "show that", the substitution x=0x=0 must be shown clearly at each step.

    c=1, d=1c = 1,\ d = 1
Answer
c=1, d=1c=1,\ d=1

VCAA 2025 Mathematical Methods Exam 2, Section B Q4gi

Worked Example 2A basic quadratic

Differentiate f(x)=x2f(x) = x^2 from first principles.

  1. 1

    Write f(x+h)f(x+h) by putting x+hx+h everywhere you see xx.

    (x+h)2=x2+2xh+h2(x+h)^2 = x^2 + 2xh + h^2
  2. 2

    Subtract f(x)f(x) and tidy the top.

    f(x+h)−f(x)=2xh+h2f(x+h) - f(x) = 2xh + h^2
  3. 3

    Divide every term by hh so it cancels.

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

    Let the gap hh shrink to zero.

    f′(x)=lim⁡h→0(2x+h)=2xf'(x) = \lim_{h \to 0} (2x + h) = 2x
Answer
f′(x)=2xf'(x) = 2x
Worked Example 3With a linear term

Differentiate f(x)=x2−3xf(x) = x^2 - 3x from first principles.

  1. 1

    Expand f(x+h)f(x+h).

    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. 2

    Subtract f(x)=x2−3xf(x) = x^2 - 3x.

    f(x+h)−f(x)=2xh+h2−3hf(x+h) - f(x) = 2xh + h^2 - 3h
  3. 3

    Divide by hh, then let h→0h \to 0.

    2xh+h2−3hh=2x+h−3\frac{2xh + h^2 - 3h}{h} = 2x + h - 3
Answer
f′(x)=2x−3f'(x) = 2x - 3
Worked Example 4A fraction, a little trickier

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

  1. 1

    Combine f(x+h)−f(x)f(x+h) - f(x) over a common denominator.

    1x+h−1x=x−(x+h)x(x+h)=−hx(x+h)\frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}
  2. 2

    Divide by hh.

    1h⋅−hx(x+h)=−1x(x+h)\frac{1}{h} \cdot \frac{-h}{x(x+h)} = \frac{-1}{x(x+h)}
  3. 3

    Let h→0h \to 0.

    f′(x)=−1x2f'(x) = -\frac{1}{x^2}
Answer
f′(x)=−1x2f'(x) = -\dfrac{1}{x^2}

Practice questions

Practice test

Try it yourself

4 questions, 6 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.Using first principles, the derivative of f(x)=x2f(x)=x^2 is:

1mark
Need a hint?
Expand (x+h)2(x+h)^2, subtract x2x^2, divide every term by hh, then let h→0h \to 0 so any leftover hh vanishes.

Q2.From first principles, the gradient function of the straight line f(x)=5xf(x)=5x is:

1mark
Need a hint?
A straight line has the same steepness everywhere. Work through the limit and think about what gradient a line of the form y=mxy = mx should have.

Q3.Using first principles, the derivative of f(x)=x2+2xf(x)=x^2+2x is:

1mark
Need a hint?
Differentiate term by term in your head as a check, but show the full limit. Remember to cancel the hh before letting h→0h \to 0.

Q4.Find, from first principles, the derivative of f(x)=x2−4xf(x)=x^2-4x. Show every step.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

What does differentiation from first principles actually mean?
It means finding the derivative directly from the limit definition, rather than using a shortcut rule like the power rule. You measure the gradient of a line through two points and then shrink the gap between those points to zero.
Why does the h cancel before you take the limit?
While the gap is still a real number you are allowed to divide top and bottom by it, so the h cancels cleanly. You must cancel first, because once you set h to zero you would otherwise be dividing by zero, which is undefined.
When am I expected to use first principles instead of just differentiating?
Use it whenever a VCE question explicitly says from first principles or by the limit definition. In that case you must show the full limit working, since a shortcut rule earns no method marks even if the final answer is correct.