BY THE END, YOU CAN
- Differentiate a simple polynomial.
- Find the gradient at a given x-value.
- Use the gradient to write a tangent equation.
Before you start: be comfortable with powers, substitution and the equation of a straight line. This is an introductory lesson, not the whole calculus syllabus.
UNDERSTAND IT
What does a derivative tell you?
On a straight line, the gradient stays the same. On a curve, it can change as you move along. The derivative gives the gradient of the tangent at a point: the straight line that matches the curve’s direction there.
If the function is written as f(x), its derivative is written as f′(x), read “f prime of x”. If the function is written as y, you may see dy/dx instead. Both notations describe a rate of change with respect to x.
The triangle belongs to the straight tangent, not the curve. Its gradient is rise ÷ run = 4 ÷ 1 = 4.
Height and slope are different. f(2) tells you the height of the curve at x = 2. f′(2) tells you its gradient there. They happen to both be 4 for y = x² at x = 2, but that is not generally true.
LEARN THE METHOD
The power rule, without the mystery
For a term axⁿ, where a is a constant and n is a positive integer, multiply the coefficient by the power, then reduce the power by one. Differentiate a sum term by term.
axn → an xn − 1
The arrow means “differentiate with respect to x”.3 × 2 = 6; the new power is 2 − 1 = 1.
x means x¹. The derivative is −4 × 1 × x⁰ = −4.
A constant does not change when x changes, so its derivative is zero.
Keep the sign attached to each term. A negative coefficient does not become positive just because you differentiate it.
WORKED EXAMPLE
First find the rule. Then find the number.
The question
Given f(x) = 3x² − 4x + 5, find f′(x). Then find the gradient of the curve at x = 2.
Differentiate each term
3x² becomes 6x, −4x becomes −4, and 5 becomes 0.
f′(x) = 6x − 4Substitute x = 2 into the derivative
Use the derivative because the question asks for the gradient, not the height.
f′(2) = 6(2) − 4 = 12 − 4 = 8State what your answer means
The tangent has gradient 8 at x = 2. The positive sign tells you the curve is increasing at that point.
Putting 2 into the original function gives f(2) = 12 − 8 + 5 = 9. That is the y-coordinate. It is not the gradient. Differentiate first when the question asks for a slope.
TAKE IT ONE STEP FURTHER
Turn the gradient into a tangent equation
A straight line is determined by a point and a gradient. You now have both: the point (2, 9) from f(2), and the gradient 8 from f′(2).
y − y₁ = m(x − x₁)
m is the gradient; (x₁, y₁) is a point on the line.Put m = 8, x₁ = 2 and y₁ = 9 into the formula:
y − 9 = 8x − 16
y = 8x − 7
Check it: at x = 2, the line gives y = 16 − 7 = 9. It passes through the correct point, and its coefficient of x is 8, the correct gradient.
YOUR TURN
Try it before opening the solution
Write your working on paper. If you get stuck, open the hint first. These are original CollegeRoute practice questions, not reproduced State Examinations Commission questions.
Find the gradient
g(x) = 2x² + 3x − 7. Find g′(x), then find the gradient at x = 2.
Show a hint
Differentiate the three terms separately. The −7 disappears; the +3x contributes +3.
Show the worked solution
g′(2) = 4(2) + 3 = 11
The gradient is positive, so the curve is increasing at x = 2.
What does a negative answer mean?
h(x) = x² − 6x + 4. Find h′(1). Is the curve increasing or decreasing at x = 1?
Show a hint
A negative derivative means a downward slope as you move from left to right.
Show the worked solution
h′(1) = 2(1) − 6 = −4
The curve is decreasing at x = 1. A negative result is meaningful, not an error to remove.
Find a tangent
For y = x² + x + 1, find the equation of the tangent at x = 1.
Show a hint
Find the y-coordinate from the original function. Find the gradient from its derivative. Then use y − y₁ = m(x − x₁).
Show the worked solution
dy/dx = 2x + 1
m = 2(1) + 1 = 3
y − 3 = 3(x − 1)
y = 3x
Check: when x = 1, y = 3. The tangent passes through (1, 3) and has gradient 3.
TAKE THIS INTO YOUR NEXT QUESTION
A three-question checklist
- Asked for a derivative? Give a function such as f′(x) = 6x − 4.
- Asked for a gradient at a point? Substitute into the derivative.
- Asked for a tangent? Use the original function for the point and the derivative for the gradient.
When you can do all three without the solutions, try a fresh question from your class work. Write each algebraic step clearly; a bare final answer does not show your method.
Scope and sources
This introduction covers simple polynomial derivatives, gradients and tangents. It does not cover first principles, the product, quotient or chain rules, or the full range of Higher Level applications.
The scope follows the NCCA Leaving Certificate Mathematics syllabus (opens in a new tab) and the learning outcomes listed in Scoilnet / PDST: Teaching Calculus for Understanding (opens in a new tab). Source pages checked 8 September 2026. The explanations, questions and diagram here are CollegeRoute’s own; neither organisation has endorsed this lesson.