🌌 Calculus Kingdom · Calculus
Derivatives & the Power Rule
Define the derivative as an instantaneous rate of change and compute it for polynomials with the power rule.
In short
- The derivative f'(x) gives the slope of the tangent at each x, defined as a limit of average slopes.
- Power rule: bring the exponent down as a multiplier and subtract one from it.
- Constants differentiate to 0, constant multipliers stay put, and sums differentiate term by term.
- Roots and reciprocals must be rewritten as powers before the rule can be applied.
The derivative is a slope function
The slope of a straight line is the same everywhere. A curve is different: its steepness changes from point to point.
The derivative f'(x) is a new function whose output at each x is the slope of the tangent to the curve at that point.
It is built from a limit. Take two points on the curve a distance h apart, compute the slope of the line joining them, and let h shrink to zero:
f'(x) = limit as h -> 0 of (f(x + h) - f(x)) / h
For f(x) = x2, expanding gives ((x + h)2 - x2)/h = (2xh + h2)/h = 2x + h, and letting h go to 0 leaves f'(x) = 2x.
The derivative answers "how fast is this changing right now?", which is why it appears everywhere: velocity is the derivative of position, and marginal cost is the derivative of cost.
The power rule
Doing that limit for every function would be exhausting. The power rule does it once and for all:
d/dx (xn) = n xn-1
Bring the exponent down as a multiplier, then subtract one from the exponent.
d/dx (x5) = 5x4 d/dx (x) = 1 (since x = x1, this is 1 x0 = 1) d/dx (7) = 0 (a constant is a flat line, with no slope)
Two more rules let you handle any polynomial:
- a constant multiplier stays put: d/dx (6x3) = 6 x 3x2 = 18x2
- a sum differentiates term by term
So for f(x) = 4x3 - 5x2 + 7x - 9:
f'(x) = 12x2 - 10x + 7
The -9 vanishes entirely. Sliding a graph up or down does not change any of its slopes.
Negative and fractional exponents
The power rule does not care what kind of number the exponent is — but it must actually be written as a power first.
1/x3 = x-3 -> derivative -3x-4 = -3/x4 sqrt(x) = x1/2 -> derivative (1/2)x-1/2 = 1/(2 sqrt(x))
Subtracting 1 always moves the exponent down: -3 becomes -4, and 1/2 becomes -1/2. Going the other way is a very common slip.
The rewriting step is where most errors happen, so do it explicitly rather than in your head. A fraction is not differentiated by differentiating the top and the bottom separately, and a root is not differentiated by taking the root of the derivative.
Using the derivative
Slope at a point. Differentiate first, substitute afterwards. For f(x) = x3 - 4x, f'(x) = 3x2 - 4, so the slope at x = 2 is 3(4) - 4 = 8.
Doing it the other way round is fatal: substituting first turns the function into a plain number, and the derivative of a number is 0.
Note also that f(2) and f'(2) are different things. f(2) = 0 is the height of the curve; f'(2) = 8 is its steepness.
Where the slope is zero. Setting f'(x) = 0 finds the flat spots — the tops of hills and the bottoms of valleys. For f(x) = x2 - 6x + 5, f'(x) = 2x - 6, which is zero at x = 3, and that is exactly the vertex of the parabola.
The second derivative. Differentiating again gives f''(x), the rate at which the slope itself is changing. If position is s(t), then s'(t) is velocity and s''(t) is acceleration.
Worked examples
Example 1
Find the derivative of f(x) = 6x4 - 3x2 + 8x - 11.
- Differentiate term by term with the power rule.
- 6x4 gives 6 x 4x3 = 24x3; -3x2 gives -3 x 2x = -6x.
- 8x gives 8, and the constant -11 gives 0.
- f'(x) = 24x3 - 6x + 8.
Example 2
f(x) = x3 - 12x. Find the values of x where the tangent is horizontal.
- A horizontal tangent means the slope is zero, so solve f'(x) = 0.
- f'(x) = 3x2 - 12.
- Set it to zero: 3x2 - 12 = 0, so x2 = 4.
- x = 2 or x = -2 — the curve is flat at both of these points.
Practice problems, with solutions
Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.
Problem 1
Difficulty 1 of 5f(x) = 4x2 + 3 Find f'(x). Give your answer as an expression in x.
Answer: 8x
- d/dx (4x2) = 4 x 2 x1 = 8x
- d/dx (3) = 0 (a constant has no slope)
- f'(x) = 8x
Problem 2
Difficulty 3 of 5f(x) = 5x2 - 8x + 8 Find f'(0) — the slope of the curve at x = 0.
Answer: -8
- f'(x) = 10x - 8
- f'(0) = 10(0)1 + -8(0)0
- f'(0) = -8
Problem 3
Difficulty 4 of 5f(x) = 8x3/2 Find f'(x). Give your answer as an expression in x.
Answer: 12x1/2
- f'(x) = 8 x (3/2) x^(3/2 - 1)
- 3/2 - 2/2 = 1/2
- f'(x) = 12x1/2
Common mistakes
- Substituting the point before differentiating, which always gives 0.
- Multiplying by the exponent but forgetting to reduce it.
- Raising a negative exponent instead of lowering it: -3 goes to -4, not -2.
- Confusing f(a) (the height) with f'(a) (the slope).
What you should be able to do
- Interpret the derivative as a slope of a tangent line.
- Differentiate a power of x with the power rule.
- Differentiate a polynomial term by term.
- Evaluate a derivative at a point to get a slope.
Where this fits in the curriculum
Common Core
- CHA-2.A
AP Calculus AB, Unit 2 — Determine average rates of change using difference quotients.
The Common Core has no calculus standards; this is the AP Calculus AB learning objective.
- CHA-2.B
AP Calculus AB, Unit 2 — Represent the derivative of a function as the limit of a difference quotient.
The Common Core has no calculus standards; this is the AP Calculus AB learning objective.
- CHA-2.C
AP Calculus AB, Unit 2 — Determine the equation of a line tangent to a curve at a given point.
The Common Core has no calculus standards; this is the AP Calculus AB learning objective.
- FUN-3.A
AP Calculus AB, Unit 2 — Calculate derivatives of familiar functions (the power rule, sums, polynomials).
The Common Core has no calculus standards; this is the AP Calculus AB learning objective.