In Lesson 0.1, grinding through the limit definition by hand gave two results: the derivative of x² is 2x, and the derivative of x³ is 3x². Look closely and a pattern jumps out — in both cases, the exponent drops down in front as a multiplier, and the new exponent is one less than the old one. That pattern isn't a coincidence, and it doesn't stop at x³. It holds for any real number exponent n:
This is the power rule. It's provable in general using the exact same limit-definition process from Lesson 0.1 (just with a more general binomial expansion) — but you don't need to re-derive it every time you use it, any more than you re-derive long division every time you divide two numbers.
Real position functions aren't a single power of t sitting alone — they're sums of several terms, each with its own coefficient, like x(t) = x₀ + v₀t + ½at². Two more rules let you handle that directly, differentiating term by term:
A constant multiplying a function just rides along for the ride — differentiate the function, then multiply by the same constant. And the derivative of a sum is just the sum of the derivatives, term by term. One more useful fact falls out of this immediately: the derivative of a plain constant, with no variable at all, is zero — a constant never changes, so its rate of change is nothing.
Find g'(x) for g(x) = 3x⁴ − 5x² + 7.
The power rule doesn't care what an expression looks like on the page — only what exponent it's actually hiding. A reciprocal like 1/x² is really x⁻², and a square root like √x is really x^(1/2). Rewrite first, using the rules of exponents, and the power rule applies exactly as before.
Find h'(x) for h(x) = 1/x² + √x.
This shows up constantly in physics — gravity's inverse-square law is built on 1/r², and a pendulum's period formula, T = 2π√(l/g), has a square root sitting right in it. Both are just power-rule problems in disguise.
Here's the payoff. A position function like x(t) = x₀ + v₀t + ½at² is nothing more than a polynomial in t — every tool from this lesson applies directly, and differentiating it twice takes you from position, to velocity, to acceleration.
An object's position is given by x(t) = 5 + 12t − 4.9t² (in meters, with t in seconds). Find v(t) and a(t).
Try it yourself below — adjust the starting position, initial velocity, and acceleration, and watch the differentiated functions and their graphs update together in real time.
Adjust x₀, v₀, and a below. x(t) is differentiated once to get v(t), and v(t) is differentiated again to get a(t) — watch all three update together.
Notice a(t) is always a flat, constant line — differentiating a constant gives zero, so once you've taken the derivative of v(t), there's nothing left to change over time.