The Power Rule for Differentiating Polynomials
Differentiate any sum of powers of x with the rule that xⁿ becomes nxⁿ⁻¹, including negative and fractional powers once the term is rewritten as a power.
What a learner can do afterwards
- Differentiate y = 4x³ - 7x + 2 term by term
- Rewrite 1/x² and √x as powers before differentiating
- Evaluate the derivative at a given x to get the gradient there
1 · Read
Last stop you ground out four derivatives by hand: x² gave 2x, x³ gave 3x², 4x² gave 8x and 2x³ gave 6x². Line those up and the pattern is hard to miss. Each power drops by one, and the old power comes down to the front as a multiplier. That pattern is the power rule, and it is not a guess. Every whole-power case is provable by the first-principles calculation you already did, which is why you may now use it without repeating the work. The Ridgeway ramp profile can finally be differentiated in one line.
Stated properly: if y = xⁿ then dy/dx = nxⁿ⁻¹, where dy/dx is another way of writing f'(x). Three companions come with it. A constant multiplier rides along untouched, so y = kxⁿ gives knxⁿ⁻¹. A sum is differentiated one term at a time. And a constant on its own differentiates to zero, because its graph is a flat line with no steepness anywhere. A product of brackets is not a sum of powers, so multiply it out before you start. That means expanding (x + 1)(x + 4) first.
Differentiate y = 4x³ - 7x + 2. Take it a term at a time. For 4x³, bring the 3 down to multiply the 4, giving 12, and drop the power to 2: that is 12x². For -7x, remember it is -7x¹, so the 1 comes down and the power becomes 0, leaving -7x⁰, and x⁰ is 1, so the term is just -7. The +2 is a constant, so it goes. That gives dy/dx = 12x² - 7. To use it, feed it an x: at x = 2 the gradient is 48 - 7 = 41.
The rule holds for every n, not only whole numbers, but the term has to be written as a power of x before it can be applied. That is what index work is for. A reciprocal moves the power upstairs with a sign change, so 1/x² is x⁻². A root becomes a fraction, so √x is x^(1/2) and ∛x is x^(1/3). Once rewritten, nothing about the method changes. Subtracting 1 from a negative power makes it more negative, which is where most slips happen.
Differentiate y = 1/x². Rewrite it as x⁻². Bringing -2 down gives -2, and one off the power gives -3, so dy/dx = -2x⁻³, which is -2/x³. The minus sign is right, since this curve falls as x grows. Now y = √x, which is x^(1/2). Bringing 1/2 down gives 1/2, and 1/2 - 1 is -1/2, so dy/dx = (1/2)x^(-1/2), which is 1/(2√x). At x = 4 that gives 1 divided by 2 times 2, which is 1/4.
The power rule says xⁿ differentiates to nxⁿ⁻¹: the power comes down in front and one is taken off it. A constant multiplier rides along, a sum is done term by term, and a constant on its own gives zero. The rule works for negative and fractional powers too, but only once the term is written as a power, so rewrite 1/x² as x⁻² and √x as x^(1/2) first. Feeding a number into dy/dx gives the gradient at that point.
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24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.