Every antiderivative technique in this course is a derivative rule read backward. Before that reversal makes sense, the forward direction has to be automatic, the way 2+2 is automatic, not something you look up. This section restates the four rules that carry the entire weight of what follows: power, product, quotient, and chain. It's a fast, deliberately dense pass, not a from-scratch derivation of each one.
Picture a rectangle whose width f(x) and height g(x) both change as x moves. A growing width alone adds area. A growing height alone adds area too. The total change has to count both effects: one side's motion times the other side's current size, added up once for each side. That sum is exactly the product rule's two terms, not a formula to memorize but a picture of two things growing at once.
The power rule, \frac{d}{dx}[x^n] = nx^{n-1}, holds for EVERY real exponent n: negative, fractional, irrational, not just positive whole numbers. That generality matters immediately. \frac{1}{x^2} is x^{-2}, and \sqrt{x} is x^{1/2}, so both differentiate by the same rule as x^5 does. No separate reciprocal or root rule is needed. The product rule, \frac{d}{dx}[fg] = f'g + fg', and the quotient rule, \frac{d}{dx}\!\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2}, exist because differentiation does NOT distribute over multiplication or division the way it does over addition: \frac{d}{dx}[fg] \neq f'g'. That's worth naming explicitly. It's the single most common algebra error this course will keep running into. The chain rule, \frac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x), differentiates a function built by feeding one function's output into another: an outer derivative times an inner one, never the other way around.
Compare a function to its derivative
Set a — watch how the derivative's own graph (dashed) tracks where the curve is rising, falling, or flat.
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A derivative as the limit of a slope
Drag both points to x=1 -- watch the secant's slope settle at the tangent slope there, f'(1)=3(1)^2-3=0.
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What this task checks
Structure checked automatically. Whether your ideas are convincing isn't graded, so reread with the checklist.
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Worked example
Differentiating a product and a quotient together
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1
Find \dfrac{d}{dx}\left[\dfrac{x^2\sin x}{1}\right], written this way on purpose to force the product rule first. Differentiate f(x) = x^2\sin x.
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2name the two factors before touching the product rule
Let u = x^2 and v = \sin x, so u' = 2x and v' = \cos x.
name the two factors before touching the product rule
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3
By the product rule, f'(x) = u'v + uv' = 2x\sin x + x^2\cos x.
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Neither term simplifies further. This is the final derivative: \boxed{f'(x) = 2x\sin x + x^2\cos x}.
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Where this goes next
Every technique ahead
u-substitution, two sections from now, is nothing but the chain rule read in reverse. Integration by parts, several chapters ahead, is the product rule read in reverse. Neither reversal is learnable without the forward rule already being automatic.
Check your understanding (counts toward your progress)
Find \dfrac{d}{dx}\left[x^{6}\sin(2x)\right].
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Product rule, u=x^{6}, v=\sin(2x): u'v+uv' = 6x^{5}\sin(2x) + 2x^{6}\cos(2x).
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For f(x) = -3x^4 -2x, find f'(1).
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Why
f'(x) = -12x^{3} -2, so f'(1) = -12\cdot(1)^{3} -2 = -14.
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For f(x) = x^3 - 2x, find f'(2).
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Why
f'(x) = 3x^2-2, so f'(2) = 12-2 = 10.
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Practice: The four rules you must have cold (4 problems · doesn’t affect your progress)
Which is the correct derivative of f(x) = x^2 e^x?
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This only differentiates the x^2 factor and drops the e^x's own contribution entirely.
This only differentiates the e^x factor and drops the x^2's own contribution entirely.
This treats the product as a sum — the product rule never distributes the derivative across factors that way.
Why
Product rule — u'v+uv' = 2x\cdot e^x + x^2\cdot e^x.
Product rule: differentiate x^2 and e^x separately, then combine as u'v + uv'.
f'(x) = 2xe^x + x^2e^x
Differentiate the fifth-power term: find \dfrac{d}{dx}\left[x^5\right].
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Power rule — bring the exponent down, subtract 1 from it — 5x^{5-1} = 5x^4.
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Find \dfrac{d}{dx}\left[(3x+1)^4\right].
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
Chain rule — outer derivative 4(3x+1)^3 times inner derivative 3, giving 12(3x+1)^3.
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Use the chain rule: differentiate the outer power first, then multiply by the derivative of 3x+1.
Find \dfrac{d}{dx}\left[\dfrac{x^2}{x+1}\right].
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\dfrac{2x(x+1) - x^2(1)}{(x+1)^2} = \dfrac{x^2+2x}{(x+1)^2}.
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Why
u = x^2, v = x+1 — apply (u'v - uv')/v^2.