Derivative Rules Practice

Skill: derivative-rules

Derivative Rules Practice

The three rules that differentiate almost everything: the product rule for multiplied chunks, the quotient rule for fractions, and the chain rule for functions inside functions. Master these and every derivative becomes bookkeeping.

1 Practice

Type your answer and press Check answer (or Enter). A wrong answer earns a hint; a second miss earns a stronger hint; a third miss walks you through the full solution. Scores and streaks are session-only.

Score 0 · Streak 0 · Problem 0
Score counts first-try correct answers. Streak counts consecutive correct answers.
Problem
Enter integers, decimals, or fractions like 3/4. For two-part answers fill both boxes. Multiple choice: click an option.

2 What you’ll practice

Every problem is generated fresh from templates across several question types — a problem never repeats until its whole pool is used up.

Product rule

Differentiate (ax+b)(cx+d) and read off the linear result.

Ex: f(x) = (2x+1)(3x−4) → f′(x) = 12x − 5

Quotient rule

Find the numerator of the derivative of (ax+b)/(cx+d).

Ex: (3x+2)/(x−1) → N = −5

Chain rule: powers

Peel (ax+b)^n into outer coefficient and new exponent.

Ex: (2x+1)^4 → K = 8, m = 3

Chain rule at a point

Evaluate a chained derivative at a given x.

Ex: f(x) = (x²+1)³, f′(1) = ?

Chain rule with trig

sin(ax), cos(ax) at special angles.

Ex: f(x) = sin(3x), f′(π/3) = ?

Product + chain combo

Both rules in one problem, evaluated at a point.

Ex: f(x) = (2x+1)²(x−3), f′(0) = ?

Spot the right rule

Multiple choice targeting the classic traps.

Ex: d/dx[(3x+1)^4] — where is the ×3?

3 Watch out for these

The mistakes students make most often on these problems.

Forgetting the inner derivative

d/dx[(3x+1)^4] is 12(3x+1)³, not 4(3x+1)³. The chain rule always has two multiplications: outer′ × inner′.

Flipping the quotient numerator

It is low d-high MINUS high d-low. Reversing the order flips the sign of the whole derivative.

“Derivative of a product” shortcut

There is none: (fg)′ = f′g + fg′. Differentiating each factor and multiplying gives the wrong answer.

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