Units 4 and 5 of AP Calculus AB are where the derivative stops being a limit to compute and becomes a tool to use. This set works through ten of those uses, one question each, deliberately without repeats — because a bank that asks the same thing three times teaches one thing three times.
The ten: implicit differentiation at a stated point; a related-rates problem; the equation of a tangent line; finding critical points; the first derivative test read off a factored derivative; concavity and every point of inflection; a constrained optimisation; the Mean Value Theorem applied to a named function and interval; linearisation, together with whether the estimate comes out high or low; and motion, where the difference between velocity and speed is the entire question.
Three of these carry the distinctions that separate a reliable score from a shaky one. The first derivative test question gives : both roots are critical points, but a squared factor cannot change sign, so the graph flattens at without turning. Finding a critical point and finding an extremum are not the same task, and the squared factor is how the exam says so. The inflection question asks for every inflection point of , not just one — vanishes twice and changes sign twice, and a candidate who solves and stops at the first root has done the algebra and skipped the test. The linearisation question asks for the number and the direction: , and because the square root is concave down its tangent line lies above the curve, so the estimate is too large. Concavity, not the size of the step, decides that.
Every distractor is a specific error rather than a random number. A sign slip on the product rule turns $-4/5$ into $-4/3$; solving for inverts it again. In the related-rates problem, exchanging the two distances gives $24/5$ and using the ladder's length in place of the height gives $10/13$. Each explanation names which mistake produces which wrong answer, so a wrong click tells you what to fix.
What this cannot train is the free-response half of the exam, where roughly half the marks live. There, the answer is the smaller part of the credit: points are awarded for setting up the relation before substituting, for naming the theorem that justifies a conclusion, and for units. Use this to make the underlying moves automatic, and write out full solutions separately.
Built against the published Course and Exam Description structure for AP Calculus AB: Unit 4 (Contextual Applications of Differentiation — related rates, linearisation, motion) and Unit 5 (Analytical Applications of Differentiation — the Mean Value Theorem, the first and second derivative tests, optimisation). The exam runs 3 hours 15 minutes in two sections, multiple choice and free response, with calculator and no-calculator parts in each; this set is no-calculator throughout. Every function, number and scenario here is invented for this material — nothing is reproduced from any College Board publication, released exam or scoring guideline. Zestly is an independent study tool. It is not affiliated with the College Board, which owns the AP Calculus AB exam, and it is not an exam centre.
Given the relation $x^2 + xy + y^2 = 7$, what is the value of $\frac{dy}{dx}$ at the point $(1, 2)$?
$-4/5$
Differentiate both sides with respect to $x$, treating $y$ as a function of $x$: $2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. At $(1, 2)$ this becomes $2 + 2 + \frac{dy}{dx} + 4\frac{dy}{dx} = 0$, so $5\frac{dy}{dx} = -4$ and $\frac{dy}{dx} = -4/5$. A sign slip on the product-rule term, writing $-x\frac{dy}{dx}$ instead of $+x\frac{dy}{dx}$, leaves $3\frac{dy}{dx} = -4$ and gives $-4/3$. Inverting the fraction at the last step — solving for $\frac{dx}{dy}$ rather than $\frac{dy}{dx}$ — turns those two answers into $-5/4$ and $-3/4$.