Conditions before computation
When taking √((ax+b)²), the result is |ax+b|, not automatically ax+b; a positivity condition justifies removing the absolute value. Dividing by a dependent-variable expression during separation can exclude equilibrium solutions, which should be considered separately when relevant. In the reciprocal-growth family the positive initial condition rules out y=0, but the finite denominator zero still bounds the solution interval. No Euler-method or logistic-model simulation is included in this AB lesson.
Worked example with a method check
For x>0, y=(3x+6)² satisfies y′=k√y. Find the constant k. On this domain 3x+6>0, so √y=3x+6. Differentiation gives y′=6(3x+6). Matching the nonzero factor gives k=6.
Compare two methods
For y′=y² and y(0)=1/3, separation gives y=1/(3−x) on the interval x<3 containing zero. At x=1 the exact solution is 1/2. A tangent estimate from zero would give 1/3+1/9=4/9; it is an approximation, not a competing exact solution. The equation also supplies concavity by differentiation: y″=2yy′, which is positive along this positive solution.
Transfer: solve before reading
Check that y=(2x+5)² satisfies y′=4√y for x>0. Then find y″ at (2,5) for a solution of y′=x−y, explaining why differentiating −y produces −y′.
Transfer reasoning and answer
For x>0, 2x+5>0, so √y=2x+5 and y′=4(2x+5), matching the equation. For the second task, y′=−3 and y″=1−y′=4. Along a solution, y depends on x; it cannot be differentiated as an independent coordinate or held constant.
Practice the next representation
Work all four task families linked to this lesson. For the illustrated items, state which numerical labels and relationships you used. For a written extension, compare each line of your solution with its original self-assessment criteria; these criteria are practice feedback, not an official AP score.
For x>0, y=(3x+6)² satisfies y′=k√y. Find the constant k.
- On this domain 3x+6>0, so √y=3x+6.
- Differentiation gives y′=6(3x+6).
- Matching the nonzero factor gives k=6.
Compare the other answer choices
Omit the outer factor 2 when differentiating the square.
Square the inner coefficient instead of differentiating the square.
Choose a negative square root despite the stated positive domain.
Explain your thinking.
Optional written extension. Self-assessed only; it does not change your automatic practice result. These are teaching criteria, not official awarded marks.
(a) State the relationship or theorem that makes this calculation valid, including its relevant conditions. (b) Show the intermediate computation and give the requested result. (c) Explain why one plausible competing method would fail for this problem.
This draft stays on this page only. Use paper if you want to keep your working.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original Slope explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
AP Calculus AB and BC Course and Exam DescriptionCalculus Volume 1: Fundamental Theorem of Calculus