Differentiation: AP Calculus BC Practice Problems

Since x=0.10 is the leftmost value in the table, there's no data point to its left — the best estimate uses a forward difference quotient with the next available point, x=0.30.

f′(0.10) ≈ [f(0.30)−f(0.10)] / (0.30−0.10) = (0.320−0.140)/0.20.

= 0.180/0.20 = 0.90.

The correct answer is B.

Since x=3.00 is the last (rightmost) value in the table, use a backward difference quotient with the closest available point, x=2.97.

f′(3) ≈ [f(3.00)−f(2.97)] / (3.00−2.97) = (6.20−6.50)/0.03.

= −0.30/0.03 = −10.

The correct answer is D.

The inverse function derivative formula is (f⁻¹)′(b) = 1/f′(a), where a is the input satisfying f(a)=b.

From the table, f(3.00) = 6.20, so a=3.00. From Question 2, f′(3) ≈ −10.

(f⁻¹)′(6.20) ≈ 1/(−10) = −0.1.

The correct answer is A.

Since f is differentiable (hence continuous) on [1,9] and f(1) = f(9) = 40, Rolle's Theorem guarantees at least one c in (1,9) with f′(c) = 0. This is the only choice that's guaranteed by the given information.

Choice A assumes the true minimum occurs exactly at the tabulated point x=5, but the actual function could dip even lower somewhere between the sampled points — the table alone doesn't guarantee f′(5)=0 specifically.

Choice B isn't guaranteed either; the function's behavior between x=5 and x=7 isn't fully determined by just two table values. Choice C isn't guaranteed since the function's true maximum between sample points could exceed 40.

The correct answer is D.

The difference quotient is [f(a+h)−f(a)]/h. Using a Taylor expansion, this is approximately f′(a) + (1/2)f″(a)h for small h.

For the difference quotient to be less than f′(a) (an underestimate) for every h>0, the correction term (1/2)f″(a)h must be negative for h>0, which means f″(a) ≤ 0.

This is exactly the condition for f to be concave down: when a function curves downward, secant lines drawn to the right of a point lie below the tangent line's direction, making the secant's slope smaller than the true derivative.

The correct answer is D.

Only continuity is given here — not differentiability. Rolle's Theorem requires differentiability to guarantee a point where f′=0, so choice C cannot be concluded, and choice D goes even further by claiming the derivative must exist everywhere, which also isn't guaranteed.

Choice A is too strong: f could rise above zero (or dip below) throughout the open interval and only return to zero exactly at the two endpoints — it doesn't have to be zero everywhere in between.

A continuous function satisfying f(p)=f(q)=0 could easily be nonzero at every point strictly between p and q (like a simple bump or dip that touches zero only at the endpoints) — this is entirely possible.

The correct answer is B.

Use the inverse function derivative formula: (f⁻¹)′(b) = 1/f′(a), where a satisfies f(a)=b.

Since f(2)=5, we know f⁻¹(5)=2, so the a we need is 2, not 5. That means f′(5)=7 is extra information that isn't actually needed here.

(f⁻¹)′(5) = 1/f′(2) = 1/4.

The correct answer is C.

This expression is exactly the definition of the derivative of h at x=3: h′(3) = lim_x→3 [h(x)−h(3)]/(x−3). Since this limit is given to equal 5, it follows directly that h′(3)=5.

For the expression h(x)−h(3) to even make sense, h(3) must be defined — ruling out choice A. Since h is differentiable at 3, it must also be continuous there — ruling out choice B.

Choice C is a trap: differentiability at 3 implies lim_x→3h(x) = h(3), but nothing here tells us h(3) itself equals 5 — that numerical value belongs to the derivative, not to h(3) or the ordinary limit of h(x).

The correct answer is D.

Estimate two ordinary derivatives from consecutive intervals, treating each as centered at its interval's midpoint. From x=1.0 to x=1.2 (centered at x=1.1): f′ ≈ (10−6)/(1.2−1.0) = 4/0.2 = 20.

From x=1.4 to x=1.6 (centered at x=1.5): f′ ≈ (28−16)/(1.6−1.4) = 12/0.2 = 60.

Now estimate the second derivative as the rate of change between these two derivative estimates, using the distance between their centers (1.1 and 1.5): f″(1.3) ≈ (60−20)/(1.5−1.1) = 40/0.4 = 100.

The correct answer is D.

For a cone, the radius at the water's surface is proportional to the depth (by similar triangles): r=kh for some constant k. So the volume is V = (1/3)πr²h = (1/3)πk²h³.

Since water drains at a constant rate, dV/dt is a negative constant. Differentiating, dV/dt = πk²h² · dh/dt, so dh/dt = (dV/dt)/(πk²h²), which is inversely proportional to h² — a nonlinear relationship, not linear.

As the tank drains, h decreases, which makes h² smaller, which makes the magnitude of dh/dt (still negative) grow larger — meaning dh/dt itself is getting more negative, i.e. decreasing, as time goes on.

The correct answer is D.

Find the secant line's slope: [f(7)−f(−7)] / 14 = [(7+cos7)−(−7+cos7)]/14 = 14/14 = 1 (the cosine terms cancel since cosine is even).

A tangent is parallel to this secant wherever f′(x) = 1−sin(x) = 1, which happens exactly when sin(x) = 0, i.e. at x = 0, ±π, ±2π, …

Check which of these fall inside [−7,7]: π ≈ 3.14 and 2π ≈ 6.28 both fit (since 6.28<7), but 3π ≈ 9.42 does not.

That gives x = 0, π, −π, 2π, −2π — five points total. The correct answer is D.

Use the standard limit lim_u→0 sin(u)/u = 1. Rewrite the expression to expose this pattern: sin(5x)/sin(2x) = [sin(5x)/(5x)] · [5x/(2x)] · [(2x)/sin(2x)].

As x→0, the first and third bracketed factors both approach 1, leaving just the middle ratio: 5x/(2x) = 5/2.

The correct answer is B.

Use the standard limit lim_u→0 [1−cos(u)]/u² = 1/2. Rewrite with u=3x: [1−cos(3x)]/x² = [1−cos(3x)]/(3x)² · 9 (multiplying and dividing by 9 to turn x² into (3x)²).

As x→0 (so 3x→0 too), the first factor approaches 1/2 by the standard limit.

(1/2) × 9 = 9/2. The correct answer is B.

Since tan(u) = sin(u)/cos(u) and cos(u) → 1 as u→0, tan(u) behaves just like sin(u) for small u, so lim_u→0 tan(u)/u = 1 as well.

Rewrite the expression to match this pattern: tan(2π x)/x = [tan(2π x)/(2π x)] × 2π.

As x→0, the bracketed factor approaches 1, leaving 1 × 2π = 2π.

The correct answer is D.

Let u=1/x. As x→∞, u→0⁺, and the expression becomes x· sin(1/x) = sin(u)/u.

This is exactly the standard limit lim_u→0 sin(u)/u = 1.

The correct answer is A.

Differentiate each equation with respect to t: dx/dt = 1+cos(t) and dy/dt = −sin(t).

Use dy/dx = (dy/dt)/(dx/dt): dy/dx = −sin(t)/(1+cos(t)).

The correct answer is B.

Use the chain rule for each: dx/dθ = 3sin²(θ)cos(θ) and dy/dθ = −3cos²(θ)sin(θ).

dy/dx = [−3cos²(θ)sin(θ)] / [3sin²(θ)cos(θ)]. The factor of 3 cancels, along with one power each of sin and cos.

This leaves −cos(θ)/sin(θ) = −cot(θ).

The correct answer is A.

Differentiate each: dx/dt = −e^−t and dy/dt = 1−e^t.

dy/dx = (1−e^t) / (−e^−t) = −(1−e^t) · e^t (multiplying numerator and denominator by e^t to clear the negative exponent).

Distribute: −(1−e^t)e^t = (e^t−1)e^t = e^2t−e^t.

The correct answer is B.

Differentiate each: dx/dt = 1/(3−t)² (using the power/chain rule on (3−t)⁻¹), and dy/dt = 1/(3−t).

dy/dx = [1/(3−t)] ÷ [1/(3−t)²] = (3−t)²/(3−t) = (3−t).

Since x = 1/(3−t), it follows that (3−t) = 1/x. Substituting gives a clean expression purely in terms of x.

dy/dx = 1/x. The correct answer is C.