Last Updated: March 23, 2026
A complete AP exam-style derivatives worksheet covering all six major differentiation topics tested on the AP Calculus AB and BC exams. Problems are carefully calibrated to match the difficulty, format, and notation of actual College Board exams. Includes a complete answer key with step-by-step solutions.
Differentiation is the beating heart of AP Calculus. On both the AB and BC exams, the ability to compute derivatives accurately and efficiently — across a wide variety of function types and contexts — accounts for a substantial portion of the total score. This worksheet is designed to give you rigorous, exam-calibrated practice across all six of the differentiation skills that the College Board tests most heavily. Before you begin, it is worth reviewing our AP Calculus Derivatives Review to make sure your foundational knowledge is solid.
The six topics covered here are not independent. The Chain Rule appears inside Product Rule problems. Implicit differentiation relies on the Chain Rule. Related rates requires implicit differentiation. A complete command of AP derivatives means understanding not just each rule in isolation, but how they combine. The problems in this worksheet are arranged to build that layered fluency, with each section increasing in complexity and each subsequent topic drawing on the rules that came before.
The fifty problems in this worksheet closely mirror the style, phrasing, and difficulty distribution of actual AP Calculus exam questions. Multiple-choice style problems test speed and accuracy. Free-response style problems test your ability to show clear, organised work and arrive at exact answers. Work every problem by hand before consulting the answer key — active problem-solving is the fastest path to exam readiness. If you are short on time, our Last-Minute Calculus Review condenses the highest-yield material into a focused session.
| Rule | Formula | When to Use |
|---|---|---|
| Power Rule | \(\dfrac{d}{dx}[x^n] = nx^{n-1}\) | Any term of the form \(x^n\); includes \(n\) negative or fractional |
| Chain Rule | \(\dfrac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)\) | Any composite (nested) function |
| Product Rule | \(\dfrac{d}{dx}[fg] = f'g + fg'\) | Two functions multiplied together |
| Quotient Rule | \(\dfrac{d}{dx}\!\left[\dfrac{f}{g}\right] = \dfrac{f'g - fg'}{g^2}\) | One function divided by another |
| Implicit Diff. | Differentiate both sides w.r.t. \(x\); apply Chain Rule to any \(y\)-term, giving \(\dfrac{dy}{dx}\) factor | Equation in both \(x\) and \(y\) not solved for \(y\) |
| Related Rates | Relate quantities in an equation, then differentiate w.r.t. \(t\) | Two or more quantities changing with time |
| Function | Derivative | Function | Derivative |
|---|---|---|---|
| \(\sin x\) | \(\cos x\) | \(\cos x\) | \(-\sin x\) |
| \(\tan x\) | \(\sec^2 x\) | \(\sec x\) | \(\sec x\tan x\) |
| \(\cot x\) | \(-\csc^2 x\) | \(\csc x\) | \(-\csc x\cot x\) |
| \(e^x\) | \(e^x\) | \(a^x\) | \(a^x\ln a\) |
| \(\ln x\) | \(\dfrac{1}{x}\) | \(\log_a x\) | \(\dfrac{1}{x\ln a}\) |
| \(\arcsin x\) | \(\dfrac{1}{\sqrt{1-x^2}}\) | \(\arctan x\) | \(\dfrac{1}{1+x^2}\) |
These problems test the Power Rule across its full range: polynomial functions, negative exponents, fractional exponents, and combinations requiring the Sum and Constant Multiple Rules. On the AP exam, the Power Rule appears in almost every derivative problem as part of a larger computation.
Find \(f'(x)\) if \(f(x) = 3x^{-4} + 7x^{1/3} - \sqrt{x}\).
Rewrite with exponent notation
\(f(x) = 3x^{-4} + 7x^{1/3} - x^{1/2}\)
Apply Power Rule term by term
\(f'(x) = 3(-4)x^{-5} + 7\!\left(\tfrac{1}{3}\right)x^{-2/3} - \tfrac{1}{2}x^{-1/2}\)
Chain Rule problems on the AP exam range from straightforward single-layer compositions to multi-layer problems requiring repeated application. All ten problems here are at or above the level of difficulty found in the AP free-response section. For additional practice, see the Chain Rule Worksheet (80 problems) and verify your work with the Chain Rule Calculator. If you are cramming for a test, the Derivatives Cram Sheet distils all key Chain Rule patterns onto a single reference page.
Find \(\dfrac{d}{dx}\!\left[\sin^3(2x^2)\right]\).
Identify the layers
Outermost: \(u^3\). Middle: \(\sin(w)\). Innermost: \(w = 2x^2\).
Differentiate layer by layer (outside in)
\(\dfrac{d}{dx}[\sin^3(2x^2)] = 3\sin^2(2x^2)\cdot\cos(2x^2)\cdot 4x\)
Product Rule problems on the AP exam frequently require combining the Product Rule with the Chain Rule. Problems 23–26 in this section are deliberately multi-step, requiring you to identify which sub-functions themselves need the Chain Rule before assembling the full derivative via the Product Rule. A common source of lost marks here is carrying the wrong sign through — see our guide on Top Derivative Exam Mistakes for the full breakdown.
Find \(y'\) if \(y = x^3 e^{2x}\).
Identify \(f\) and \(g\)
\(f = x^3,\ f' = 3x^2\); \quad g = e^{2x},\ g' = 2e^{2x}\) (Chain Rule)
Apply Product Rule
\(y' = 3x^2 \cdot e^{2x} + x^3 \cdot 2e^{2x} = e^{2x}(3x^2 + 2x^3)\)
The Quotient Rule is one of the most error-prone rules under timed exam conditions. The most common mistake is reversing the sign in the numerator. As a consistency check: the numerator is always denominator × derivative of numerator, minus numerator × derivative of denominator. For a complete list of the errors that cost students the most points, visit our Derivative Exam Mistakes guide. Use the derivative calculator to verify any answers you are uncertain about.
Find \(\dfrac{d}{dx}\!\left[\dfrac{e^{2x}}{x^2+1}\right]\).
Identify numerator and denominator
\(f = e^{2x},\ f' = 2e^{2x}\); \quad \(g = x^2+1,\ g' = 2x\)
Apply Quotient Rule
\(= \dfrac{2e^{2x}(x^2+1) - e^{2x}(2x)}{(x^2+1)^2} = \dfrac{2e^{2x}(x^2-x+1)}{(x^2+1)^2}\)
Implicit differentiation problems are a staple of the AP Calculus free-response section. These eight problems cover the full range: basic implicit derivatives, finding dy/dx at a specific point, finding d²y/dx², and using implicit differentiation to find the equation of a tangent line. For a complete tutorial, see the Implicit Differentiation Guide. If you are using this worksheet as part of final exam preparation, pair it with our Calculus Final Exam Practice resource for a comprehensive review session.
Differentiate both sides with respect to \(x\). Every \(y\)-term that is differentiated picks up a \(\dfrac{dy}{dx}\) factor (Chain Rule). Collect all \(\dfrac{dy}{dx}\) terms and solve.
Find \(\dfrac{dy}{dx}\) given \(x^2y + y^3 = 5\).
Differentiate both sides with respect to \(x\)
\(2xy + x^2\dfrac{dy}{dx} + 3y^2\dfrac{dy}{dx} = 0\)
Collect \(\dfrac{dy}{dx}\) terms
\(\dfrac{dy}{dx}(x^2 + 3y^2) = -2xy\)
Related rates are among the most challenging AP free-response problems because they require you to model a situation geometrically or algebraically, then differentiate implicitly with respect to time. The key skill is setting up the correct equation before differentiating. For each problem below, clearly define all variables and explicitly state the equation you are differentiating. Our AP Calculus Derivatives Review includes additional worked examples of related rates from past AP exams.
Step 1: Draw and label a diagram. Assign variables to all changing quantities.
Step 2: Write an equation relating the variables.
Step 3: Differentiate both sides with respect to \(t\), using the Chain Rule.
Step 4: Substitute the given values and solve for the unknown rate.
A circular oil spill is expanding. Its radius is increasing at \(3\) m/s. How fast is the area increasing when the radius is \(20\) m?
Equation relating area and radius
\(A = \pi r^2\)
Differentiate with respect to \(t\)
\(\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}\)
Substitute \(r = 20\), \(\dfrac{dr}{dt} = 3\)
\(\dfrac{dA}{dt} = 2\pi(20)(3) = 120\pi\) m²/s
Use these to check your final answers. Click the button below for full step-by-step solutions.
\(f(x) = 4x^5 - 7x^3 + 2x - 9\)
Apply Power Rule term by term: \(\dfrac{d}{dx}[4x^5] = 20x^4\), \(\dfrac{d}{dx}[-7x^3] = -21x^2\), \(\dfrac{d}{dx}[2x] = 2\), \(\dfrac{d}{dx}[-9] = 0\).
Rewrite: \(6x^{-3} + 5x^{-2} - 8x\). Apply Power Rule: \(-18x^{-4} - 10x^{-3} - 8\).
\(g(x) = x^{7/3} - 4x^{5/2} + x^{-1/4}\)
\(g'(x) = \dfrac{7}{3}x^{4/3} - 4\cdot\dfrac{5}{2}x^{3/2} + \left(-\dfrac{1}{4}\right)x^{-5/4} = \dfrac{7}{3}x^{4/3} - 10x^{3/2} - \dfrac{1}{4}x^{-5/4}\)
Rewrite: \(y = 3x^{-4} - \dfrac{1}{2}x^{-1/3} + 5\)
\(y' = -12x^{-5} - \dfrac{1}{2}\cdot\left(-\dfrac{1}{3}\right)x^{-4/3} = -12x^{-5} + \dfrac{1}{6}x^{-4/3}\)
Expand: \(h(x) = (2x^3-1)(x+4) = 2x^4 + 8x^3 - x - 4\)
\(h'(x) = 8x^3 + 24x^2 - 1\)
Note: Several textbook editions distribute differently; expanding first avoids Product Rule and is faster under exam conditions.
Simplify first: \(f(x) = \dfrac{x^4 - 3x^2 + x}{x} = x^3 - 3x + 1\)
\(f'(x) = 3x^2 - 3\)
\(s(t) = t^4 - 8t^2 + 3t\)
Velocity: \(v(t) = s'(t) = 4t^3 - 16t + 3\)
Acceleration: \(a(t) = v'(t) = 12t^2 - 16\)
\(y = x^3 - 4x + 1\), so \(y' = 3x^2 - 4\).
At \(x = 2\): slope \(= 3(4) - 4 = 8\); point \(= (2, 2^3 - 8 + 1) = (2, 1)\).
Tangent line: \(y - 1 = 8(x - 2) \Rightarrow y = 8x - 15\)
Outer: \(u^6\), inner: \(u = 3x^2 - 5x + 1\), \(u' = 6x - 5\).
\(f'(x) = 6(3x^2-5x+1)^5(6x-5)\)
Outer: \(e^u\), inner: \(u = 4x^3 - x\), \(u' = 12x^2 - 1\).
\(y' = e^{4x^3-x}(12x^2 - 1)\)
Outer: \(\ln(u)\), inner: \(u = x^2 + \sin x\), \(u' = 2x + \cos x\).
\(\dfrac{d}{dx} = \dfrac{2x + \cos x}{x^2 + \sin x}\)
Write as \((5x^3 - 2x + 7)^{1/2}\). Outer: \(u^{1/2}\), inner: \(u = 5x^3 - 2x + 7\), \(u' = 15x^2 - 2\).
\(g'(x) = \dfrac{1}{2}(5x^3-2x+7)^{-1/2}(15x^2-2) = \dfrac{15x^2-2}{2\sqrt{5x^3-2x+7}}\)
Write as \([\cos(3x)]^4\). Outer: \(u^4\), middle: \(\cos(v)\), inner: \(v = 3x\).
\(\dfrac{d}{dx} = 4\cos^3(3x)\cdot(-\sin(3x))\cdot 3 = -12\cos^3(3x)\sin(3x)\)
Outer: \(\arctan(u)\), inner: \(u = e^{2x}\), \(u' = 2e^{2x}\).
\(\dfrac{d}{dx} = \dfrac{2e^{2x}}{1+(e^{2x})^2} = \dfrac{2e^{2x}}{1+e^{4x}}\)
Outer: \(\sin(u)\), inner: \(u = (x^2+1)^{-1}\), \(u' = -2x(x^2+1)^{-2}\).
\(y' = \cos\!\left(\dfrac{1}{x^2+1}\right)\cdot\dfrac{-2x}{(x^2+1)^2}\)
Outer: \(e^u\), middle: \(\sin(v)\), inner: \(v = x^2\).
\(\dfrac{d}{dx}[e^{\sin(x^2)}] = e^{\sin(x^2)}\cdot\cos(x^2)\cdot 2x\)
Use log property: \(\ln(\cos^3 x) = 3\ln(\cos x)\).
\(\dfrac{d}{dx}[3\ln(\cos x)] = 3\cdot\dfrac{-\sin x}{\cos x} = -3\tan x\)
Let \(u = \dfrac{x^2+1}{x-3}\). Outer: \(u^5\), \(u' = \dfrac{2x(x-3)-(x^2+1)}{(x-3)^2} = \dfrac{x^2-6x-1}{(x-3)^2}\).
\(y' = 5\!\left(\dfrac{x^2+1}{x-3}\right)^4\cdot\dfrac{x^2-6x-1}{(x-3)^2}\)
\(f = x^4,\ f' = 4x^3\); \(g = \sin x,\ g' = \cos x\).
\(y' = 4x^3\sin x + x^4\cos x\)
\(f = 3x-1,\ f' = 3\); \(g = \ln x,\ g' = 1/x\).
\(f'(x) = 3\ln x + (3x-1)\cdot\dfrac{1}{x} = 3\ln x + 3 - \dfrac{1}{x}\)
\(f = x^2,\ f' = 2x\); \(g = e^{-x},\ g' = -e^{-x}\).
\(\dfrac{d}{dx}[x^2 e^{-x}] = 2xe^{-x} + x^2(-e^{-x}) = e^{-x}(2x - x^2) = xe^{-x}(2-x)\)
\(f = \sqrt{x} = x^{1/2},\ f' = \dfrac{1}{2\sqrt{x}}\); \(g = \cos(2x),\ g' = -2\sin(2x)\).
\(g'(x) = \dfrac{\cos(2x)}{2\sqrt{x}} - 2\sqrt{x}\sin(2x)\)
\(f = (x^2+3)^4,\ f' = 8x(x^2+3)^3\); \(g = e^{3x},\ g' = 3e^{3x}\).
\(h'(x) = 8x(x^2+3)^3 e^{3x} + (x^2+3)^4\cdot 3e^{3x} = e^{3x}(x^2+3)^3[8x + 3(x^2+3)]\)
\(f = x,\ f' = 1\); \(g = \ln(x^2+1),\ g' = \dfrac{2x}{x^2+1}\).
\(y' = \ln(x^2+1) + x\cdot\dfrac{2x}{x^2+1} = \ln(x^2+1) + \dfrac{2x^2}{x^2+1}\)
\(f = \sin(3x),\ f' = 3\cos(3x)\); \(g = \cos(5x),\ g' = -5\sin(5x)\).
\(f'(x) = 3\cos(3x)\cos(5x) - 5\sin(3x)\sin(5x)\)
\(f = e^{x^2},\ f' = 2xe^{x^2}\); \(g = \arctan x,\ g' = \dfrac{1}{1+x^2}\).
\(\dfrac{d}{dx} = 2xe^{x^2}\arctan x + \dfrac{e^{x^2}}{1+x^2}\)
\(f = x^3,\ f' = 3x^2\); \(g = x^2+1,\ g' = 2x\).
\(f'(x) = \dfrac{3x^2(x^2+1) - x^3(2x)}{(x^2+1)^2} = \dfrac{3x^4+3x^2-2x^4}{(x^2+1)^2} = \dfrac{x^4+3x^2}{(x^2+1)^2}\)
\(f = \sin x,\ f' = \cos x\); \(g = x,\ g' = 1\).
\(y' = \dfrac{x\cos x - \sin x}{x^2}\)
Simplify: \(\dfrac{x^2-4}{x+2} = \dfrac{(x-2)(x+2)}{x+2} = x - 2\) (for \(x \neq -2\)).
\(\dfrac{d}{dx}[x-2] = 1\)
\(f = e^x,\ f' = e^x\); \(g = \ln x,\ g' = 1/x\).
\(g'(x) = \dfrac{e^x \ln x - e^x\cdot(1/x)}{(\ln x)^2} = \dfrac{e^x(\ln x - 1/x)}{(\ln x)^2}\)
\(f = \cos(3x),\ f' = -3\sin(3x)\); \(g = x^2+5,\ g' = 2x\).
\(y' = \dfrac{-3\sin(3x)(x^2+5) - \cos(3x)(2x)}{(x^2+5)^2}\)
\(f = \arctan x,\ f' = \dfrac{1}{1+x^2}\); \(g = x^3,\ g' = 3x^2\).
\(\dfrac{d}{dx} = \dfrac{\dfrac{x^3}{1+x^2} - 3x^2\arctan x}{x^6} = \dfrac{x^3 - (1+x^2)\cdot 3x^2\arctan x}{x^6(1+x^2)}\)
\(f = (2x+1)^3,\ f' = 6(2x+1)^2\); \(g = e^x,\ g' = e^x\).
\(f'(x) = \dfrac{6(2x+1)^2 e^x - (2x+1)^3 e^x}{e^{2x}} = \dfrac{(2x+1)^2[6-(2x+1)]}{e^x} = \dfrac{(2x+1)^2(5-2x)}{e^x}\)
\(f = \ln(x^2) = 2\ln x,\ f' = 2/x\); \(g = x^2+x,\ g' = 2x+1\).
\(\dfrac{d}{dx} = \dfrac{(2/x)(x^2+x) - \ln(x^2)(2x+1)}{(x^2+x)^2} = \dfrac{2(x+1)/x - 2\ln(x)(2x+1)}{x^2(x+1)^2}\)
Differentiate \(x^2 + y^2 = 25\): \(2x + 2y\dfrac{dy}{dx} = 0\).
Solve: \(\dfrac{dy}{dx} = -\dfrac{x}{y}\)
Differentiate \(x^3 + 3xy - y^3 = 7\):
\(3x^2 + 3y + 3x\dfrac{dy}{dx} - 3y^2\dfrac{dy}{dx} = 0\)
\(\dfrac{dy}{dx}(3x - 3y^2) = -3x^2 - 3y\)
Differentiate \(e^{xy} = x + y\) using Chain and Product Rule on left side:
\(e^{xy}\!\left(y + x\dfrac{dy}{dx}\right) = 1 + \dfrac{dy}{dx}\)
\(\dfrac{dy}{dx}(xe^{xy} - 1) = 1 - ye^{xy}\)
Differentiate \(\sin(x+y) = x^2 y\):
\(\cos(x+y)\!\left(1+\dfrac{dy}{dx}\right) = 2xy + x^2\dfrac{dy}{dx}\)
\(\dfrac{dy}{dx}[\cos(x+y) - x^2] = 2xy - \cos(x+y)\)
Differentiate \(x^2 + 2xy + y^2 = 9\): \(2x + 2y + 2x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0\).
\(\dfrac{dy}{dx} = -\dfrac{2x+2y}{2x+2y} = -1\)
(Note: \((x+y)^2 = 9\), so \(x+y = \pm3\); the slope is always \(-1\).)
At \((1,2)\): tangent \(y - 2 = -1(x-1) \Rightarrow y = -x + 3\).
Differentiate \(\ln(xy) + y^2 = 3x\): write \(\ln x + \ln y + y^2 = 3x\).
\(\dfrac{1}{x} + \dfrac{1}{y}\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 3\)
\(\dfrac{dy}{dx}\!\left(\dfrac{1}{y} + 2y\right) = 3 - \dfrac{1}{x}\)
From \(x^2 + y^2 = r^2\): first derivative: \(\dfrac{dy}{dx} = -\dfrac{x}{y}\).
Second derivative: \(\dfrac{d^2y}{dx^2} = -\dfrac{y - x\dfrac{dy}{dx}}{y^2} = -\dfrac{y - x(-x/y)}{y^2} = -\dfrac{y^2 + x^2}{y^3} = -\dfrac{r^2}{y^3}\)
Differentiate \(x^2\sin y + y\cos x = 1\):
\(2x\sin y + x^2\cos y\dfrac{dy}{dx} + \dfrac{dy}{dx}\cos x - y\sin x = 0\)
\(\dfrac{dy}{dx}(x^2\cos y + \cos x) = y\sin x - 2x\sin y\)
Let \(x\) = distance from wall, \(y\) = height on wall. \(x^2 + y^2 = 100\).
Differentiate: \(2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0\).
When \(x = 6\): \(y = \sqrt{100-36} = 8\). Given \(\dfrac{dx}{dt} = 2\):
\(2(6)(2) + 2(8)\dfrac{dy}{dt} = 0 \Rightarrow \dfrac{dy}{dt} = -\dfrac{12}{8} = -\dfrac{3}{2}\) ft/s
\(V = \dfrac{4}{3}\pi r^3\). Differentiate: \(\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}\).
Given \(\dfrac{dV}{dt} = 24\pi\), \(r = 3\): \(24\pi = 4\pi(9)\dfrac{dr}{dt} \Rightarrow \dfrac{dr}{dt} = \dfrac{24\pi}{36\pi} = \dfrac{2}{3}\) cm/s.
Similar triangles: \(\dfrac{r}{h} = \dfrac{4}{12} = \dfrac{1}{3}\), so \(r = h/3\).
\(V = \dfrac{1}{3}\pi\!\left(\dfrac{h}{3}\right)^2 h = \dfrac{\pi h^3}{27}\).
\(\dfrac{dV}{dt} = \dfrac{\pi h^2}{9}\dfrac{dh}{dt}\). Given \(\dfrac{dV}{dt} = -2\), \(h = 6\):
\(-2 = \dfrac{\pi(36)}{9}\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = -\dfrac{2}{4\pi} = -\dfrac{1}{2\pi}\approx -\dfrac{8}{3\pi}\) m/min
Let \(x\) = distance east, \(y\) = distance north, \(D\) = distance between. \(D^2 = x^2 + y^2\).
After 0.5 hr: \(x = 30\) mi, \(y = 40\) mi, \(D = 50\) mi.
\(2D\dfrac{dD}{dt} = 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} \Rightarrow 50\dfrac{dD}{dt} = 30(60)+40(80) = 5000\).
Let \(a = 8,\ b = 6,\ c = \sqrt{100} = 10\). \(c^2 = a^2 + b^2\).
\(2c\dfrac{dc}{dt} = 2a\dfrac{da}{dt} + 2b\dfrac{db}{dt} = 2(8)(3)+2(6)(-2) = 48 - 24 = 24\).
\(\dfrac{dc}{dt} = \dfrac{24}{20} = \dfrac{6}{5}\) cm/s.
\(y = x^3\). Differentiate with respect to \(t\): \(\dfrac{dy}{dt} = 3x^2\dfrac{dx}{dt}\).
At \(x = 2\), \(\dfrac{dx}{dt} = 4\): \(\dfrac{dy}{dt} = 3(4)(4) = 48\) units/s.
Given \(h = 3r\), so \(r = h/3\). \(V = \dfrac{1}{3}\pi(h/3)^2 h = \dfrac{\pi h^3}{27}\).
\(\dfrac{dV}{dt} = \dfrac{\pi h^2}{9}\dfrac{dh}{dt}\). At \(h=6\), \(\dfrac{dV}{dt}=12\):
\(12 = \dfrac{36\pi}{9}\dfrac{dh}{dt} = 4\pi\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = \dfrac{3}{\pi}\approx\dfrac{4}{3\pi}\) ft/min. (Exact: \(3/\pi\) ft/min.)
Let \(x\) = person's distance from base, \(s\) = shadow length. By similar triangles: \(\dfrac{15}{x+s} = \dfrac{6}{s}\).
Solving: \(15s = 6x + 6s \Rightarrow 9s = 6x \Rightarrow s = \dfrac{2x}{3}\).
\(\dfrac{ds}{dt} = \dfrac{2}{3}\dfrac{dx}{dt} = \dfrac{2}{3}(5) = \dfrac{10}{3}\) ft/s (shadow length growing).
Tip of shadow position \(= x + s = x + \dfrac{2x}{3} = \dfrac{5x}{3}\), so \(\dfrac{d}{dt}(x+s) = \dfrac{5}{3}\cdot 5 = \dfrac{25}{3}\) ft/s.
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