Applications of Derivatives Practice Questions with Answers and a Guided Lesson
Use the question set below to practice applications of derivatives — the most practical real-world skills in Calculus. You will work with the derivative as an instantaneous rate of change and as the slope of a tangent line, compute velocity and acceleration from position functions, solve classic related rates problems using implicit differentiation (ladders, circles, spheres, cylinders), and master optimization problems (maximize revenue, minimize cost, maximize area with fixed perimeter). You will also use critical points and derivative tests (increasing/decreasing, first derivative test), and apply linear approximation (tangent line approximation / differentials) to estimate values quickly. If you want a refresher, click Start lesson to open a step-by-step guide with worked examples and quick checks.
Answer the question set and review your mistakes at the end.
How this applications of derivatives practice works
1. Take the practice set: answer the applications of derivatives questions below.
2. Open the lesson (optional): review related rates, optimization, motion (velocity/acceleration), derivative tests, and linear approximation with clear examples.
3. Retry: return to the question set and apply the derivative tools immediately.
What you will learn in the applications of derivatives lesson
Rates of change & motion
Derivative meaning: instantaneous rate of change and tangent slope
Purpose: Build strong skills in applications of derivatives so you can use derivatives as rates of change and slopes, solve motion problems (velocity and acceleration), set up and solve related rates using implicit differentiation, handle optimization (maximize revenue, minimize cost, maximize area with constraints), analyze functions with critical points and increasing/decreasing tests, and use linear approximation (tangent line approximation) to estimate values quickly.
Success criteria
Interpret the derivative as instantaneous rate of change and slope of the tangent line.
Compute velocity and acceleration from a position function \(s(t)\).
Use the chain rule for rates: \(\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}\).
Set up related rates with a geometry equation and differentiate with respect to \(t\).
Compute rates for circles, spheres, cylinders, and ladders (Pythagorean theorem).
Write an objective function and a constraint for optimization problems.
Find critical points where \(f'(x)=0\) or \(f'(x)\) does not exist.
Determine where a function is increasing or decreasing from the sign of \(f'(x)\).
Use linear approximation: \(f(x)\approx f(a)+f'(a)(x-a)\).
Apply the Mean Value Theorem idea: average rate vs instantaneous rate.
Key vocabulary
Derivative: \(f'(x)\), the slope of the tangent line and instantaneous rate of change.
Related rates: using a relationship among variables to connect rates like \(dx/dt\), \(dy/dt\), \(dr/dt\).
Optimization: maximizing or minimizing a quantity subject to constraints.
Critical point: a point where \(f'(x)=0\) or \(f'(x)\) is undefined.
Linear approximation: using the tangent line to estimate function values.
Mean Value Theorem: there exists \(c\) where \(f'(c)\) equals the average slope on \([a,b]\).
Quick pre-check
Pre-check 1: An object’s position is \(s(t)=5t^2\). What is its velocity at \(t=3\)?
Hint: Velocity is \(v(t)=s'(t)\). Differentiate \(5t^2\) to get \(10t\).
Pre-check 2: What is the slope of the tangent to \(y=\frac{1}{x}\) at \(x=2\)?
Hint: \(y=x^{-1}\Rightarrow y'=-x^{-2}=-\dfrac{1}{x^2}\). Evaluate at \(x=2\).
Rates of Change
Derivatives as rates: velocity, acceleration, and chain rule with time
Learning goal: Use derivatives to convert position into velocity and acceleration, and connect rates using the chain rule.
Key idea
In applications, the derivative often means an instantaneous rate of change. If position is \(s(t)\), then: \[ v(t)=s'(t)\quad \text{(velocity)}, \qquad a(t)=v'(t)=s''(t)\quad \text{(acceleration)}. \] When variables depend on time, the chain rule connects rates: \[ \frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}. \]
Worked example
Example: What is the instantaneous velocity of \(s(t)=3t^2+2t\) at \(t=1\)?
Differentiate: \[ v(t)=s'(t)=6t+2. \] Evaluate at \(t=1\): \[ v(1)=6(1)+2=8. \]
Try it
Try it 1: What is the instantaneous velocity of \(s(t) = 3t^2 + 2t\) at \(t = 1\)?
Hint: Velocity is \(s'(t)\). Differentiate, then plug in \(t=1\).
Try it 2: If \(y=\tan(x)\) and \(\frac{dx}{dt}=1\) at \(x=0\), what is \(\frac{dy}{dt}\)?
Hint: \(\dfrac{dy}{dt}=\sec^2(x)\dfrac{dx}{dt}\). At \(x=0\), \(\sec^2(0)=1\).
Summary
Velocity is \(s'(t)\); acceleration is \(s''(t)\).
For time-dependent variables, use \(\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}\).
Related Rates
Related rates: set up an equation, then differentiate with respect to time
Learning goal: Translate a word problem into an equation and connect rates like \(dx/dt\) and \(dy/dt\).
Key idea
Related rates problems follow a consistent process:
Step 1: Draw a picture and write a relationship among variables (often geometry).
Step 2: Differentiate both sides with respect to time \(t\).
Step 3: Substitute the values at the instant and solve for the unknown rate.
Worked example
Example: A ladder \(10\) ft long leans against a wall. If the bottom slides away at \(1\) ft/s, how fast is the top sliding down when the bottom is \(6\) ft from the wall?
Let \(x\) be the bottom distance from the wall and \(y\) be the height of the top. Then: \[ x^2+y^2=10^2=100. \] Differentiate with respect to \(t\): \[ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \quad \Rightarrow \quad \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}. \] When \(x=6\), \(y=\sqrt{100-36}=8\). With \(\frac{dx}{dt}=1\), \[ \frac{dy}{dt}=-\frac{6}{8}(1)=-\frac{3}{4}. \] So the top is sliding down at \(\frac{3}{4}\) ft/s.
Try it
Try it 1: If a circle’s radius grows at \(2\) units/sec, how fast is its area increasing when \(r=3\)?
Hint: \(A=\pi r^2\Rightarrow \dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}\). Plug in \(r=3\), \(dr/dt=2\).
Try it 2: If the volume of a sphere is increasing at \(100\) units\(^3\)/sec, how fast is its radius increasing when \(r=5\)?
Hint: \(V=\dfrac{4}{3}\pi r^3\Rightarrow \dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}\). Solve for \(dr/dt\) at \(r=5\).
Summary
Write a relationship first (geometry), then differentiate with respect to time.
Plug in the instant values last and solve for the unknown rate.
Optimization
Optimization: maximize or minimize with derivatives
Learning goal: Turn a word problem into a one-variable function, then use derivatives to find maxima/minima.
Key idea
Optimization problems follow a reliable checklist:
Define variables and write the quantity you want to optimize (objective function).
Use a constraint to rewrite the objective using one variable.
Differentiate and solve \(A'(x)=0\) (or \(R'(p)=0\), etc.) to find critical points.
Confirm max/min (use a test or the problem context).
Worked example
Example: Which price \(p\) maximizes revenue given \(R(p)=p(100-p)\)?
Expand: \[ R(p)=100p-p^2. \] Differentiate and set equal to zero: \[ R'(p)=100-2p=0 \Rightarrow p=50. \] Because \(R(p)\) is a downward-opening parabola, \(p=50\) gives the maximum revenue.
Try it
Try it 1: A rectangle has perimeter \(40\) units. Which dimensions maximize its area?
Hint: For a fixed perimeter, the rectangle with maximum area is a square.
Try it 2: For a rectangle with fixed perimeter \(P\), which shape maximizes its area?
Hint: Symmetry wins: equal sides maximize area for a fixed perimeter.
Summary
Write the objective, use the constraint, then differentiate and solve for critical points.
Use context or a test to confirm you found a maximum or minimum.
Derivative Tests
Critical points, increasing/decreasing, and the first derivative test
Learning goal: Use \(f'(x)\) to locate critical points and decide where a function increases or decreases.
Key idea
Derivative tests turn graphs into algebra:
Critical points: solve \(f'(x)=0\) (and include where \(f'(x)\) is undefined).
Increasing/decreasing: if \(f'(x)>0\), \(f\) increases; if \(f'(x)<0\), \(f\) decreases.
First derivative test: if \(f'\) changes from \(+\) to \(-\), you have a local maximum; from \(-\) to \(+\), a local minimum.
Worked example
Example: Find the critical points of \(f(x)=x^3-3x^2+4\).
Differentiate: \[ f'(x)=3x^2-6x=3x(x-2). \] Set \(f'(x)=0\): \[ 3x(x-2)=0 \Rightarrow x=0 \text{ or } x=2. \] So the critical points occur at \(x=0\) and \(x=2\).
Try it
Try it 1: Find the critical points of \(f(x)=x^3-3x^2+4\).
Hint: Differentiate and solve \(f'(x)=0\).
Try it 2: Find where \(f(x)= x^5-5x^4\) is decreasing.
Hint: Compute \(f'(x)\), factor it, then use a sign chart for \(f'(x)\).
Summary
Critical points come from \(f'(x)=0\) (and where \(f'\) is undefined).
Sign of \(f'(x)\) tells where \(f\) increases/decreases; sign changes locate local extrema.
Linear Approximation
Linear approximation: the tangent line as a fast estimator
Learning goal: Use the tangent line \(L(x)=f(a)+f'(a)(x-a)\) to estimate values near \(a\).
Key idea
Near a point \(x=a\), a differentiable function behaves almost like its tangent line: \[ f(x)\approx f(a)+f'(a)(x-a). \] This is powerful for quick mental estimates and for understanding error sensitivity.
Worked example
Example: Use linear approximation to estimate \(\sqrt{9.1}\).
Let \(f(x)=\sqrt{x}\) and choose \(a=9\) since \(\sqrt{9}=3\). Then \(f(9)=3\) and \(f'(x)=\dfrac{1}{2\sqrt{x}}\), so \(f'(9)=\dfrac{1}{6}\). With \(x=9.1\), \(x-a=0.1\): \[ \sqrt{9.1}\approx 3+\frac{1}{6}(0.1)=3+\frac{0.1}{6}\approx 3.0167. \]
Try it
Try it 1: Using linear approximation at \(a=9\), estimate \(\sqrt{9.1}\).
Hint: Use \(f(x)=\sqrt{x}\), \(f(9)=3\), \(f'(9)=1/6\), and \(\Delta x=0.1\).
Try it 2: Using linear approximation at \(x=1\), estimate \(\ln(1.02)\).
Hint: \(f(x)=\ln x\), \(f(1)=0\), \(f'(1)=1\), and \(\Delta x=0.02\).
Summary
Linear approximation uses the tangent line: \(f(x)\approx f(a)+f'(a)(x-a)\).
Pick \(a\) where \(f(a)\) and \(f'(a)\) are easy to compute.
Mean Value Theorem
Average vs instantaneous change: the Mean Value Theorem idea
Learning goal: Connect average rate of change to an instantaneous derivative value.
Key idea
If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then the Mean Value Theorem says there exists at least one number \(c\in(a,b)\) such that \[ f'(c)=\frac{f(b)-f(a)}{b-a}. \] This is the formal bridge from an average slope to a tangent slope.
Worked example
Example: For \(f(x)=x^2\) on \([1,3]\), find a \(c\) that satisfies the Mean Value Theorem.
Average slope: \[ \frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4. \] Derivative: \[ f'(x)=2x. \] Set \(2c=4\Rightarrow c=2\), which lies in \((1,3)\).
Try it
Try it 1: For \(f(x)=x^2\) on \([0,2]\), a value \(c\in(0,2)\) that satisfies MVT is:
Hint: Compute the average slope \(\dfrac{f(2)-f(0)}{2-0}\), then solve \(f'(c)=\) that slope.
Try it 2: If \(y=\ln(x)\) and \(\frac{dx}{dt}=2\) at \(x=1\), what is \(\frac{dy}{dt}\)?
Hint: \(\dfrac{dy}{dt}=\dfrac{1}{x}\dfrac{dx}{dt}\). Evaluate at \(x=1\).
Summary
MVT guarantees at least one point where the tangent slope equals the average slope on an interval.
Chain rule rates often appear alongside applications: \(\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}\).
Applications & Big Picture
Why applications of derivatives matter
Learning goal: Connect derivative tools to real modeling — and finish with a final check.
Where applications of derivatives show up
Physics: velocity, acceleration, and rates of change in time.
Derivative tests: critical points and sign of \(f'(x)\) reveal where a function increases/decreases.
Linear approximation: \(f(x)\approx f(a)+f'(a)(x-a)\) near \(a\).
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the application skill you need (rates, related rates, optimization, tests, or approximation).
