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📝 The Chain Rule in calculus (14 MCQs)

📖 From Calculus • 14. Partial Derivatives Calculus • 14 questions available

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What is The Chain Rule in calculus?

Definition:
Multivariable chain rule computes $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$ for composite functions $z=f(x(t),y(t))$ .

Example:
If $z=x^2+y^2$ , $x=t^2$ , $y=\sin t$ , then $\frac{dz}{dt} = 2x(2t) + 2y(\cos t) = 4t^3 + 2\sin t \cos t$ .

Reason:
This rule propagates rates through dependent variable chains, essential for related rates, parametric differentiation, and transforming coordinates in physics.

Easy

Medium

Hard

📝 All The Chain Rule in calculus MCQs

Q1. A weather model defines temperature $T$ as a function of pressure $p$ and altitude $h$ . Both $p$ and $h$ depend on time $t$ along a balloon's path. If $\frac{\partial T}{\partial p} = -2$ , $\frac{\partial T}{\partial h} = 5$ , $\frac{dp}{dt} = 3$ , and $\frac{dh}{dt} = -1$ , what is the physical interpretation of $\frac{dT}{dt} = -11$ ?

A.Temperature decreases at 11 units per second due to combined effects of rising altitude and increasing pressure.

B.Temperature increases at 11 units per second because the partial derivative with respect to altitude dominates.

C.The rate of change is -11, but this only accounts for pressure changes; altitude effects must be added separately.

D.The total derivative equals -11, representing the instantaneous rate of temperature change experienced by the balloon as both variables evolve simultaneously. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: This application question requires interpreting the multivariable chain rule in a real-world context. Students must understand that $\frac{dT}{dt}$ combines all pathways of change. The correct answer emphasizes simultaneous evolution of variables, distinguishing total from partial derivatives and avoiding misattribution of causality to individual terms.

Q2. Given $z = f(x,y)$ where $x = r\cos\theta$ and $y = r\sin\theta$ , a student computes $\frac{\partial z}{\partial r} = \frac{\partial f}{\partial x}\cos\theta + \frac{\partial f}{\partial y}\sin\theta$ . They then claim $\frac{\partial^2 z}{\partial r^2} = \frac{\partial^2 f}{\partial x^2}\cos^2\theta + \frac{\partial^2 f}{\partial y^2}\sin^2\theta$ . What is the fundamental error?

A.They forgot to apply the product rule when differentiating the first-order expression with respect to

r

B.They incorrectly assumed

\theta

is constant when taking the second derivative.

C.They omitted cross-partial terms like

2\frac{\partial^2 f}{\partial x \partial y}\cos\theta\sin\theta

D.Both A and C are correct; the error involves missing product rule application and neglecting mixed partial derivatives. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: This error analysis question targets a common misconception in higher-order chain rule applications. Students often forget that $\frac{\partial f}{\partial x}$ itself depends on $r$ through $x$ and $y$ , requiring another chain rule application. The correct identification of multiple simultaneous errors demonstrates deep conceptual understanding beyond mechanical computation.

Q3. Consider a contour plot of $f(x,y)$ with level curves spaced evenly. A particle moves along a path $\vec{r}(t)$ that crosses these contours. At point $P$ , the path is tangent to a level curve. Without computing derivatives, what can you conclude about $\frac{df}{dt}$ at $P$ ?

A.It is zero because motion along a level curve implies no change in function value. ✅

B.It is maximal because tangency indicates steepest ascent direction.

C.It cannot be determined without knowing the parameterization speed.

D.It equals the magnitude of the gradient vector at

P

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This graph-based question tests conceptual understanding of the chain rule’s geometric meaning. When velocity is tangent to a level curve, the directional derivative vanishes since $\nabla f \cdot \vec{v} = 0$ . Students must connect visual contour information to the analytical expression $\frac{df}{dt} = \nabla f \cdot \vec{r}'(t)$ without explicit formulas.

Q4. Let $w = g(u,v)$ where $u = x^2 - y^2$ and $v = 2xy$ . If $\frac{\partial w}{\partial x} = 4x$ and $\frac{\partial w}{\partial y} = -4y$ at point $(1,1)$ , and given $\frac{\partial g}{\partial u}(0,2) = 2$ , find $\frac{\partial g}{\partial v}(0,2)$ .

A.0

B.1 ✅

C.-1

D.2

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This multi-step reasoning problem reverses the typical chain rule application. Students must set up the system $\frac{\partial w}{\partial x} = g_u u_x + g_v v_x$ and $\frac{\partial w}{\partial y} = g_u u_y + g_v v_y$ , substitute known values at $(1,1)$ where $u=0, v=2$ , and solve for the unknown partial. It tests algebraic manipulation within the chain rule framework.

Q5. In thermodynamics, entropy $S$ depends on temperature $T$ and volume $V$ . However, in an adiabatic process, $V$ is constrained by $V = kT^{-\gamma}$ . Which expression correctly represents $\left(\frac{dS}{dT}\right)_{\text{adiabatic}}$ ?

\frac{\partial S}{\partial T}

evaluated at constant

V

\frac{\partial S}{\partial T} + \frac{\partial S}{\partial V}\frac{dV}{dT}

with

\frac{dV}{dT} = -\gamma k T^{-\gamma-1}

✅

\frac{\partial S}{\partial T} \frac{dT}{dV} + \frac{\partial S}{\partial V}

D.Zero, since entropy is constant in adiabatic processes

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This scenario-based question integrates physics constraints with the chain rule. Students must recognize that even though $S$ is naturally expressed in $(T,V)$ , the adiabatic constraint makes $V$ dependent on $T$ , requiring the full chain rule. Option D is a tempting misconception confusing reversible adiabatic processes with general cases.

Q6. Suppose $f(x,y) = x^2y$ and a student attempts to compute $\frac{d}{dt}f(t,t^2)$ by first substituting to get $t^4$ , yielding $4t^3$ . Another uses the chain rule: $f_x \cdot 1 + f_y \cdot 2t = 2t \cdot t^2 + t^2 \cdot 2t = 4t^3$ . Why do both methods agree?

A.They always agree; substitution and chain rule are equivalent for smooth functions. ✅

B.Agreement occurs only because

f

is polynomial; transcendental functions would differ.

C.The chain result is coincidental; substitution is fundamentally more reliable.

D.They agree here but diverge when intermediate variables are not independent.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This conceptual understanding question addresses method equivalence. While direct recall suggests simple verification, HOTS requires recognizing that both approaches are mathematically identical for differentiable compositions. Distractors exploit doubts about generality or independence assumptions. The explanation reinforces that the chain rule formalizes what substitution accomplishes implicitly.

Q7. A manufacturing cost $C$ depends on labor $L$ and materials $M$ . Due to supply chain issues, $M = 100 - 2L$ . At current operating point, $\frac{\partial C}{\partial L} = 50$ and $\frac{\partial C}{\partial M} = 30$ . Management wants to reduce costs. Should they increase or decrease labor?

A.Increase labor, since

\frac{dC}{dL} = 50 + 30(-2) = -10 < 0

✅

B.Decrease labor, since partial derivative

\frac{\partial C}{\partial L} = 50 > 0

C.Cannot decide without knowing second derivatives.

D.Maintain current levels; the constraint makes optimization impossible.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This applied modeling question requires synthesizing economic reasoning with constrained chain rule. Students must compute total derivative $\frac{dC}{dL} = C_L + C_M \frac{dM}{dL}$ and interpret its sign for decision-making. Option B represents the critical misconception of ignoring constraints, while C and D test overcomplication tendencies.

Q8. Given $z = f(x,y)$ with $x = s+t$ , $y = s-t$ , and knowing $\frac{\partial z}{\partial s} = 6$ , $\frac{\partial z}{\partial t} = 2$ at a point, determine $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at the corresponding $(x,y)$ .

\frac{\partial z}{\partial x} = 4

\frac{\partial z}{\partial y} = 2

✅

\frac{\partial z}{\partial x} = 8

\frac{\partial z}{\partial y} = 4

\frac{\partial z}{\partial x} = 2

\frac{\partial z}{\partial y} = 4

D.Insufficient information without explicit form of

f

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This mixed concepts problem inverts the standard chain rule direction. Students must solve the linear system $z_s = z_x x_s + z_y y_s$ and $z_t = z_x x_t + z_y y_t$ , recognizing $x_s=y_t=1$ , $x_t=y_s=-1$ . It tests understanding that partial derivatives transform via Jacobian inversion, not just forward propagation.

Q9. A student claims that if $\frac{\partial f}{\partial x} = 0$ everywhere, then $\frac{df}{dt} = 0$ for any parametric path $x(t), y(t)$ . Under what condition is this claim valid?

A.Always true; zero partial implies constant function.

B.Only if

\frac{\partial f}{\partial y} = 0

as well.

C.Only if the path has

\frac{dy}{dt} = 0

. ✅

D.Never true; partial derivatives don't constrain total derivatives.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: This error analysis question probes misunderstanding of partial versus total derivatives. The claim fails unless the path doesn't vary in directions where $f$ changes. Since $f_x=0$ means $f$ depends only on $y$ , $df/dt = f_y \cdot dy/dt$ , which vanishes iff $dy/dt=0$ or $f_y=0$ . Option C captures the necessary path constraint.

Q10. Consider $F(t) = \int_{a(t)}^{b(t)} g(x,t) dx$ . Using Leibniz rule and chain rule concepts, which term accounts for the movement of integration limits?

\int_{a(t)}^{b(t)} \frac{\partial g}{\partial t} dx

g(b(t),t) \cdot b&#039;(t) - g(a(t),t) \cdot a&#039;(t)

✅

\frac{\partial}{\partial t} \int_{a(t)}^{b(t)} g(x,t) dx

g(b(t),t) + g(a(t),t)

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This challenging Olympiad-style question extends chain rule thinking to integral operators. Students must recognize Leibniz rule as a generalized chain rule where boundary terms arise from limit dependence. The explanation connects differentiation under integral sign to multivariable chain rule intuition, emphasizing that moving boundaries contribute flux-like terms absent in static integrals.

Q11. On a topographic map, elevation $z = f(x,y)$ . A hiker walks northeast at constant speed. At location $P$ , the gradient points north. Compare $\frac{dz}{dt}$ for northeast versus northward paths at same speed.

A.Northeast path has smaller

\frac{dz}{dt}

due to angular deviation from gradient. ✅

B.Both paths yield identical rates since speed is constant.

C.Northward path has smaller rate because it aligns with contour lines.

D.Cannot compare without knowing gradient magnitude.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This graph-based comparison question links directional derivatives to chain rule. Since $dz/dt = \nabla f \cdot \vec{v} = |\nabla f||\vec{v}|\cos\theta$ , maximum occurs when $\vec{v}$ parallels $\nabla f$ . Northeast deviates 45° from north-pointing gradient, reducing cosine factor. Tests spatial reasoning beyond formula memorization.

Q12. Let $u = f(r,\theta)$ in polar coordinates. A student writes $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}$ , omitting the $\theta$ -term. What specific scenario makes this omission accidentally correct?

A.When

f

is radially symmetric (

\partial u/\partial \theta = 0

) ✅

B.When evaluating at the origin where

\theta

is undefined

C.When

x = 0

\partial \theta / \partial x = 0

D.Never correct; the term is always required

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This error analysis identifies conditional validity of incomplete chain rule. Radial symmetry nullifies the missing term regardless of coordinate singularity. Options B and C reference technicalities but don’t guarantee correctness universally. Tests nuanced understanding that mathematical errors can be contextually masked by special function properties.

Q13. In neural networks, loss $L$ depends on weights $w$ through nested activations. During backpropagation, why is the chain rule applied recursively rather than computing $dL/dw$ directly via finite differences?

A.Finite differences fail for non-differentiable activation functions.

B.Recursive chain rule exploits computational graph structure, reusing intermediate gradients efficiently. ✅

C.Direct computation is theoretically impossible for deep networks.

D.Chain rule provides exact derivatives while finite differences introduce bias.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This mixed concepts question bridges abstract calculus with modern applications. Students must recognize algorithmic efficiency as the primary motivation, not theoretical limitations. Distractors include plausible but secondary concerns like differentiability or bias. Explanation emphasizes reuse of computed partials across layers, demonstrating practical HOTS beyond pure mathematics.

Q14. Suppose $h(t) = f(g(t), k(t))$ where $g(t) = t^2$ and $k(t) = \sin t$ . At $t=0$ , $f_x(0,0)=3$ , $f_y(0,0)=0$ , but $f_{xy}(0,0)=5$ . Does $f_{xy}$ affect $h'(0)$ ?

A.Yes, because mixed partials contribute to first-order chain rule.

B.No,

h&#039;(0)

depends only on first partials of

f

and first derivatives of

g,k

. ✅

C.Yes, since

k&#039;(0)=1 \neq 0

activates the cross term.

D.Only if

g&#039;(0) \neq 0

, which it isn’t at

t=0

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This challenging question separates first-order chain rule from higher-order effects. Despite nonzero mixed partial, $h'(t) = f_x g' + f_y k'$ , and at $t=0$ , $g'(0)=0$ , $k'(0)=1$ , so $h'(0)=f_y(0,0)\cdot1=0$ . Mixed partials influence $h''(t)$ but not $h'(t)$ . Tests precise scope awareness of chain rule orders.

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