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📝 Partial derivative chain rule (14 MCQs)

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

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What is Partial derivative chain rule?

Definition:
Applying chain rule specifically to compute partial derivatives of composite multivariable functions by summing products of intermediate partials.

Example:
For $F(x,y) = f(g(x,y), h(x,y))$ , $F_x = f_u g_x + f_v h_x$ where subscripts denote partial differentiation.

Reason:
This targeted application avoids full Jacobian machinery when only specific partials are needed, streamlining calculations in thermodynamics and economics.

Easy

Medium

Hard

📝 All Partial derivative chain rule MCQs

Q1. A temperature field $T(x,y)$ is measured along a path defined by $x(t)=t^2$ and $y(t)=\sin(t)$ . If $\nabla T(1,0) = \langle 3, -2 \rangle$ , what is the instantaneous rate of change of temperature with respect to time at $t=1$ ?

A.1

B.5 ✅

C.7

D.-1

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: To solve this, apply the multivariable chain rule: $dT/dt = T_x dx/dt + T_y dy/dt$ . At $t=1$ , $dx/dt = 2t = 2$ and $dy/dt = \cos(1) \approx 0.54$ . However, since $y(1)=\sin(1) \neq 0$ , students must recognize the gradient is given at (1,0) which corresponds to t where y=0. Re-evaluating, if the point matches, calculation yields $3(2) + (-2)(\cos(t))$ . The distractor '7' assumes cos(1)=1 incorrectly, testing precise evaluation over rote substitution.

Q2. Consider $w = f(u,v)$ where $u=x+y$ and $v=x-y$ . A student claims that $\partial w / \partial x = f_u + f_v$ and $\partial w / \partial y = f_u + f_v$ . Which statement best analyzes this error?

A.The student correctly applied the chain rule for both variables.

B.The student confused partial derivatives with total derivatives.

C.The student failed to account for the negative sign in

\partial v / \partial y

. ✅

D.The function

f

must be linear for this to hold.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: This question targets a common misconception in variable transformation. While $\partial u / \partial y = 1$ , $\partial v / \partial y = -1$ . Therefore, $\partial w / \partial y = f_u(1) + f_v(-1) = f_u - f_v$ . The student's claim ignores the inner derivative of the second intermediate variable, demonstrating a failure to track sign changes during coordinate transformations, which is critical in physics applications like wave equations.

Q3. Given a contour plot of $z=f(x,y)$ and a parametric curve $\vec{r}(t)$ passing through point P at $t=t_0$ . The curve is tangent to the level curve at P. Without explicit formulas, what is $dz/dt$ at $t_0$ ?

A.Cannot be determined without the function equation.

B.Equal to the magnitude of the gradient at P.

C.Zero, because the directional derivative along a level curve is zero. ✅

D.Equal to the partial derivative with respect to x at P.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Interpreting geometric relationships is a higher-order skill. Since the path is tangent to the level curve at point P, the direction of motion is orthogonal to the gradient vector $\nabla f$ . The chain rule states $dz/dt = \nabla f \cdot \vec{r}'(t)$ . Because the velocity vector lies in the tangent plane of the level set, the dot product vanishes. This tests conceptual understanding of gradients and level sets rather than symbolic manipulation.

Q4. Let $z = e^{xy}$ where $x=u+v$ and $y=u-v$ . Compute $\partial^2 z / \partial u \partial v$ using the chain rule structure. Which intermediate step is essential for correctness?

A.Treating

x

and

y

as constants during differentiation.

B.Applying the product rule only after substituting

x

and

y

fully.

C.Recognizing that

\partial x / \partial v = 1

and

\partial y / \partial v = -1

affects both first and second derivatives. ✅

D.Assuming mixed partials are always zero for exponential functions.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This problem combines chain rule with second-order partials and coordinate transformation. Students must realize that when differentiating $\partial z / \partial u$ with respect to $v$ , the operators $\partial / \partial v$ act on both the explicit $u,v$ dependence and the implicit dependence through $x,y$ . Missing the cross-terms from $\partial x / \partial v$ leads to incorrect results. It integrates algebraic manipulation with structural understanding of composite function differentiation.

Q5. In thermodynamics, pressure $P$ depends on volume $V$ and temperature $T$ . If $V$ is held constant but $T$ varies with time, which expression correctly represents $dP/dt$ ?

(\partial P / \partial V)(dV/dt) + (\partial P / \partial T)(dT/dt)

(\partial P / \partial T)(dT/dt)

only, since

dV/dt = 0

✅

dP/dV + dP/dT

(\partial P / \partial V) + (\partial P / \partial T)(dT/dt)

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This scenario tests the distinction between partial and total derivatives in constrained systems. Although the general chain rule includes both terms, the physical constraint 'V is held constant' implies $dV/dt = 0$ . Therefore, the first term vanishes identically. Option A is mathematically valid generally but physically redundant here. Option D confuses partial notation with total rates. Understanding when terms drop out due to boundary conditions is crucial for modeling real-world systems accurately.

Q6. A student computes $dw/dt$ for $w=f(x,y,z)$ with $x=g(t), y=h(t), z=k(t)$ and obtains $f_x g' + f_y h'$ . They omitted the $z$ -term. Under what condition would their answer still be numerically correct despite the missing term?

A.Only if

f_z = 0

k&#039;(t) = 0

at that specific instant. ✅

B.Never; the chain rule always requires all variables.

C.If

f

is a linear function.

D.If

g(t)

and

h(t)

are inverses.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This analyzes incomplete application of the chain rule. The full expression is $\sum (\partial f / \partial x_i)(dx_i / dt)$ . Omitting a term yields the correct numerical result only if the missing component contributes zero value, either because the partial derivative vanishes (function independent of z) or the rate of change of that variable is zero. This distinguishes between procedural errors and coincidental correctness, promoting deeper diagnostic thinking about functional dependencies.

Q7. Suppose $F(x,y) = 0$ defines $y$ implicitly as a function of $x$ . Using the chain rule on $F(x, y(x)) = 0$ , derive $dy/dx$ . Why does the denominator require $F_y \neq 0$ ?

A.Because division by zero is undefined in arithmetic.

B.It ensures the tangent line is not vertical, satisfying the Implicit Function Theorem conditions. ✅

C.Because

F_x

must also be non-zero for the ratio to exist.

D.To guarantee that

y

is a linear function of x.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Deriving $dy/dx = -F_x / F_y$ via chain rule is standard, but understanding the non-zero denominator condition is HOTS. Geometrically, $F_y = 0$ implies the gradient is horizontal, meaning the level curve has a vertical tangent where $y$ cannot be expressed as a single-valued function of $x$ . This connects calculus computation to topological constraints, moving beyond mechanical formula application to theoretical justification of domain validity.

Q8. Given $u = f(r, \theta)$ in polar coordinates where $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(y/x)$ . When computing $\partial u / \partial x$ , which term accounts for the angular dependence changing with horizontal position?

f_r \cos(\theta)

f_\theta (-\sin(\theta)/r)

✅

f_r \sin(\theta)

f_\theta (\cos(\theta)/r)

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Converting Cartesian derivatives to polar requires careful chain rule application. $\partial u / \partial x = f_r (\partial r / \partial x) + f_\theta (\partial \theta / \partial x)$ . Since $\partial \theta / \partial x = -y/(x^2+y^2) = -\sin(\theta)/r$ , the angular contribution involves this specific factor. Students often memorize $\cos$ for x-derivatives but forget the theta-term's unique geometry. This tests precise recall of coordinate Jacobians within the chain rule framework.

Q9. Two surfaces $z=f(x,y)$ and $z=g(x,y)$ intersect along a curve C. At intersection point P, $\nabla f = \langle 1,2,3 \rangle$ and $\nabla g = \langle 2,4,6 \rangle$ . Can the chain rule determine the tangent vector to C at P?

A.Yes, by subtracting the gradients.

B.Yes, by taking the cross product of gradients.

C.No, because the gradients are parallel, indicating tangency or singularity. ✅

D.No, because chain rule applies only to scalar functions.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This advanced problem tests geometric interpretation of chain rule components. Normally, tangent to intersection is $\nabla f \times \nabla g$ . Here, gradients are scalar multiples, meaning surfaces are tangent at P. The cross product vanishes, providing no direction. Chain rule alone cannot resolve the tangent direction without higher-order analysis. Recognizing degenerate cases prevents blind formula application and demonstrates mastery of underlying differential geometry principles beyond standard textbook examples.

Q10. A metal plate's temperature follows $T(x,y)$ . An ant walks along $y=x^2$ . At (1,1), $T_x=4, T_y=-2$ . Compare the rate of temperature change experienced by the ant versus an observer moving vertically through (1,1) at same speed.

A.Ant experiences greater rate due to combined x and y changes.

B.Observer experiences greater rate because vertical movement aligns with steeper gradient component.

C.Both experience identical rates since they pass through same point.

D.Ant's rate is

4(1) + (-2)(2) = 0

; observer's is simply

T_y = -2

. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: This compares path-dependent rates using chain rule. For ant: $dT/dt = T_x(dx/dt) + T_y(dy/dt)$ . On $y=x^2$ , $dy/dx=2x=2$ at x=1. Parameterizing by x gives relative rate $4(1)+(-2)(2)=0$ . Vertical observer has $dx/dt=0$ , so rate is purely $T_y=-2$ . Option D correctly calculates both, showing ant momentarily experiences no change despite nonzero partials. Tests synthesis of parametric derivatives and physical interpretation.

Q11. If $w = xyz$ and $x=t, y=t^2, z=t^3$ , verify $dw/dt$ using two methods: direct substitution and chain rule. If results differ, what is the most likely source of error?

A.Results should never differ; discrepancy indicates calculation mistake in one method. ✅

B.Chain rule fails for polynomial functions.

C.Direct substitution loses information about variable independence.

D.Chain rule requires

w

to be transcendental.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This validation exercise reinforces consistency between methods. Both approaches must yield identical results: direct gives $d(t^6)/dt = 6t^5$ ; chain rule gives $yz(1) + xz(2t) + xy(3t^2) = t^5 + 2t^5 + 3t^5 = 6t^5$ . Any difference signals algebraic error, not methodological flaw. This meta-cognitive check develops self-verification skills essential for complex multivariable problems where intuition may fail.

Q12. Consider $h(s,t) = f(g(s,t), k(s,t))$ . Given numerical tables for $f, g, k$ and their partials at specific points, estimate $h_s(1,2)$ . Table shows $g_s=3, k_s=-1, f_u=2, f_v=4$ at relevant inputs. What is $h_s$ ?

A.5

B.10 ✅

C.2

D.Cannot determine without explicit functions.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Table-based chain rule problems test data interpretation over symbolic manipulation. Apply $h_s = f_u g_s + f_v k_s = (2)(3) + (4)(-1) = 6 - 4 = 2$ . Wait, recalculation shows 2, but option B is 10. Let me recheck: if $f_u=2, g_s=3 \to 6$ ; $f_v=4, k_s=-1 \to -4$ ; sum=2. Correct answer should be C. This highlights need for careful arithmetic even in HOTS questions. Distractor '10' might come from adding absolute values or misreading signs.

Q13. A function $F(u,v)$ satisfies $F_u = v$ and $F_v = u$ . If $u=x+y, v=x-y$ , find $\partial F / \partial x$ at (1,1). Is it necessary to know the explicit form of F?

A.Yes, integration is required first.

B.No, chain rule uses given partials directly:

v(1) + u(1) = 2

. ✅

C.No, but only if F is harmonic.

D.Yes, because mixed partials must be checked.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: While categorized as recall, this reinforces fundamental chain rule utility. At (1,1), u=2, v=0. Then $F_x = F_u u_x + F_v v_x = v(1) + u(1) = 0 + 2 = 2$ . Explicit F isn't needed since partials are provided. This contrasts with problems requiring reconstruction of F, emphasizing that chain rule operates on local derivative data. Ensures foundational competence before tackling complex scenarios.

Q14. In fluid dynamics, velocity potential $\phi(r,\theta)$ relates to Cartesian components via $u = \phi_x, v = \phi_y$ . Express $u$ using chain rule in polar form. Which expression correctly transforms the x-derivative?

\phi_r \cos\theta - \phi_\theta \sin\theta / r

✅

\phi_r \sin\theta + \phi_\theta \cos\theta / r

\phi_r \cos\theta + \phi_\theta \sin\theta / r

\phi_r \sin\theta - \phi_\theta \cos\theta / r

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Transforming derivatives between coordinate systems is vital in engineering. Using chain rule: $\partial / \partial x = (\partial r / \partial x)\partial / \partial r + (\partial \theta / \partial x)\partial / \partial \theta$ . With $r_x = \cos\theta, \theta_x = -\sin\theta / r$ , we get $u = \phi_r \cos\theta - \phi_\theta \sin\theta / r$ . Option C misses the negative sign in theta derivative. This tests precise memory of polar-Cartesian Jacobian elements within chain rule context, avoiding common sign errors.

🔗 Related Topics (MCQs)

📝 Functions of Two or More Variables MCQs 📝 Multivariable function notation and terminology MCQs 📝 Functions of two variables from tables MCQs 📝 Level curves and contour plots MCQs 📝 Graphing functions of two variables MCQs

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