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Question 1. What does the notation ∂f/∂x represent for a function f(x,y)? A) The total derivative of f with respect to x. B) The rate of change of f when only y varies. C) The partial derivative of f holding y constant. D) The second derivative of f with respect to x. Answer: C Explanation: ∂f/∂x is defined as the limit of the difference quotient with y kept fixed, giving the slope of f in the x‑direction. Question 2. Which of the following limits correctly defines ∂f/∂y at (x₀,y₀)? A) limₕ→ 0 [f(x₀+h,y₀)−f(x₀,y₀)]/h B) limₕ→ 0 [f(x₀,y₀+h)−f(x₀,y₀)]/h C) limₕ→ 0 [f(x₀+h,y₀+h)−f(x₀,y₀)]/h D) limₕ→ 0 [f(x₀,y₀)−f(x₀,y₀−h)]/h Answer: B Explanation: Holding x fixed and varying y by h gives the definition of the partial derivative with respect to y. Question 3. For f(x,y)=x²y+3y³, what is fₓ? A) 2xy+3y³ B) 2xy C) 2x y+9y² D) 2x y+9y² Answer: B Explanation: Differentiate with respect to x treating y as constant: ∂/∂x (x²y)=2xy, ∂/∂x (3y³)=0.
Question 4. If f(x,y)=e^{xy}, then f_y equals: A) x e^{xy} B) y e^{xy} C) e^{xy} D) xy e^{xy} Answer: A Explanation: Treat x as constant; derivative of e^{xy} with respect to y is x e^{xy}. Question 5. Which statement about continuity and partial derivatives is true? A) Existence of all partial derivatives guarantees continuity. B) Continuity guarantees existence of all partial derivatives. C) A function can have partial derivatives at a point yet be discontinuous there. D) Neither continuity nor partial derivatives imply the other. Answer: C Explanation: Partial derivatives may exist even when the function has a jump discontinuity at the point. Question 6. The geometric interpretation of fₓ(a,b) is: A) Slope of the curve obtained by intersecting the surface z=f(x,y) with the plane y=b. B) Slope of the curve obtained by intersecting the surface with the plane x=a. C) Gradient magnitude at (a,b). D) Directional derivative in the direction ⟨1,0⟩. Answer: A Explanation: Holding y=b produces a slice; the slope of that slice at x=a is fₓ(a,b).
Explanation: dz/dt = fₓ·dx/dt + f_y·dy/dt = fₓ·2t + f_y·e^t. Question 10. For z = f(u,v) with u = xy and v = x²−y², the mixed partial ∂z/∂x is: A) f_u·y + f_v·2x B) f_u·y + f_v·(-2y) C) f_u·(y) + f_v·(2x) D) Both A and C (they are identical) Answer: D Explanation: ∂u/∂x = y, ∂v/∂x = 2x, so ∂z/∂x = f_u·y + f_v·2x. Question 11. Implicit differentiation of F(x,y)=x³+y³−6xy=0 gives dy/dx equal to: A) (6y−3x²)/(3y²−6x) B) (3x²−6y)/(6x−3y²) C) (6x−3y²)/(3x²−6y) D) (3y²−6x)/(6y−3x²) Answer: A Explanation: Differentiate: 3x² + 3y²·dy/dx −6y −6x·dy/dx =0 → collect dy/dx: (3y²−6x)dy/dx = 6y−3x² → dy/dx = (6y−3x²)/(3y²−6x). Question 12. The second‑order partial f_{xy} of f(x,y)=x e^{y} is: A) e^{y} B) x e^{y} C) 0 D) y e^{y} Answer: A
Explanation: fₓ = e^{y}; then differentiate with respect to y: (e^{y})y = e^{y}. Question 13. For a C² function, Clairaut’s theorem guarantees that: A) f{xx}=f_{yy} B) f_{xy}=f_{yx} C) f_{x}=f_{y} D) All mixed third‑order derivatives are equal. Answer: B Explanation: Equality of mixed second‑order partials holds when the function’s second partials are continuous. Question 14. Compute f_{xxx} for f(x,y)=x³y+ sin x. A) 6xy + −sin x B) 6xy + −cos x C) 6y + −sin x D) 6y + −cos x Answer: D Explanation: First, fₓ=3x²y+cos x; second, f_{xx}=6xy−sin x; third, f_{xxx}=6y−cos x. Question 15. The gradient vector ∇f at a point gives: A) The direction of steepest descent. B) The magnitude of the steepest ascent. C) A vector orthogonal to level curves. D) Both B and C. Answer: D
Answer: A Explanation: ∇f = ⟨2x, 2y⟩ → at (1,2) = ⟨2, 4⟩. Plane: z−5 = 2(x−1)+4(y−2). Question 19. A critical point of f(x,y)=x³−3xy² occurs at: A) (0,0) only B) (1,1) and (−1,−1) C) (0,0), (1,0), (−1,0) D) (0,0), (1,1), (−1,−1) Answer: A Explanation: Set fₓ=3x²−3y²=0 and f_y=−6xy=0 → solutions give x=±y and xy=0 → only (0,0) satisfies both. Question 20. For the function in Question 19, the discriminant D = f_{xx}f_{yy}−(f_{xy})² at the critical point (0,0) is: A) 0 B) 9 C) − D) − Answer: A Explanation: Compute second derivatives: f_{xx}=6x, f_{yy}=−6x, f_{xy}=−6y. At (0,0) all are zero → D = 0·0−0² = 0. Question 21. Using the second‑derivative test, the critical point (0,0) of f(x,y)=x³−3xy² is classified as: A) Local maximum B) Local minimum
C) Saddle point D) Inconclusive Answer: D Explanation: D=0 makes the test inconclusive; higher‑order analysis shows (0,0) is a saddle. Question 22. Applying Lagrange multipliers, maximize f(x,y)=xy subject to g(x,y)=x²+y²=1. The optimal points are: A) (±1/√2, ±1/√2) with same signs B) (±1,0) and (0,±1) C) (±1/2, ±√3/2) D) No maximum exists. Answer: A Explanation: ∇f = ⟨y, x⟩, ∇g = ⟨2x, 2y⟩ → y = λ2x, x = λ2y → leads to x=±y and x²+y²=1 → x=±1/√2, y=±1/√ with same sign. Max value = 1/2. Question 23. In the heat equation ∂u/∂t = α ∇²u, the term ∇²u represents: A) First‑order spatial derivative. B) Laplacian, sum of second‑order partials. C) Directional derivative in time. D) Gradient magnitude. Answer: B Explanation: ∇²u = u_{xx}+u_{yy}+u_{zz} is the Laplacian, a second‑order spatial operator. Question 24. For a fluid with pressure p(x,y,z), the force per unit volume is given by: A) −∇p
A) 6x y² B) 6x² y C) 6x y D) 6y² Answer: A Explanation: fₓ = 2xy³; f_{xx}=2y³; then f_{xxy}=∂/∂y (2y³)=6y²·x? Wait compute correctly: f_{xx}=2y³, differentiate w.r.t y: 6y² → then multiply by x? Actually f = x²y³, fₓ =2x y³, f_{xx}=2 y³, f_{xxy}=∂/∂y ( y³)=6y². No x factor. So answer D (6y²). But option D is 6y². Hence answer D. Question 28. For f(x,y,z)=xyz, the gradient at (1,2,3) is: A) ⟨6,3,2⟩ B) ⟨6,3,1⟩ C) ⟨6,2,3⟩ D) ⟨2,3,6⟩ Answer: A Explanation: ∇f = ⟨yz, xz, xy⟩ → at (1,2,3): ⟨2·3, 1·3, 1·2⟩ = ⟨6,3,2⟩. Question 29. The directional derivative of f(x,y)=x²y in the direction of the vector v = ⟨4,−3⟩ (at (1,1)) is: A) −2/ B) 2/ C) −10/ D) 10/ Answer: B Explanation: Unit vector u = (1/5)⟨4,−3⟩. ∇f = ⟨2xy, x²⟩ → at (1,1) = ⟨2,1⟩. Dot: (2,1)·(4/5,−3/5)= (8−3)/ =5/5=1. Wait compute: 24/5 + 1(-3/5)=8/5−3/5=5/5=1. So directional derivative =1. None of options
match; need adjust. Choose nearest? Actually answer should be 1. Since not provided, we must change options; but we cannot. We'll replace answer with “None of the above”. However format expects A‑D. We'll correct options: A) 1, B) −1, C) 2, D) −2. Then answer A. Let's redo. Question 29. The directional derivative of f(x,y)=x²y in the direction of v = ⟨4,−3⟩ (at (1,1)) is: A) 1 B) − C) 2 D) − Answer: A Explanation: Unit vector u = (1/5)⟨4,−3⟩, ∇f(1,1)=⟨2,1⟩, dot product = 2·4/5 + 1·(−3/5)=8/5−3/5=1. Question 30. The total differential dz of z = ln(x² + y²) is: A) (2x dx + 2y dy)/(x² + y²) B) (2x dx − 2y dy)/(x² + y²) C) (2x dx + 2y dy)·(x² + y²) D) (2x dx + 2y dy)·ln(x² + y²) Answer: A Explanation: dz = (1/(x²+y²))(2x dx + 2y dy). Question 31. If f(x,y)=x e^{y²}, then f_{yy} equals: A) 2x e^{y²} + 4x y² e^{y²} B) 2x e^{y²} + 4x y e^{y²} C) 4x y e^{y²} D) 2x e^{y²}
B) 2
C) −
D) 4
Answer: A Explanation: f_{xx}=2, f_{yy}=−2, sum =0. Question 34. Which of the following functions is NOT differentiable at (0,0) despite having partial derivatives there? A) f(x,y)=x y/(x²+y²) for (x,y)≠(0,0), 0 at (0,0) B) f(x,y)=x² + y² C) f(x,y)=|x| + |y| D) f(x,y)=e^{x+y} Answer: A Explanation: The limit defining differentiability does not exist, though both partials at (0,0) equal 0. Question 35. For the implicit surface F(x,y,z)=x²+y²+z²−9=0, the normal vector at (2,2,1) is: A) ⟨2,2,1⟩ B) ⟨4,4,2⟩ C) ⟨2,2,−1⟩ D) ⟨4,4,−2⟩ Answer: B Explanation: ∇F = ⟨2x, 2y, 2z⟩ → at (2,2,1) gives ⟨4,4,2⟩. Question 36. The directional derivative of f(x,y,z)=xyz in the direction of the unit vector u = (1/√3)⟨1,1,1⟩ at (1,2,3) is:
A) 6/√
B) 12/√
C) 18/√
D) 24/√
Answer: B Explanation: ∇f = ⟨yz, xz, xy⟩ → at (1,2,3) = ⟨6,3,2⟩. Dot with u: (6+3+2)/√3 =11/√3 ≈6.35, not matching options. Re‑evaluate: Actually yz=2·3=6, xz=1·3=3, xy=1·2=2 → sum=11. None of the options. Choose the closest: B (12/√3). But not exact. We'll adjust options. Question 36. The directional derivative of f(x,y,z)=xyz in the direction of u = (1/√3)⟨1,1,1⟩ at (1,2,3) is: A) 11/√ B) 6/√ C) 9/√ D) 12/√ Answer: A Explanation: Dot product of ∇f(1,2,3)=⟨6,3,2⟩ with u gives (6+3+2)/√3 =11/√3. Question 37. For f(x,y)=x³y−y³x, the mixed partial f_{xy} at (1,1) equals: A) 0 B) 2 C) − D) 4 Answer: A Explanation: Compute fₓ = 3x²y − y³; then f_{xy}=3x² − 3y². At (1,1) gives 3−3=0.
Question 41. The Jacobian determinant of the transformation x = u²−v², y = 2uv is: A) 4uv B) −4uv C) 2(u²+v²) D) −2(u²+v²) Answer: A Explanation: Compute ∂(x,y)/∂(u,v) = |∂x/∂u ∂x/∂v; ∂y/∂u ∂y/∂v| = |2u −2v; 2v 2u| = (2u)(2u) − (−2v)(2v) =4u²+4v²? Wait compute determinant: (2u)(2u) - (−2v)(2v) =4u² - (−4v²)=4u²+4v² =4(u²+v²). But answer options not matching. Let's recalc: x_u = 2u, x_v = - 2v, y_u = 2v, y_v = 2u. Determinant = (2u)(2u)
- (-2v)(2v) =4u² - (-4v²)=4u²+4v² =4(u²+v²). Not in options. Choose A as 4uv? Not correct. We'll modify options. Question 41. The Jacobian determinant of the transformation x = u²−v², y = 2uv is: A) 4(u²+v²) B) −4(u²+v²) C) 4uv D) −4uv Answer: A Explanation: Determinant = (2u)(2u) − (−2v)(2v) =4u²+4v² =4(u²+v²). Question 42. The total differential of w = x²y + yz³ is: A) 2xy dx + (x² + 3z²y) dy + 3yz² dz B) 2xy dx + (x² + z³) dy + 3yz² dz C) 2xy dx + (x² + 3y z²) dy + 3y z² dz D) 2xy dx + (x² + 3y z²) dy + 3y z² dz
Answer: D Explanation: Compute partials: wₓ =2xy, w_y = x² + 3yz², w_z = 3y z². Then dz = wₓdx + w_ydy + w_zdz. Question 43. For the function f(x,y)=e^{x y}, the second‑order mixed derivative f_{xy} equals: A) e^{xy} B) x e^{xy} C) y e^{xy} D) xy e^{xy} Answer: A Explanation: fₓ = y e^{xy}; differentiate w.r.t y: f_{xy}=e^{xy}+ y·x e^{xy}=e^{xy}(1+xy). Wait compute correctly: fₓ = y e^{xy}. Then f_{xy}=∂/∂y (y e^{xy}) = e^{xy} + y·x e^{xy}=e^{xy}(1+xy). None of options. Adjust. Question 43. For f(x,y)=e^{xy}, the mixed partial f_{xy} equals: A) e^{xy}(1+xy) B) e^{xy}(1−xy) C) x e^{xy} D) y e^{xy} Answer: A Explanation: Differentiate fₓ = y e^{xy} with respect to y: gives e^{xy}+xy e^{xy}=e^{xy}(1+xy). Question 44. The limit lim_{(x,y)→(0,0)} (x²y)/(x⁴+y⁴) exists? A) Yes, equals 0 B) Yes, equals 1 C) No, limit depends on path
C) −
D) 8
Answer: A Explanation: Second derivatives: f_{xx}=2, f_{yy}=2, f_{xy}=−2. D = (2)(2)−(−2)² =4−4=0. Question 48. For the constrained problem maximize f(x,y)=x y subject to x²+y²=4, the Lagrange multiplier λ equals: A) 1/ B) −1/ C) 1 D) − Answer: C Explanation: ∇f = ⟨y, x⟩, ∇g = ⟨2x, 2y⟩ → y = λ2x, x = λ2y → substitute → y = 2λx and x = 2λy → combine → y = 4λ²y → either y=0 (gives x=±2, product 0) or 4λ²=1 → λ = ±1/2. But need max product; solving yields λ = 1/2 gives x=y=√2, product 2. However λ=−1/2 gives x=−y=√2 leading product −2 (minimum). So λ for maximum is 1/2. But answer options list 1/2? Option A is 1/2. So answer A. Question 48 Revised Answer: A Explanation: λ = 1/2 gives the maximum product. Question 49. The gradient of f(x,y,z)=x²y+yz³ at the point (1,−1,2) is: A) ⟨2y, x²+3z²y, 3yz²⟩ → ⟨−2, 1+12(−1), 3(−1)(4)⟩ = ⟨−2, −11, −12⟩ B) ⟨2y, x²+3z²y, 3yz²⟩ → ⟨−2, 1−12, −12⟩ = ⟨−2, −11, −12⟩ C) ⟨2xy, z³, 3yz²⟩ D) ⟨2x y, z³, 3y z²⟩ Answer: B
Explanation: Compute partials: fₓ = 2xy, f_y = x² + z³, f_z = 3yz². At (1,−1,2): fₓ = 2·1·(−1)=−2, f_y = 1² + 8 =9, f_z = 3(−1)(4)=−12. Wait correction: f_y = x² + 3y z²? Actually original function f = x²y + y z³. Then f_y = x² + z³, f_z = 3 y z². So at (1,−1,2): fₓ = 2·1·(−1)=−2, f_y = 1 + 8 =9, f_z = 3(−1)(4)=−12. None of options match; choose correct vector ⟨−2, 9, −12⟩. Let's adjust. Question 49. The gradient of f(x,y,z)=x²y+yz³ at (1,−1,2) is: A) ⟨−2, 9, −12⟩ B) ⟨−2, −11, −12⟩ C) ⟨2, 9, 12⟩ D) ⟨2, −9, 12⟩ Answer: A Explanation: fₓ=2xy → −2; f_y = x²+z³ → 1+8=9; f_z = 3y z² → 3(−1)(4)=−12. Question 50. If f(x,y)=x⁴−y⁴, then the Hessian determinant at any point (x,y) is: A) 16(x²−y²)² B) −16(x²−y²)² C) 0 D) 16(x²+y²)² Answer: A Explanation: f_{xx}=12x², f_{yy}=−12y², f_{xy}=0 → Hessian determinant = (12x²)(−12y²)−0 = −144 x² y². Wait that's not matching options. Let's compute correctly: f = x⁴−y⁴ → fₓ =4x³, f_{xx}=12x²; f_y = −4y³, f_{yy}=−12y²; f_{xy}=0. Determinant = (12x²)(−12y²)−0 = −144 x² y². None of options. We'll adjust. Question 50. For f(x,y)=x⁴−y⁴, the Hessian determinant H at (x,y) equals: A) −144 x² y² B) 144 x² y²
