250+ TOP MCQs on Partial Differentiation and Answers

Engineering Mathematics Multiple Choice Questions on “Partial Differentiation – 1”.

1. f(x, y) = x² + xyz + z Find f_x at (1,1,1)
a) 0
b) 1
c) 3
d) -1
Answer: c
Explanation: f_x = 2x + yz
Put (x,y,z) = (1,1,1)
f_x = 2 + 1 = 3.

2. f(x, y) = sin(xy) + x² ln(y) Find f_yx at (0, ^π⁄₂)
a) 33
b) 0
c) 3
d) 1
Answer: d
Explanation: f_y = xcos(xy) + ^x2⁄_y
f_yx = cos(xy) – xysin(xy) + ^2x⁄_y
Put (x,y) = (0, ^π⁄₂)
= 1.

3. f(x, y) = x² + y³ ; X = t² + t³; y = t³ + t⁹ Find ^df⁄_dt at t=1.
a) 0
b) 1
c)-1
d) 164
Answer: d
Explanation: Using chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dt}+f_y.frac{dy}{dt})
=(2x).(2t + 3t²) + (3y²).(3t² + 9t⁸)
Put t = 1; we have x = 2; y = 2
=4.(5) + 12.(12) = 164.

4. f(x, y) = sin(x) + cos(y) + xy²; x = cos(t); y = sin(t) Find ^df⁄_dt at t = ^π⁄₂
a) 2
b)-2
c) 1
d) 0
Answer: b
Explanation:Using chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dt}+f_y.frac{dy}{dt})
= (cos(x) + y²).(-sin(t)) + (-sin(y) + 2xy).(cos(t))
Put t= ^π⁄₂; we have x=0; y=1
=(1 + 1).(-1) + 0 = -2.

5. f(x, y, z, t) = xy + zt + x² yzt; x = k³ ; y = k²; z = k; t = √k
Find ^df⁄_dt at k = 1
a) 34
b) 16
c) 32
d) 61
Answer: b
Explanation: Using Chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dk}+f_y.frac{dy}{dk}+f_z.frac{dz}{dk}+f_t.frac{dt}{dk})
= (y + 2xyzt).(3k²) + (x + x²zt).(2k) + (t + x²yt).(1) + (z + x²yz).((frac{1}{2sqrt{k}})
Put k=1; we have x=y=z=t=1
9 + 4 + 2 + 1 = 16.

6. The existence of first order partial derivatives implies continuity.
a) True
b) False
Answer: b
Explanation: The mere existence cannot be declared as a condition for contnuity because the second order derivatives should also be continuous.

7. The gradient of a function is parallel to the velocity vector of the level curve.
a) True
b) False
Answer: b
Explanation: The gradient is perpendicular and not parallel to the velocity vector of the level curve.

8. f(x, y) = sin(y + yx²) / 1 + x² Value of f_xy at (0,1) is
a) 0
b) 1
c) 67
d) 90
Answer: a
Explanation: First find
f_y = cos(y + yx²)
Hence
f_yx = f_xy = – (2xy).sin(y + yx²)
Now put (x,y) = (0,1)
= 0.

9. f(x, y) = sin(xy + x³y) / x + x³ Find f_xy at (0,1).
a) 2
b) 5
c) 1
d) undefined
Answer: c
Explanation: First find
f_y = sin(xy + x³y)
Hence
f_yx = f_xy = (cos(xy + x³y)) . (y + 3x²³y)
Now put (x,y) = (0,1)
= 1.