# Engineering Mathematics Questions and Answers – Partial Differentiation – 1

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Partial Differentiation – 1”.

1. f(x, y) = x2 + xyz + z Find fx at (1,1,1)
a) 0
b) 1
c) 3
d) -1

Explanation: fx = 2x + yz
Put (x,y,z) = (1,1,1)
fx = 2 + 1 = 3.

2. f(x, y) = sin(xy) + x2 ln(y) Find fyx at (0, π2)
a) 33
b) 0
c) 3
d) 1

Explanation: fy = xcos(xy) + x2y
fyx = cos(xy) – xysin(xy) + 2xy
Put (x,y) = (0, π2)
= 1.

3. f(x, y) = x2 + y3 ; X = t2 + t3; y = t3 + t9 Find dfdt at t=1.
a) 0
b) 1
c)-1
d) 164

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 dfdt at t = π2
a) 2
b)-2
c) 1
d) 0

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= π2; 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 dfdt at k = 1
a) 34
b) 16
c) 32
d) 61

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).$$(\frac{1}{2\sqrt{k}}$$
Put k=1; we have x=y=z=t=1
9 + 4 + 2 + 1 = 16.
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6. The existence of first order partial derivatives implies continuity.
a) True
b) False

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

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 fxy at (0,1) is
a) 0
b) 1
c) 67
d) 90

Explanation: First find
fy = cos(y + yx2)
Hence
fyx = fxy = – (2xy).sin(y + yx2)
Now put (x,y) = (0,1)
= 0.

9. f(x, y) = sin(xy + x3y) / x + x3 Find fxy at (0,1).
a) 2
b) 5
c) 1
d) undefined

Explanation: First find
fy = sin(xy + x3y)
Hence
fyx = fxy = (cos(xy + x3y)) . (y + 3x23y)
Now put (x,y) = (0,1)
= 1.

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