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

1. f(x, y) = x^{2} + xyz + z Find f_{x} at (1,1,1)

a) 0

b) 1

c) 3

d) -1

View Answer

Explanation: f

_{x}= 2x + yz

Put (x,y,z) = (1,1,1)

f_{x} = 2 + 1 = 3.

2. f(x, y) = sin(xy) + x^{2} ln(y) Find f_{yx} at (0, ^{π}⁄_{2})

a) 33

b) 0

c) 3

d) 1

View Answer

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) = x^{2} + y^{3} ; X = t^{2} + t^{3}; y = t^{3} + t^{9} Find ^{df}⁄_{dt} at t=1.

a) 0

b) 1

c)-1

d) 164

View Answer

Explanation: Using chain rule we have

= (2x).(2t + 3t

^{2}) + (3y

^{2}).(3t

^{2}+ 9t

^{8})

Put t = 1; we have x = 2; y = 2

= 4.(5) + 12.(12) = 164.

4. f(x, y) = sin(x) + cos(y) + xy^{2}; x = cos(t); y = sin(t) Find ^{df}⁄_{dt} at t = ^{π}⁄_{2}

a) 2

b)-2

c) 1

d) 0

View Answer

Explanation:Using chain rule we have

= (cos(x) + y

^{2}) . (-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 + x^{2} yzt; x = k^{3} ; y = k^{2}; z = k; t = √k

Find ^{df}⁄_{dt} at k = 1

a)34

b) 16

c) 32

d) 61

View Answer

6. The existence of first order partial derivatives implies continuity

a) True

b) False

View Answer

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

View Answer

Explanation: The gradient is perpendicular and not parallel to the velocity vector of the level curve.

8. f(x, y) = sin(y + yx^{2}) / 1 + x^{2} Value of f_{xy} at (0,1) is

a) 0

b) 1

c) 67

d) 90

View Answer

Explanation: First find

f

_{y}= cos(y + yx

^{2})

Hence

f

_{yx}= f

_{xy}= – (2xy).sin(y + yx

^{2})

Now put (x,y) = (0,1)

= 0.

9. f(x, y) = sin(xy + x^{3}y) / x + x^{3} Find f_{xy} at (0,1)

a) 2

b) 5

c) 1

d) undefined

View Answer

Explanation: First find

f

_{y}= sin(xy + x

^{3}y)

Hence

f

_{yx}= f

_{xy}= (cos(xy + x

^{3}y)) . (y + 3x

^{23}y)

Now put (x,y) = (0,1)

= 1.

**Sanfoundry Global Education & Learning Series – Engineering Mathematics.**

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