MTH281: Mathematical Method I TMA3

MTH281: Mathematical Method I TMA3

Question 1 : Evaluate the second partial derivative of the functonf(x,y)=2x3y2+y3f(x,y)=2x3y2+y3
A.∂2f∂x2=12xy2,∂2f∂y2=4×3+6y,∂2f∂x∂y=12x2y∂2f∂x2=12xy2,∂2f∂y2=4×3+6y,∂2f∂x∂y=12x2y
B.∂2f∂x2=12x2y2,∂2f∂y2=4x+6y,∂2f∂x∂y=10x2y∂2f∂x2=12x2y2,∂2f∂y2=4x+6y,∂2f∂x∂y=10x2y
C.∂2f∂x2=12xy,∂2f∂y2=x3+y,∂2f∂x∂y=2x2y∂2f∂x2=12xy,∂2f∂y2=x3+y,∂2f∂x∂y=2x2y
D.∂2f∂x2=5x3y2,∂2f∂y2=6×3+6y,∂2f∂x∂y=2x2y2∂2f∂x2=5x3y2,∂2f∂y2=6×3+6y,∂2f∂x∂y=2x2y2

 

Question 2 : Compute the third derivative ofsinxInxsin⁡xInxusing Leibnitz theorem
A.(2x−2−3x−2)cosx−(3x−3+In2x)sinx
B.(x−3−x−2)cosx−(x−2+Inx)cosx
C.(2x−3−3x−1)sinx−(3x−2+Inx)cosx
D.(3x−3−4x−1)sinx−(3x−2+Inx)sinx

 

Question 3 : Use Leibnitz theorem to evaluate the fourth derivative of(2×3+3×2+x+2)e2x(2×3+3×2+x+2)e2x
A.8(x2+5×2+3x+14)e2x8(x2+5×2+3x+14)e2x
B.16(3×2+5×2+2x+3)e2x16(3×2+5×2+2x+3)e2x
C.10(3×2+10×2+3x+15)e2x10(3×2+10×2+3x+15)e2x
D.16(2×3+15×2+31x+19)e2x16(2×3+15×2+31x+19)e2x

 

Question 4 : Find the total differential of the functionf(x,y)=x2+3xyf(x,y)=x2+3xywth respect to x, given thaty=sin−1xy=sin−1x.
A.2x+2sin−1x+x(2−2x2122x+2sin−1x+x(2−2×212
B.2x+3sin−1x+3x(1−x2122x+3sin−1x+3x(1−x212
C.x+sin−1x+2x(1−x212x+sin−1x+2x(1−x212
D.2x+sin−1x+3x(1−x3122x+sin−1x+3x(1−x312

 

Question 5 : Evaluate the stationary points of the functionf(x,y)=xy(x2+y2−1)f(x,y)=xy(x2+y2−1)
A.(0,0),(0,0),(±1,0),±(12,12),±(12,0)(0,0),(0,0),(±1,0),±(12,12),±(12,0)
B.c=3±√(3)c=3±(3)
C.(0,0),(0,0),(0,0),±(0,12),±(0,−12)(0,0),(0,0),(0,0),±(0,12),±(0,−12)
D.(0,0),(0,±1),(±1,0),±(12,12),±(12,−12)(0,0),(0,±1),(±1,0),±(12,12),±(12,−12)

 

Question 6 : Compute the n-th differential coefficient ofy=xlogexy=xloge⁡x
A.(−1)n−1(n−1)!xn−2(n2−2)(−1)n−1(n−1)!xn−2(n2−2)
B.(−1)n−2(n−2)!xn−1(n3−2)(−1)n−2(n−2)!xn−1(n3−2)
C.(−1)n+1(n+1)!xn+2(n2+2)(−1)n+1(n+1)!xn+2(n2+2)
D.(−1)n−2(n+2)!xn+1(n3+2)(−1)n−2(n+2)!xn+1(n3+2)

 

Question 7 : Obtain the n-th differential coefficient ofy=(x2+1)e2xy=(x2+1)e2x
A.2nex(4x2x+4nx−n+4)2nex(4x2x+4nx−n+4)
B.2n−2e2x(4x2x+4nx+n2−n+4)2n−2e2x(4x2x+4nx+n2−n+4)
C.2n−2e2x(4x3x+5nx+n3−n+4)2n−2e2x(4x3x+5nx+n3−n+4)
D.2n−3e4x(x2x+nx+n3−n+4)2n−3e4x(x2x+nx+n3−n+4)

 

Question 8 : Find the first partial derivative of the functonf(x,y)=2x3y2+y3f(x,y)=2x3y2+y3
A.∂f∂x=6x3y3,∂f∂y=4x4y+y2∂f∂x=6x3y3,∂f∂y=4x4y+y2
B.∂f∂x=x2y,∂f∂y=2x3y+y∂f∂x=x2y,∂f∂y=2x3y+y
C.∂f∂x=x2y2,∂f∂y=x3y+y2∂f∂x=x2y2,∂f∂y=x3y+y2
D.∂f∂x=6x2y2,∂f∂y=4x3y+y2∂f∂x=6x2y2,∂f∂y=4x3y+y2

 

Question 9 : Use Leibnitz theorem to find the second derivative ofcosxsin2xcos⁡xsin⁡2x
A.3sinx(2−9sin2x)3sin⁡x(2−9sin2⁡x)
B.2cosx(3−5cos2x)2cos⁡x(3−5cos2⁡x)
C.2sinx(1−5cos3x)2sin⁡x(1−5cos3⁡x)
D.2sinx(2−9cos2x)2sin⁡x(2−9cos2⁡x)

 

Question 10 : Find the total differential of the functionf(x,y)=yex+yf(x,y)=yex+y
A.df=[yex−y]dx+[(1+y)ex−y]dydf=[yex−y]dx+[(1+y)ex−y]dy
B.df=[yex−y]dx−[(1+y)ex−y]dydf=[yex−y]dx−[(1+y)ex−y]dy
C.df=[yex+y]dx+[(1+y)ex+y]dydf=[yex+y]dx+[(1+y)ex+y]dy
D.df=[yex+y]dx−[(1+y)ex+y]dy

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