Chapter 18 Partial Differentiation-I

18.1 Exercise 10(i)

18.1.1 Question 1

Find fxfx and fyfy of the following functions f(x,y)f(x,y)

  1. 3x24xy+2y23x24xy+2y2

fx=6x4yfy=4y4x

  1. x2y2x2+y2

fx=(x2+y2).2x(x2y2).2x(x2+y2)2=2x(x2+y2x2+y2)(x2+y2)2=4xy2(x2+y2)2

fy=(x2+y2)×(2y)(x2y2)×2y(x2+y2)2=2y(x2y2x2+y2)(x2+y2)2=4x2y(x2+y2)2

  1. sin1(ax+by)

fx=a1(ax+by)2fy=b1(ax+by)2

  1. log(x3+y2)

fx=3x2x3+y2fy=2yx3+y2

18.1.2 Question 2

Find the partial derivatives of second order of the following functions of x amd y.

  1. ax2+2hxy+by2+2gx+2fy+c

fx=fx=2ax+2hy+2gfxx=2afy=2hx+2by+2ffyy=2bfyx=y(fx)=2hfxy=x(fy)=2h

  1. xcosy+ycosx

fx=fx=cosyysinxfxx=2fx2=ycosxfy=xsiny+cosxfyy=xcosyfyx=y(fx)=sinysinxfxy=x(fy)=sinysinx

  1. tan12xyx2y2

fx=11+(2xyx2y2)2×(x2y2).2y2xy.2x(x2y2)2=(x2y2)2(x2y2)2+4x2y2×2x2y2y34x2y(x2y2)2=2y32x2y(x2+y2)2=2y(y2+x2)(x2+y2)2fx=2yx2+y2

fxx=(x2+y2).0(2y).2x(x2+y2)2=4xy(x2+y2)2

fyx=(x2+y2).(2)(2y).2y(x2+y2)2=2x22y2+4y2(x2+y2)2=2(y2x2)(x2+y2)2

fy=11+(2xyx2y2)2×(x2y2).2x2xy.(2y)(x2y2)2=2x32xy2+4xy2(x2y2)2+4x2y2=2x(x2+y2)(x2+y2)2fy=2xx2+y2

fyy=(x2+y2).02x.2y(x2+y2)2=4xy(x2+y2)2

fxy=2(x2+y2)2x.2x(x2+y2)2=2x2+2y24x2(x2+y2)2

18.1.3 Question 3

Verify that 2uxy=2uyx when u is

  1. ex2+xy+y2

Here u=ex2+xy+y2.

2uxy=x(uy)=x(ex2+xy+y2.(x+2y))=ex2+xy+y2.1+(x+2y)[ex2+xy+y2.(2x+y)]=ex2+xy+y2(1+(x+2y)(2x+y))

2uyx=y(ux)=y(ex2+xy+y2)(2x+y)=ex2+xy+y2.1+(2x+y).ex2+xy+y2.(x+2y)=ex2+xy+y2(1+(x+2y)(2x+y))

Thus 2uxy=2uyx. Hence, verified.

  1. sin1yx

Here, u=sin1yx.

2uxy=x(uy)=x(11(y/x)2×1x)=x(xx2y2×1x)=x(1x2y2)=01×12x2y2×2x(x2y2)2uxy=x(x2y2)3/2

2uyx=y(ux)=y(11(y/x)2×(yx2))=y(yxx2y2)=xx2y2.(1)(y)[x2x2y2×(2y)+0]x2(x2y2)=xx2y2xy2x2y2x2(x2y2)=(x2y2)y2x(x2y2)3/2=x2x(x2y2)3/22uyx=x(x2y2)3/2

Thus 2uxy=2uyx. Hence, verified.

18.1.4 Question 4

  1. If u=x2+y2+z2, show that xux+yuy+zuz=2u

Here,

ux=2xuy=2yuz=2z

Thus,

xux+yuy+zuz=x.2x+y.2y+z.2z=2(x2+y2+z2)=2u

  1. If u=x2y+y2z+z2x, show that ux+uy+uz=(x+y+z)2

Here,

ux=2xy+z2uy=x2+2yzuz=y2+2zx

Thus,

ux+uy+uz=2xy+z2+x2+2yz+y2+2zx=x2+2xy+y2+z2+2(x+y)z=(x+y)2+2(x+y)z+z2=(x+y+z)2

  1. If u=yz+zx+xy, prove that xux+yuy+zuz=0

Taking partial derivatives,

ux=zx2+1yuy=1zxy2uz=yz2+1x

So,

xux+yuy+zuz=x(zx2+1y)+y(1zxy2)+z(yz2+1x)=zx+xy+yzxyyz+zx=0

  1. If f(x,y,z)=ex/y+ey/z+ez/x, show that xfx+yfy+zfz=0

Here, f(x,y,z)=ex/y+ey/z+ez/x. Taking partial derivatives,

fx=ex/y.1y+0+ez/x.(zx2)=ex/yyez/x.zx2fy=ex/y.(xy2)+ey/z.1z+0=xex/yy2+ey/zzfz=0+ey/z.(yz2)+ez/x.1x=yey/zz2+ez/xx

So,

xfx+yfy+zfz=x(ex/yyez/x.zx2)+y(xex/yy2+ey/zz)+z(yey/zz2+ez/xx)=xex/yyez/xzxxex/yy+yey/zzyey/zz+zez/xx=0

18.1.5 Question 5

If z=tan(y+ax)+(yax)3/2, find the value of 2zx2a22zy2.

2zx2=x(zx)=x(sec2(y+ax).a+32(yax)1/2.(a))=2sec(y+ax)sec(y+ax)tan(y+ax).a.a32a×12yax×(a)2zx2=2a2sec2(y+ax)tan(y+ax)+3a24yax

2zy2=y(zy)=y(sec2(y+ax).1+32(yax)1/2.1)=2sec(y+ax)sec(y+ax)tan(y+ax).1+32×12yax.12zy2=2sec2(y+ax)tan(y+ax)+34yax

2zx2a22zy2=2a2sec2(y+ax)tan(y+ax)+3a24yaxa2(2sec2(y+ax)tan(y+ax)+34yax)2zx2a22zy2=0

18.1.6 Question 6

  1. If u=logx2+y2+z2, show that (x2+y2+z2)(2ux2+2uy2+2uz2)=1

2ux2=x(ux)=x(1x2+y2+z2×12x2+y2+z2×2x)=x(xx2+y2+z2)=(x2+y2+z2).1x.2x(x2+y2+z2)2=x2+y2+z2(x2+y2+z2)2

Similarly,

2uy2=x(ux)=x2y2+z2(x2+y2+z2)22uz2=z(uz)=x2+y2z2(x2+y2+z2)2

So,

(x2+y2+z2)(2ux2+2uy2+2uz2)=(x2+y2+z2)(x2+y2+z2+x2y2+z2+x2+y2z2(x2+y2+z2)2)=(x2+y2+z2)(x2+y2+z2)(x2+y2+z2)2=1

  1. If u=log(x2+y2+z2), prove that x2uyz=y2uzx=z2uxy

2uyz=y(uz)y(1x2+y2+z2×2z)=02x.2y(x2+y2+z2)2=4yz(x2+y2+z2)2

Similarly,

2uzx=z(ux)=4zx(x2+y2+z2)22uxy=x(uy)=4xy(x2+y2+z2)2

Thus,

x2uyz=y2uzx=z2uxy=4xyz(x2+y2+z2)2

  1. If V=f(xyz), show that xVx=yVy=zVz and x2Vxx=y2Vyy=z2Vzz

Vx=f(xyz).yzVy=f(xyz).xzVz=f(xyz).xy

Again,

Vxx=f(xyz).yz.yz+0=f(xyz)y2z2

Similarly,

Vyy=f(xyz)x2z2Vzz=f(xyz)x2y2

Thus,

xVx=yVy=zVz=f(xyz)xyz

and,

x2Vxx=y2Vyy=z2Vzz=f(xyz)x2y2z2

18.1.7 Question 7

If u=exyz, prove that 3uxyz=(1+3xyz+x2y2z2)exyz

3uxyz=2xy(uz)=2xy(exyz.xy)=xy(exyz.xy)=x(exyz.x+xyexyz.xz)=x(exyz.x)+x(x2yz.exyz)=exyz.1+xexyz.yz+exyz.2xyz+x2yz.exyz.yz=exyz+3xyzexyz+x2y2z2exyz3uxyz=(1+3xyz+x2y2z2)exyz

Hence, proved.

18.1.8 Question 8

If u=log(x3+y3+z33xyz), show that

  1. ux+uy+uz=3x+y+z

ux=1x3+y3+z33xyz×(3x23yz)uy=1x3+y3+z33xyz×(3y23xz)uz=1x3+y3+z33xyz×(3z23xy)

So,

ux+uy+uz=3(x2+y2+z2)3(xy+yz+xz)x3+y3+z33xyz=3(x2+y2+z2xyyzxz)x3+y3+z33xyz

The expansion of (x+y+z)(x2+y2+z2xyyzxz) yields x3+y3+z33xyz. Hence, above equation becomes

=3(x2+y2+z2xyyzxz)(x+y+z)(x2+y2+z2xyyzxz)

ux+uy+uz=3x+y+z

  1. 2ux2+2uy2+2uz2=3(x+y+z)2

2ux2=x(ux)=x(3x23yzx3+y3+z33xyz)=(x3+y3+z33xyz).6x(3x23yz)(3x23yz)(x3+y3+z33xyz)2=6x4+6xy3+6xz318x2yz9x4+18x2yz9y2z2(x3+y3+z33xyz)2=3x4+6xy3+6xz39y2z2(x3+y3+z33xyz)2

2uy2=y(uy)=y(3y23xzx3+y3+z33xyz)=(x3+y3+z33xyz).6y(3y23xz)(3y23xz)(x3+y3+z33xyz)2=6x3y+6y4+6yz318xy2z9y4+18xy2z9x2z2(x3+y3+z33xyz)2=6x3y3y4+6yz39x2z2(x3+y3+z33xyz)2

2uz2=z(uz)=(x3+y3+z33xyz).6z(3z23xy)(3z23xy)(x3+y3+z33xyz)2=6x3z+6y3z+6z418xyz29z4+18xyz29x2y2(x3+y3+z33xyz)2=6x3z+6y3z3z49x2y2(x3+y3+z33xyz)2

Now,

2ux2+2uy2+2uz2=

3x4+6xy3+6xz39y2z2+6x3y3y4+6yz39x2z2+6x3z+6y3z3z49x2y2(x3+y3+z33xyz)2=3x43y43z4+6xy3+6xz3+6x3y+6yz3+6x3z+6y3z9y2z29x2z29x2y2(x3+y3+z33xyz)2=3(x4+y4+z42xy32xz32x3y2yz32x3z2y3z+3y2z2+3x2z2+3x2y2)(x3+y3+z33xyz)2

The expansion of (x+y+z)(x2+y2+z2xyyzxz) yields x3+y3+z33xyz. And the expansion of (x2+y2+z2xyyzxz)2 yields x4+y4+z42xy32xz32x3y2yz32x3z2y3z+3y2z2+3x2z2+3x2y2.

So, substituting these values in above equation we get,

2ux2+2uy2+2uz2=3(x2+y2+z2xyyzxz)2[(x+y+z)(x2+y2+z2xyyzxz)]2=3(x+y+z)2

  1. (x+y+z)2u=9(x+y+z)2

We know,

(x+y+z)2u=(x+y+z)(x+y+z)u=(x+y+z)(ux+uy+uz)

From (18.1), we get

=(x+y+z)(3x+y+z)=x(3x+y+z)+y(3x+y+z)+z(3x+y+z)=3(x+y+z)2+3(x+y+z)2+3(x+y+z)2=9(x+y+z)2

Hence, proved.

18.1.9 Question 9

Show that 2ux2+2uy2=0 if

  1. u=log(x2+y2)

2ux2=x(ux)=x(1x2+y2×2x)=(x2+y2).22x.2x(x2+y2)2=2y22x2(x2+y2)2

2uy2=y(uy)=y(1x2+y2×2y)=(x2+y2).22y.2y(x2+y2)2=2x22y2(x2+y2)2

So, now

2ux2+2uy2=2y22x2+2x22y2(x2+y2)2=0

  1. u=tan1(yx)

2ux2=x(ux)=x[11+(y/x)2×yx2]=x(yx2+y2)=0(y).2x(x2+y2)2=2xy(x2+y2)2

2uy2=y(uy)=y[11+(y/x)2×1x]=y(xx2+y2)=0x.2y(x2+y2)2=2xy(x2+y2)2

So, now

2ux2+2uy2=2xy2xy(x2+y2)2=0

  1. u=ex(xcosyysiny)

2ux2=x(ux)=x(ex(cosy)+(xcosyysiny).ex)=x[ex(cosy+xcosyysiny)]=ex(cosy)+(cosy+xcosyysiny).ex=ex(2cosy+xcosyysiny)

2uy2=y(uy)=y(ex(xsiny(ycosy+siny))+0)=y(ex(xsinyycosysiny))=ex(xcosy(ysiny+cosy)cosy)+0=ex(xcosy+ysiny2cosy)

So, now

2ux2+2uy2=ex(2cosy+xcosyysinyxcosy+ysiny2cosy)=0

18.1.10 Question 10

If x=rcosθ,y=rsinθ, prove that

2rx2+2ry2=1r[(rx)2+(ry)2]

Here,

x2+y2=r2cos2θ+r2sin2θx2+y2=r2r2=x2+y2

Now taking partial derivative w.r.t x,

2rrx=2xrx=xr2rx2=r.1x.rxr2=rx.xrr2=r2x2r32rx2=y2r3

Again taking partial derivative of r2=x2+y2 w.r.t y,

2rry=2yry=yr2ry2=r.1y.ryr2=ry2rr2=r2y2r32ry2=x2r3

Now,

2rx2+2ry2=y2r3+x2r3=x2+y2r3

Substituting r2=x2+y2, L.H.S. evaluates to

2rx2+2ry2=r2r3=1r

Now evaluating R.H.S,

(rx)2=x2r2(ry)2=y2r2

1r[(rx)2+(ry)2]=1r[x2r2+y2r2]=1r[x2+y2r2]=1rr2r2=1r

Thus, L.H.S.=R.H.S. Hence proved.