Chapter 18 Partial Differentiation-I

18.1 Exercise 10(i)

18.1.1 Question 1

Find fx$f_x$ and fy$f_y$ of the following functions f(x,y)$f(x,y)$

1. 3x2−4xy+2y2$3x^2-4xy+2y^2$

fx=6x−4yfy=4y−4x

2. x2−y2x2+y2

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

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

3. sin−1(ax+by)

fx=a√1−(ax+by)2fy=b√1−(ax+by)2

4. 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=∂f∂x=2ax+2hy+2gfxx=2afy=2hx+2by+2ffyy=2bfyx=∂∂y(∂f∂x)=2hfxy=∂∂x(∂f∂y)=2h

2. xcosy+ycosx

fx=∂f∂x=cosy−ysinxfxx=∂2f∂x2=−ycosxfy=−xsiny+cosxfyy=−xcosyfyx=∂∂y(∂f∂x)=−siny−sinxfxy=∂∂x(∂f∂y)=−siny−sinx

3. tan−12xyx2−y2

fx=11+(2xyx2−y2)2×(x2−y2).2y−2xy.2x(x2−y2)2=(x2−y2)2(x2−y2)2+4x2y2×2x2y−2y3−4x2y(x2−y2)2=−2y3−2x2y(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=−2x2−2y2+4y2(x2+y2)2=2(y2−x2)(x2+y2)2

fy=11+(2xyx2−y2)2×(x2−y2).2x−2xy.(−2y)(x2−y2)2=2x3−2xy2+4xy2(x2−y2)2+4x2y2=2x(x2+y2)(x2+y2)2fy=2xx2+y2

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

fxy=2(x2+y2)−2x.2x(x2+y2)2=2x2+2y2−4x2(x2+y2)2

18.1.3 Question 3

Verify that ∂2u∂x∂y=∂2u∂y∂x when u is

1. ex2+xy+y2

Here u=ex2+xy+y2.

∂2u∂x∂y=∂∂x(∂u∂y)=∂∂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))

∂2u∂y∂x=∂∂y(∂u∂x)=∂∂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 ∂2u∂x∂y=∂2u∂y∂x. Hence, verified.

2. sin−1yx

Here, u=sin−1yx.

∂2u∂x∂y=∂∂x(∂u∂y)=∂∂x(1√1−(y/x)2×1x)=∂∂x(x√x2−y2×1x)=∂∂x(1√x2−y2)=0−1×12√x2−y2×2x(x2−y2)∂2u∂x∂y=−x(x2−y2)3/2

∂2u∂y∂x=∂∂y(∂u∂x)=∂∂y(1√1−(y/x)2×(−yx2))=∂∂y(−yx√x2−y2)=x√x2−y2.(−1)−(−y)[x2√x2−y2×(−2y)+0]x2(x2−y2)=−x√x2−y2−xy2√x2−y2x2(x2−y2)=−(x2−y2)−y2x(x2−y2)3/2=−x2x(x2−y2)3/2∂2u∂y∂x=−x(x2−y2)3/2

Thus ∂2u∂x∂y=∂2u∂y∂x. 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

2. 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

3. If u=yz+zx+xy, prove that x∂u∂x+y∂u∂y+z∂u∂z=0

Taking partial derivatives,

∂u∂x=−zx2+1y∂u∂y=1z−xy2∂u∂z=−yz2+1x

So,

x∂u∂x+y∂u∂y+z∂u∂z=x(−zx2+1y)+y(1z−xy2)+z(−yz2+1x)=−zx+xy+yz−xy−yz+zx=0

4. If f(x,y,z)=ex/y+ey/z+ez/x, show that x∂f∂x+y∂f∂y+z∂f∂z=0

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

∂f∂x=ex/y.1y+0+ez/x.(−zx2)=ex/yy−ez/x.zx2∂f∂y=ex/y.(−xy2)+ey/z.1z+0=−xex/yy2+ey/zz∂f∂z=0+ey/z.(−yz2)+ez/x.1x=−yey/zz2+ez/xx

So,

x∂f∂x+y∂f∂y+z∂f∂z=x(ex/yy−ez/x.zx2)+y(−xex/yy2+ey/zz)+z(−yey/zz2+ez/xx)=xex/yy−ez/xzx−xex/yy+yey/zz−yey/zz+zez/xx=0

18.1.5 Question 5

If z=tan(y+ax)+(y−ax)3/2, find the value of ∂2z∂x2−a2∂2z∂y2.

∂2z∂x2=∂∂x(∂z∂x)=∂∂x(sec2(y+ax).a+32(y−ax)1/2.(−a))=2sec(y+ax)sec(y+ax)tan(y+ax).a.a−32a×12√y−ax×(−a)∂2z∂x2=2a2sec2(y+ax)tan(y+ax)+3a24√y−ax

∂2z∂y2=∂∂y(∂z∂y)=∂∂y(sec2(y+ax).1+32(y−ax)1/2.1)=2sec(y+ax)sec(y+ax)tan(y+ax).1+32×12√y−ax.1∂2z∂y2=2sec2(y+ax)tan(y+ax)+34√y−ax

∂2z∂x2−a2∂2z∂y2=2a2sec2(y+ax)tan(y+ax)+3a24√y−ax−a2(2sec2(y+ax)tan(y+ax)+34√y−ax)∂2z∂x2−a2∂2z∂y2=0

18.1.6 Question 6

1. If u=log√x2+y2+z2, show that (x2+y2+z2)(∂2u∂x2+∂2u∂y2+∂2u∂z2)=1

∂2u∂x2=∂∂x(∂u∂x)=∂∂x(1√x2+y2+z2×12√x2+y2+z2×2x)=∂∂x(xx2+y2+z2)=(x2+y2+z2).1−x.2x(x2+y2+z2)2=−x2+y2+z2(x2+y2+z2)2

Similarly,

∂2u∂y2=∂∂x(∂u∂x)=x2−y2+z2(x2+y2+z2)2∂2u∂z2=∂∂z(∂u∂z)=x2+y2−z2(x2+y2+z2)2

So,

(x2+y2+z2)(∂2u∂x2+∂2u∂y2+∂2u∂z2)=(x2+y2+z2)(−x2+y2+z2+x2−y2+z2+x2+y2−z2(x2+y2+z2)2)=(x2+y2+z2)(x2+y2+z2)(x2+y2+z2)2=1

2. If u=log(x2+y2+z2), prove that x∂2u∂y∂z=y∂2u∂z∂x=z∂2u∂x∂y

∂2u∂y∂z=∂∂y(∂u∂z)∂∂y(1x2+y2+z2×2z)=0−2x.2y(x2+y2+z2)2=−4yz(x2+y2+z2)2

Similarly,

∂2u∂z∂x=∂∂z(∂u∂x)=−4zx(x2+y2+z2)2∂2u∂x∂y=∂∂x(∂u∂y)=−4xy(x2+y2+z2)2

Thus,

x∂2u∂y∂z=y∂2u∂z∂x=z∂2u∂x∂y=−4xyz(x2+y2+z2)2

3. 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 ∂3u∂x∂y∂z=(1+3xyz+x2y2z2)exyz

∂3u∂x∂y∂z=∂2∂x∂y(∂u∂z)=∂2∂x∂y(exyz.xy)=∂∂x∂∂y(exyz.xy)=∂∂x(exyz.x+xyexyz.xz)=∂∂x(exyz.x)+∂∂x(x2yz.exyz)=exyz.1+xexyz.yz+exyz.2xyz+x2yz.exyz.yz=exyz+3xyzexyz+x2y2z2exyz∂3u∂x∂y∂z=(1+3xyz+x2y2z2)exyz

Hence, proved.

18.1.8 Question 8

If u=log(x3+y3+z3−3xyz), show that

1. ∂u∂x+∂u∂y+∂u∂z=3x+y+z

∂u∂x=1x3+y3+z3−3xyz×(3x2−3yz)∂u∂y=1x3+y3+z3−3xyz×(3y2−3xz)∂u∂z=1x3+y3+z3−3xyz×(3z2−3xy)

So,

∂u∂x+∂u∂y+∂u∂z=3(x2+y2+z2)−3(xy+yz+xz)x3+y3+z3−3xyz=3(x2+y2+z2−xy−yz−xz)x3+y3+z3−3xyz

The expansion of (x+y+z)(x2+y2+z2−xy−yz−xz) yields x3+y3+z3−3xyz. Hence, above equation becomes

=3(x2+y2+z2−xy−yz−xz)(x+y+z)(x2+y2+z2−xy−yz−xz)

∂u∂x+∂u∂y+∂u∂z=3x+y+z

2. ∂2u∂x2+∂2u∂y2+∂2u∂z2=−3(x+y+z)2

∂2u∂x2=∂∂x(∂u∂x)=∂∂x(3x2−3yzx3+y3+z3−3xyz)=(x3+y3+z3−3xyz).6x−(3x2−3yz)(3x2−3yz)(x3+y3+z3−3xyz)2=6x4+6xy3+6xz3−18x2yz−9x4+18x2yz−9y2z2(x3+y3+z3−3xyz)2=−3x4+6xy3+6xz3−9y2z2(x3+y3+z3−3xyz)2

∂2u∂y2=∂∂y(∂u∂y)=∂∂y(3y2−3xzx3+y3+z3−3xyz)=(x3+y3+z3−3xyz).6y−(3y2−3xz)(3y2−3xz)(x3+y3+z3−3xyz)2=6x3y+6y4+6yz3−18xy2z−9y4+18xy2z−9x2z2(x3+y3+z3−3xyz)2=6x3y−3y4+6yz3−9x2z2(x3+y3+z3−3xyz)2

∂2u∂z2=∂∂z(∂u∂z)=(x3+y3+z3−3xyz).6z−(3z2−3xy)(3z2−3xy)(x3+y3+z3−3xyz)2=6x3z+6y3z+6z4−18xyz2−9z4+18xyz2−9x2y2(x3+y3+z3−3xyz)2=6x3z+6y3z−3z4−9x2y2(x3+y3+z3−3xyz)2

Now,

∂2u∂x2+∂2u∂y2+∂2u∂z2=

−3x4+6xy3+6xz3−9y2z2+6x3y−3y4+6yz3−9x2z2+6x3z+6y3z−3z4−9x2y2(x3+y3+z3−3xyz)2=−3x4−3y4−3z4+6xy3+6xz3+6x3y+6yz3+6x3z+6y3z−9y2z2−9x2z2−9x2y2(x3+y3+z3−3xyz)2=−3(x4+y4+z4−2xy3−2xz3−2x3y−2yz3−2x3z−2y3z+3y2z2+3x2z2+3x2y2)(x3+y3+z3−3xyz)2

The expansion of (x+y+z)(x2+y2+z2−xy−yz−xz) yields x3+y3+z3−3xyz. And the expansion of (x2+y2+z2−xy−yz−xz)2 yields x4+y4+z4−2xy3−2xz3−2x3y−2yz3−2x3z−2y3z+3y2z2+3x2z2+3x2y2.

So, substituting these values in above equation we get,

∂2u∂x2+∂2u∂y2+∂2u∂z2=−3(x2+y2+z2−xy−yz−xz)2[(x+y+z)(x2+y2+z2−xy−yz−xz)]2=−3(x+y+z)2

3. (∂∂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)(∂u∂x+∂u∂y+∂u∂z)

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 ∂2u∂x2+∂2u∂y2=0 if

1. u=log(x2+y2)

∂2u∂x2=∂∂x(∂u∂x)=∂∂x(1x2+y2×2x)=(x2+y2).2−2x.2x(x2+y2)2=2y2−2x2(x2+y2)2

∂2u∂y2=∂∂y(∂u∂y)=∂∂y(1x2+y2×2y)=(x2+y2).2−2y.2y(x2+y2)2=2x2−2y2(x2+y2)2

So, now

∂2u∂x2+∂2u∂y2=2y2−2x2+2x2−2y2(x2+y2)2=0

2. u=tan−1(yx)

∂2u∂x2=∂∂x(∂u∂x)=∂∂x[11+(y/x)2×−yx2]=∂∂x(−yx2+y2)=0−(−y).2x(x2+y2)2=2xy(x2+y2)2

∂2u∂y2=∂∂y(∂u∂y)=∂∂y[11+(y/x)2×1x]=∂∂y(xx2+y2)=0−x.2y(x2+y2)2=−2xy(x2+y2)2

So, now

∂2u∂x2+∂2u∂y2=2xy−2xy(x2+y2)2=0

3. u=ex(xcosy−ysiny)

∂2u∂x2=∂∂x(∂u∂x)=∂∂x(ex(cosy)+(xcosy−ysiny).ex)=∂∂x[ex(cosy+xcosy−ysiny)]=ex(cosy)+(cosy+xcosy−ysiny).ex=ex(2cosy+xcosy−ysiny)

∂2u∂y2=∂∂y(∂u∂y)=∂∂y(ex(−xsiny−(ycosy+siny))+0)=∂∂y(ex(−xsiny−ycosy−siny))=ex(−xcosy−(−ysiny+cosy)−cosy)+0=ex(−xcosy+ysiny−2cosy)

So, now

∂2u∂x2+∂2u∂y2=ex(2cosy+xcosy−ysiny−xcosy+ysiny−2cosy)=0

18.1.10 Question 10

If x=rcosθ,y=rsinθ, prove that

∂2r∂x2+∂2r∂y2=1r[(∂r∂x)2+(∂r∂y)2]

Here,

x2+y2=r2cos2θ+r2sin2θx2+y2=r2r2=x2+y2

Now taking partial derivative w.r.t x,

2r∂r∂x=2x∂r∂x=xr∂2r∂x2=r.1−x.∂r∂xr2=r−x.xrr2=r2−x2r3∂2r∂x2=y2r3

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

2r∂r∂y=2y∂r∂y=yr∂2r∂y2=r.1−y.∂r∂yr2=r−y2rr2=r2−y2r3∂2r∂y2=x2r3

Now,

∂2r∂x2+∂2r∂y2=y2r3+x2r3=x2+y2r3

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

∂2r∂x2+∂2r∂y2=r2r3=1r

Now evaluating R.H.S,

(∂r∂x)2=x2r2(∂r∂y)2=y2r2

1r[(∂r∂x)2+(∂r∂y)2]=1r[x2r2+y2r2]=1r[x2+y2r2]=1rr2r2=1r

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