Chapter 19 Partial Differentiation-II

Review this video on homogeneous functions and Euler’s theorem before working on the questions.

19.1 Exercise 10(ii)

19.1.1 Question 1

Verify Euler’s theorem for the following functions:

1. u=ax2+2hxy+by2

The given equation is a homogeneous function of degree n=2, so we expect x∂u∂x+y∂u∂y=nu=2u according to Euler’s theorem.

x∂u∂x+y∂u∂y=x(2ax+2hy)+y(2hx+2by)=2ax2+2hxy+2hxy+2by2=2(ax2+2hxy+by2)=2u

Thus, Euler’s theorem is verified.

2. u=x3+y3+z3−3xyz

This is a homogeneous function of degree n=3 with three independent variables f(x,y,z). So from Euler’s theorem, we expect

x∂u∂x+y∂u∂y+z∂u∂z=nu=3u

∂u∂x=3x2−3yz∂u∂y=3y2−3xz∂u∂z=3z2−3xy

So,

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

Thus, Euler’s theorem verified.

3. u=(x2+y2)1/3

This is a homogeneous function of degree n=2/3. So we expect x∂u∂x+y∂u∂y=23u.

x∂u∂x+y∂u∂y=x13(x2+y2)−2/3×2x+y13(x2+y2)−2/3×2y=23x2(x2+y2)−2/3+23y2(x2+y2)−2/3=23(x2+y2)−2/3(x2+y2)=23(x2+y2)1/3=23u

4. u=xntan−1(yx) (TU 2061, 2065)

Here, f(x,y)=u=xntan−1(yx). Then f(λx,λy)=(λx)ntan−1(λyλx)=(λ)nxntan−1(yx)=(λ)nf(x,y).

Thus the given function is a homogeneous function of degree n. So from Euler’s theorem, we expect x∂u∂x+y∂u∂y=nu=nxntan−1(yx).

x∂u∂x+y∂u∂y=x(xn11+(y/x)2×(−y/x2)+tan−1(yx)nxn−1)+y(xn11+(y/x)2×1x)=−y1+(y/x)2×xn−1+tan−1(yx)nxn+y1+(y/x)2×xn−1=nxntan−1(yx)=nu

Hence, Euler’s theorem is verified.

22. u=xf(yx)

This is a homogeneous function of degree 1. So we expect

x∂u∂x+y∂u∂y=1u=u=xf(yx)

∂u∂x=xf′(yx)×(−yx2)+f(yx)∂u∂y=x(f′(yx)×1x)

x∂u∂x+y∂u∂y=x(xf′(yx)×(−yx2)+f(yx))+yx(f′(yx)×1x)=−yf′(yx)+xf(yx)+yf′(yx)=xf(yx)=1×u=u

6. u=x1/4+y1/4x1/5+y1/5

The function can be written as

u=x1/4+y1/4x1/5+y1/5=x1/4(1+(y/x)1/4)x1/5(1+(y/x)1/5)=x1/20(1+(y/x)1/4)(1+(y/x)1/5)u=x1/20ϕ(yx)

Thus this is a homogeneous function of degree n=1/20. So we expect

x∂u∂x+y∂u∂y=120u

∂u∂x=x1/20ϕ′(y/x)(−y/x2)+ϕ(y/x)×(120)x120−1∂u∂y=x1/20ϕ′(y/x)×1x

x∂u∂x+y∂u∂y=x(x1/20ϕ′(y/x)(−y/x2)+ϕ(y/x)×(120)x120−1)+y(x1/20ϕ′(y/x)×1x)=120x1/20ϕ(y/x)=120u

Hence, Euler’s theorem is verified.

7. u=xnsin(yx)

Here, f(x,y)=u=xnsin(yx).

f(λx,λy)=(λx)nsin(λyλx)=(λ)nxnsin(yx)=(λ)nf(x,y). Thus the given function is a homogeneous one with degree n.

So we need to prove

x∂u∂x+y∂u∂y=nu=nxnsin(yx)

∂u∂x=xncos(y/x)×(−y/x2)+sin(y/x)nxn−1∂u∂y=xncos(y/x)×(1/x)

x∂u∂x+y∂u∂y=x(xncos(y/x)×(−y/x2)+sin(y/x)nxn−1)+y(xncos(y/x)×(1/x))=−xn−1ycos(y/x)+nxnsin(y/x)+xn−1ycos(y/x)=nxnsin(yx)=nu

Thus, Euler’s theorem is verified.

19.1.2 Question 2

1. If u=tan−1x3+y3x−y, x≠y, prove that x∂u∂x+y∂u∂y=sin2u. (TU 2054)

u is not a homogeneous function of x and y.

Let z=tanu=x3+y3x−y=x3(1+(y/x)3)x(1−(y/x))=x2(1+(y/x)3)1−(y/x).

Thus z is a homogeneous function of degree n=2. According to Euler’s theorem,

x∂z∂x+y∂z∂y=2z

∂z∂x=∂z∂u×∂u∂x=sec2u∂u∂x∂z∂y=∂z∂u×∂u∂y=sec2u∂u∂y

Now,

x∂z∂x+y∂z∂y=2zx(sec2u∂u∂x)+y(sec2u∂u∂y)=2tanux∂u∂x+y∂u∂y=2tanusec2u=2sinucosu×cos2u=2sinucosux∂u∂x+y∂u∂y=sin2u

2. If u=sin−1x2+y2x+y, prove that x∂u∂x+y∂u∂y=tanu. (TU 2058, 2062, 2066)

Let z=sinu=x2+y2x+y. This function is a homogeneous one with degree 1.

Now,

x∂z∂x+y∂z∂y=1×z

∂z∂x=∂z∂u×∂u∂x=cosu∂u∂x∂z∂y=∂z∂u×∂u∂y=cosu∂u∂y

Now,

x∂z∂x+y∂z∂y=1×zxcosu∂u∂x+ycosu∂u∂y=sinux∂u∂x+y∂u∂y=tanu

3. If sinv=√x−√y√x+√y, prove that x∂v∂x+y∂v∂y=0.

Let

z=sinv=√x−√y√x+√y=√x(1−(y/x)1/2)√x(1+(y/x)1/2)=x0(1−(y/x)1/2)(1+(y/x)1/2)=x0f(yx)

Thus z is a homogeneous function of degree 0.

According to Euler’s theorem,

x∂z∂x+y∂z∂y=0.z

∂z∂x=∂z∂v×∂v∂x=cosv∂v∂x∂z∂y=∂z∂v×∂v∂y=cosv∂v∂y

Now,

x∂z∂x+y∂z∂y=0xcosv∂v∂x+ycosv∂v∂y=0x∂v∂x+y∂v∂y=0

Hence proved.

4. If v=logx2+y2x+y, prove that x∂v∂x+y∂v∂y=1.

Let z=ev=x2+y2x+y. Thus z is a homogeneous function of degree 1. Thus according to Euler’s theorem,

x∂z∂x+y∂z∂y=1.z

∂z∂x=∂z∂v×∂v∂x=ev∂v∂x∂z∂y=∂z∂v×∂v∂y=ev∂v∂y

Substituting these partial derivative values in Euler’s theorem,

x∂z∂x+y∂z∂y=1.zxev∂v∂x+yev∂v∂y=1.evx∂v∂x+y∂v∂y=1

Hence proved.

19.1.3 Question 3

1. Find dzdt if

   1. z=xlogy,x=t2,y=et

   This is a composite function. So,

   dzdt=∂z∂x×dxdt+∂z∂y×dydt

   Now,

   ∂z∂x=logy∂z∂y=xydxdt=2tdydt=et

   Now substituting these values in the derivative of composite function

   dzdt=∂z∂x×dxdt+∂z∂y×dydt=logy.2t+xy.et=t.2t+t2et.etdzdt=3t2

   2. z=x2sin(ax+by),x=2t+1,y=t2−1

   The given function is a composite function.

   ∂z∂x=x2cos(ax+by)×a+sin(ax+by)×2x∂z∂y=x2cos(ax+by)×bdxdt=2dydt=2t

   Now substituting these values in the derivative of composite function

   dzdt=∂z∂x×dxdt+∂z∂y×dydt

   dzdt=2(ax2cos(ax+by)+2xsin(ax+by))+2btx2cos(ax+by)=2x2cos(ax+by)(a+bt)+4xsin(ax+by)=2(a+bt)(2t+1)2cos{a(2t+1)+b(t2−1)}+4(2t+1)sin{a(2t+1)+b(t2−1)}

2. Find dzdx if z=(y+x)exy,y=1x2

Here z=(y+x)exy

∂z∂x=(y+x)exy.y+exy∂z∂y=(y+x)exy.x+exydydx=−2x3

So the total derivative is given by

dzdx=∂z∂x+∂z∂y×dydx=(y+x)exyy+exy+[(y+x)exyx+exy]×(−2x3)=exy(y2+xy+1−2yx2−2x−2x3)=e1/x(1x4+1x+1−2x4−2x−2x3)=e1/x(1−1x−2x3−1x4)

19.1.4 Question 4

If z=f(x,y) and if x=eu+e−v,y=e−u−ev, prove that ∂z∂u−∂z∂v=x∂z∂x−y∂z∂y.

Here, the function z is a composite function of two variables u and v. So,

∂z∂u=∂z∂x.∂x∂u+∂z∂y.∂y∂u=∂z∂x.eu−∂z∂y.e−u∂z∂v=∂z∂x.∂x∂v+∂z∂y.∂y∂v=−∂z∂x.e−v−∂z∂y.ev

Now,

Subtracting ∂z∂v from ∂z∂u,

∂z∂u−∂z∂v=∂z∂x.eu−∂z∂y.e−u+∂z∂x.e−v+∂z∂y.ev=∂z∂x(eu+e−v)−∂z∂y(e−u−ev)

Substituting x=eu+e−v,y=e−u−ev, we have

∂z∂u−∂z∂v=x∂z∂x−y∂z∂y

19.1.5 Question 5

Find dydx in the following cases

1. x3+y3=3axy

Let f(x,y)=x3+y3−3axy=0. Then,

fx=3x2−3ayfy=3y2−3ax

So,

dydx=−fxfy=−3x2−3ay3y2−3axdydx=x2−ayax−y2

2. x2/3+y2/3=a2/3

Let f(x,y)=x2/3+y2/3−a2/3=0. Then,

fx=23x−1/3fy=23y−1/3

So,

dydx=−fxfy=−23x−1/323y−1/3dydx=−(yx)1/3

3. xy=yx

Let f(x,y)=xy−yx=0. Then taking partial derivative w.r.t x taking y as constant.

fx=yxy−1−yxlogy

Also taking partial derivative w.r.t y taking x as constant.

fy=xylogx−xyx−1

So,

dydx=−fxfy=−yxy−1−yxlogyxylogx−xyx−1=−yxyx−yxlogyyxlogx−xyyx=−yx−logylogx−xy=y−xlogyxx−ylogxydydx=y(y−xlogy)x(x−ylogx)

4. xpyq=(x+y)p+q

Let f(x,y)=xpyq−(x+y)p+q=0. Then

fx=pxp−1yq−(p+q)(x+y)p+q−1fy=qxpyq−1−(p+q)(x+y)p+q−1

So,

dydx=−fxfy=−pxp−1yq−(p+q)(x+y)p+q−1qxpyq−1−(p+q)(x+y)p+q−1=−pxxpyq−(p+q)xpyqx+yqyxpyq−(p+q)xpyqx+y=−px−p+qx+yqy−p+qx+y=−px+py−px−qxx(x+y)qx+qy−py−qyy(x+y)=py−qxx×ypy−qxdydx=yx

22. (tanx)y+(y)tanx=0

Here, f(x,y)=(tanx)y+(y)tanx=0.

Taking partial derivative w.r.t x setting y as constant,

∂f∂x=fx=y(tanx)y−1sec2x+logy.ytanx.sec2x

Also taking partial derivative w.r.t y setting x as constant,

∂f∂y=fy=(tanx)ylog(tanx)+tanx.ytanx−1

So,

dydx=−fxfy=−y(tanx)y−1sec2x+logy.ytanx.sec2x(tanx)ylog(tanx)+tanxytanx−1dydx=−sec2x(y(tanx)y−1+ytanxlogy)(tanx)ylog(tanx)+tanx.ytanx−1

19.1.6 Question 6

If x√1−y2+y√1−x2=a, show that

d2ydx2=−a(1−x2)3/2

The given equation is

x√1−y2+y√1−x2=a

which can be written as f(x,y)=x√1−y2+y√1−x2−a=0. Then

fx=√1−y2+y.12√1−x2×(−2x)=√1−y2−xy√1−x2fy=x2√1−y2×(−2y)+√1−x2=−xy√1−y2+√1−x2

Then,

dydx=−fxfy=−√1−y2−xy√1−x2−xy√1−y2+√1−x2=−(√1−y2√1−x2−xy)√1−x2×√1−y2−xy+√1−x2√1−y2dydx=−√1−y2√1−x2

Now,

d2ydx2=ddx(dydx)=ddx(−√1−y2√1−x2)=−√1−x212√1−y2.(−2ydydx)−√1−y2×12√1−x2×(−2x)(1−x2)=−−2y√1−x22√1−y2×(−√1−y2√1−x2)+2x√1−y22√1−x2(1−x2)=−y+x√1−y2√1−x2(1−x2)d2ydx2=−y√1−x2+x√1−y2(1−x2)3/2

Substituting numerator from (19.1), we get

d2ydx2=−a(1−x2)3/2