Chapter 18 Partial Differentiation-I

18.1 Exercise 10(i)

18.1.1 Question 1

Find \(f_x\) and \(f_y\) of the following functions \(f(x,y)\)

  1. \(3x^2-4xy+2y^2\)

\[\begin{equation*} \begin{split} f_x &= 6x-4y\\ f_y &= 4y -4x \end{split} \end{equation*}\]

  1. \(\frac{x^2-y^2}{x^2+y^2}\)

\[\begin{equation*} \begin{split} f_x &= \frac{(x^2 + y^2).2x - (x^2 - y^2).2x}{(x^2 + y^2)^2}\\ &= \frac{2x(x^2 + y^2 - x^2 + y^2)}{(x^2+y^2)^2}\\ &= \frac{4xy^2}{(x^2 + y^2)^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} f_y &= \frac{(x^2+y^2)\times (-2y)-(x^2 - y^2)\times 2y}{(x^2+y^2)^2}\\ &= \frac{2y(-x^2-y^2-x^2+y^2)}{(x^2+y^2)^2}\\ &= -\frac{4x^2y}{(x^2+y^2)^2} \end{split} \end{equation*}\]

  1. \(\sin^{-1}(ax+by)\)

\[\begin{equation*} \begin{split} f_x &= \frac{a}{\sqrt{1-(ax+by)^2}}\\ f_y &= \frac{b}{\sqrt{1-(ax+by)^2}}\\ \end{split} \end{equation*}\]

  1. \(\log(x^3+y^2)\)

\[\begin{equation*} \begin{split} f_x &= \frac{3x^2}{x^3 + y^2}\\ f_y &= \frac{2y}{x^3 + y^2} \end{split} \end{equation*}\]

18.1.2 Question 2

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

  1. \(ax^2 + 2hxy + by^2 + 2gx + 2fy + c\)

\[\begin{equation*} \begin{split} f_x = \frac{\partial f}{\partial x} &= 2ax + 2hy + 2g\\ f_{xx} &= 2a\\ f_y &= 2hx + 2by + 2f\\ f_{yy} &= 2b\\ f_{yx} = \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x} \right) &= 2h\\ f_{xy} = \frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y} \right) &= 2h \end{split} \end{equation*}\]

  1. \(x \cos y + y \cos x\)

\[\begin{equation*} \begin{split} f_x = \frac{\partial f}{\partial x} &= \cos y - y\sin x\\ f_{xx} = \frac{\partial ^2 f}{\partial x^2} &= -y \cos x\\ f_y &= -x\sin y + \cos x\\ f_{yy} &= -x\cos y\\ f_{yx} = \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x} \right) &= -\sin y - \sin x\\ f_{xy} = \frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y} \right) &= -\sin y - \sin x \end{split} \end{equation*}\]

  1. \(\tan ^{-1} \frac{2xy}{x^2 -y^2}\)

\[\begin{equation*} \begin{split} f_x &= \frac{1}{1 + \left(\frac{2xy}{x^2-y^2}\right)^2}\times \frac{(x^2 - y^2).2y - 2xy.2x}{(x^2-y^2)^2}\\ &= \frac{(x^2-y^2)^2}{(x^2-y^2)^2 + 4x^2y^2}\times \frac{2x^2y-2y^3-4x^2y}{(x^2-y^2)^2}\\ &= \frac{-2y^3-2x^2y}{(x^2+y^2)^2}\\ &= \frac{-2y(y^2+x^2)}{(x^2+y^2)^2}\\ f_x &= \frac{-2y}{x^2 + y^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} f_{xx} &= \frac{(x^2+y^2).0 - (-2y).2x}{(x^2 + y^2)^2}\\ &= \frac{4xy}{(x^2+y^2)^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} f_{yx} &= \frac{(x^2+y^2).(-2)-(-2y).2y}{(x^2+y^2)^2}\\ &= \frac{-2x^2-2y^2 + 4y^2}{(x^2+y^2)^2}\\ &= \frac{2(y^2-x^2)}{(x^2+y^2)^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} f_y &= \frac{1}{1 + \left(\frac{2xy}{x^2-y^2}\right)^2}\times \frac{(x^2 - y^2).2x - 2xy.(-2y)}{(x^2-y^2)^2}\\ &= \frac{2x^3 - 2xy^2 + 4xy^2}{(x^2-y^2)^2 + 4x^2y^2}\\ &= \frac{2x(x^2+y^2)}{(x^2+y^2)^2}\\ f_y &= \frac{2x}{x^2 + y^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} f_{yy} &= \frac{(x^2+y^2).0 - 2x.2y}{(x^2+y^2)^2}\\ &= \frac{-4xy}{(x^2+y^2)^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} f_{xy} &= \frac{2(x^2 + y^2)-2x.2x}{(x^2+y^2)^2}\\ &= \frac{2x^2 + 2y^2 -4x^2}{(x^2+y^2)^2} \end{split} \end{equation*}\]

18.1.3 Question 3

Verify that \(\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial^2 u}{\partial y \partial x}\) when \(u\) is

  1. \(e^{x^2 + xy + y^2}\)

Here \(u=e^{x^2 + xy + y^2}\).

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x \partial y} &= \frac{\partial }{\partial x} \left(\frac{\partial u}{\partial y} \right)\\ &= \frac{\partial }{\partial x} \left(e^{x^2 + xy + y^2}.(x+2y)\right)\\ &= e^{x^2 + xy + y^2}.1 + (x+2y)[e^{x^2 + xy + y^2}.(2x+y)]\\ &= e^{x^2 + xy + y^2}(1+(x+2y)(2x+y)) \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial y \partial x} &= \frac{\partial }{\partial y} \left(\frac{\partial u}{\partial x} \right)\\ &= \frac{\partial }{\partial y} (e^{x^2 + xy + y^2})(2x+y)\\ &= e^{x^2 + xy + y^2}.1 + (2x+y).e^{x^2 + xy + y^2}.(x+2y)\\ &= e^{x^2 + xy + y^2}(1+(x+2y)(2x+y)) \end{split} \end{equation*}\]

Thus \(\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial^2 u}{\partial y \partial x}\). Hence, verified.

  1. \(\sin^{-1}\frac{y}{x}\)

Here, \(u=\sin^{-1}\frac{y}{x}\).

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x \partial y} &= \frac{\partial }{\partial x} \left(\frac{\partial u}{\partial y} \right)\\ &= \frac{\partial }{\partial x} \left( \frac{1}{\sqrt{1-(y/x)^2}}\times \frac{1}{x} \right)\\ &= \frac{\partial }{\partial x} \left(\frac{x}{\sqrt{x^2-y^2}}\times \frac{1}{x} \right)\\ &= \frac{\partial }{\partial x} \left(\frac{1}{\sqrt{x^2-y^2}} \right)\\ &= \frac{0-1\times \frac{1}{2\sqrt{x^2-y^2}}\times 2x}{(x^2-y^2)}\\ \frac{\partial^2 u}{\partial x \partial y} &= -\frac{x}{(x^2-y^2)^{3/2}} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial y \partial x} &= \frac{\partial }{\partial y} \left(\frac{\partial u}{\partial x}\right)\\ &= \frac{\partial }{\partial y} \left( \frac{1}{\sqrt{1-(y/x)^2}}\times \left(\frac{-y}{x^2}\right)\right)\\ &= \frac{\partial }{\partial y} \left(\frac{-y}{x\sqrt{x^2-y^2}}\right)\\ &= \frac{x\sqrt{x^2-y^2}.(-1)-(-y)\left[\frac{x}{2\sqrt{x^2-y^2}}\times (-2y) + 0\right]}{x^2(x^2-y^2)}\\ &= \frac{-x\sqrt{x^2-y^2 - \frac{xy^2}{\sqrt{x^2-y^2}}}}{x^2(x^2-y^2)}\\ &= \frac{-(x^2-y^2)-y^2}{x(x^2-y^2)^{3/2}}\\ &= \frac{-x^2}{x(x^2-y^2)^{3/2}}\\ \frac{\partial^2 u}{\partial y \partial x} &= -\frac{x}{(x^2-y^2)^{3/2}} \end{split} \end{equation*}\]

Thus \(\frac{\partial^2 u}{\partial x \partial y} = \frac{\partial^2 u}{\partial y \partial x}\). Hence, verified.

18.1.4 Question 4

  1. If \(u=x^2 + y^2 + z^2\), show that \(x u_x + y u_y + z u_z = 2u\)

Here,

\[\begin{equation*} \begin{split} u_x &= 2x\\ u_y &= 2y\\ u_z &= 2z \end{split} \end{equation*}\]

Thus,

\[\begin{equation*} \begin{split} x u_x + y u_y + z u_z &= x.2x+y.2y+z.2z\\ &= 2(x^2+y^2+z^2)\\ &= 2u \end{split} \end{equation*}\]

  1. If \(u=x^2 y + y^2 z + z^2 x\), show that \(u_x + u_y + u_z = (x+y+z)^2\)

Here,

\[\begin{equation*} \begin{split} u_x &= 2xy + z^2\\ u_y &= x^2 + 2yz\\ u_z &= y^2 + 2zx \end{split} \end{equation*}\]

Thus,

\[\begin{equation*} \begin{split} u_x + u_y + u_z &= 2xy + z^2 + x^2 + 2yz + y^2 + 2zx\\ &= x^2 + 2xy + y^2 + z^2 + 2(x+y)z\\ &= (x+y)^2 + 2(x+y)z + z^2\\ &= (x+y+z)^2 \end{split} \end{equation*}\]

  1. If \(u=\frac{y}{z} + \frac{z}{x} + \frac{x}{y}\), prove that \(x\frac{\partial u}{\partial x} +y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} =0\)

Taking partial derivatives,

\[\begin{equation*} \begin{split} \frac{\partial u}{\partial x} &= \frac{-z}{x^2} + \frac{1}{y}\\ \frac{\partial u}{\partial y} &= \frac{1}{z} - \frac{x}{y^2}\\ \frac{\partial u}{\partial z} &= \frac{-y}{z^2} + \frac{1}{x} \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} + z\frac{\partial u}{\partial z} &= x\left(\frac{-z}{x^2} + \frac{1}{y}\right) + y\left(\frac{1}{z} - \frac{x}{y^2}\right) + \\ z\left(\frac{-y}{z^2} + \frac{1}{x}\right)\\ &= \frac{-z}{x} + \frac{x}{y} + \frac{y}{z} - \frac{x}{y} - \frac{y}{z} + \frac{z}{x}\\ &= 0 \end{split} \end{equation*}\]

  1. If \(f(x,y,z) = e^{x/y} + e^{y/z} + e^{z/x}\), show that \(x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} + z\frac{\partial f}{\partial z}=0\)

Here, \(f(x,y,z)=e^{x/y} + e^{y/z} + e^{z/x}\). Taking partial derivatives,

\[\begin{equation*} \begin{split} \frac{\partial f}{\partial x} &= e^{x/y}.\frac{1}{y} + 0 + e^{z/x}.\left(\frac{-z}{x^2}\right)\\ &= \frac{e^{x/y}}{y} - \frac{e^{z/x}.z}{x^2}\\ \frac{\partial f}{\partial y} &=e^{x/y}.\left(\frac{-x}{y^2}\right) + e^{y/z}.\frac{1}{z} + 0\\ &= \frac{-xe^{x/y}}{y^2} + \frac{e^{y/z}}{z}\\ \frac{\partial f}{\partial z} &= 0 + e^{y/z}.\left(\frac{-y}{z^2}\right) + e^{z/x}.\frac{1}{x}\\ &= \frac{-ye^{y/z}}{z^2} + \frac{e^{z/x}}{x} \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} + z\frac{\partial f}{\partial z} &= x\left(\frac{e^{x/y}}{y} - \frac{e^{z/x}.z}{x^2}\right) + y\left(\frac{-xe^{x/y}}{y^2} + \frac{e^{y/z}}{z}\right) + \\ z\left(\frac{-ye^{y/z}}{z^2} + \frac{e^{z/x}}{x}\right)\\ &= \frac{xe^{x/y}}{y} - \frac{e^{z/x}z}{x} -\frac{xe^{x/y}}{y} + \frac{ye^{y/z}}{z}-\frac{ye^{y/z}}{z} + \\ \frac{ze^{z/x}}{x}\\ &= 0 \end{split} \end{equation*}\]

18.1.5 Question 5

If \(z=\tan (y+ax) + (y-ax)^{3/2}\), find the value of \(\frac{\partial^2 z}{\partial x^2} -a^2 \frac{\partial^2 z}{\partial y^2}\).

\[\begin{equation*} \begin{split} \frac{\partial^2 z}{\partial x^2} &= \frac{\partial }{\partial x} \left(\frac{\partial z}{\partial x} \right)\\ &= \frac{\partial }{\partial x} \left(\sec^2(y+ax).a + \frac{3}{2}(y-ax)^{1/2}.(-a)\right)\\ &= 2\sec(y+ax)\sec(y+ax)\tan(y+ax).a.a - \\ \frac{3}{2}a \times \frac{1}{2\sqrt{y-ax}}\times (-a)\\ \frac{\partial^2 z}{\partial x^2} &= 2a^2\sec^2(y+ax)\tan(y+ax) + \frac{3a^2}{4\sqrt{y-ax}} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial^2 z}{\partial y^2} &= \frac{\partial }{\partial y} \left(\frac{\partial z}{\partial y} \right)\\ &= \frac{\partial }{\partial y} \left( \sec^2(y+ax).1 + \frac{3}{2}(y-ax)^{1/2}.1\right)\\ &= 2\sec(y+ax)\sec(y+ax)\tan(y+ax).1 + \\ \frac{3}{2}\times \frac{1}{2\sqrt{y-ax}}.1\\ \frac{\partial^2 z}{\partial y^2} &= 2\sec^2(y+ax)\tan(y+ax) + \frac{3}{4\sqrt{y-ax}} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial^2 z}{\partial x^2} -a^2 \frac{\partial^2 z}{\partial y^2} &= \\ 2a^2\sec^2(y+ax)\tan(y+ax) + \frac{3a^2}{4\sqrt{y-ax}} - \\ a^2\left(2\sec^2(y+ax)\tan(y+ax) + \frac{3}{4\sqrt{y-ax}}\right)\\ \frac{\partial^2 z}{\partial x^2} -a^2 \frac{\partial^2 z}{\partial y^2} &= 0 \end{split} \end{equation*}\]

18.1.6 Question 6

  1. If \(u=\log \sqrt{x^2 + y^2 + z^2}\), show that \((x^2 + y^2 + z^2)\left(\frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2} +\frac{\partial^2 u}{\partial z^2}\right)=1\)

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} &= \frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x} \right) \\ &= \frac{\partial }{\partial x} \left(\frac{1}{\sqrt{x^2+y^2+z^2}}\times \frac{1}{2\sqrt{x^2+y^2+z^2}}\times 2x\right)\\ &= \frac{\partial }{\partial x} \left(\frac{x}{x^2+y^2+z^2}\right)\\ &= \frac{(x^2+y^2+z^2).1 - x.2x}{(x^2+y^2+z^2)^2}\\ &= \frac{-x^2 + y^2 + z^2}{(x^2+y^2+z^2)^2} \end{split} \end{equation*}\]

Similarly,

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial y^2} &= \frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x} \right)\\ &= \frac{x^2 - y^2 + z^2}{(x^2+y^2+z^2)^2}\\ \frac{\partial^2 u}{\partial z^2} &= \frac{\partial }{\partial z}\left(\frac{\partial u}{\partial z} \right)\\ &= \frac{x^2 + y^2 - z^2}{(x^2+y^2+z^2)^2} \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} (x^2 + y^2 + z^2)\left(\frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2} +\frac{\partial^2 u}{\partial z^2}\right) &= \\ (x^2+y^2+z^2)\left(\frac{-x^2 + y^2 + z^2 + x^2 - y^2 + z^2 + x^2 + y^2 - z^2}{(x^2+y^2+z^2)^2}\right)\\ &= \\ \frac{(x^2+y^2+z^2)(x^2+y^2+z^2)}{(x^2+y^2+z^2)^2}\\ &= 1 \end{split} \end{equation*}\]

  1. If \(u=\log (x^2 + y^2 + z^2)\), prove that \(x\frac{\partial^2 u}{\partial y \partial z} = y\frac{\partial^2 u}{\partial z \partial x} = z \frac{\partial^2 u}{\partial x \partial y}\)

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial y \partial z} &= \frac{\partial }{\partial y} \left(\frac{\partial u}{\partial z} \right)\\ \frac{\partial }{\partial y} \left(\frac{1}{x^2+y^2+z^2}\times 2z\right)\\ &= \frac{0-2x.2y}{(x^2+y^2+z^2)^2}\\ &= \frac{-4yz}{(x^2+y^2+z^2)^2} \end{split} \end{equation*}\]

Similarly,

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial z \partial x} &= \frac{\partial }{\partial z} \left(\frac{\partial u}{\partial x} \right)\\ &= \frac{-4zx}{(x^2+y^2+z^2)^2}\\ \frac{\partial^2 u}{\partial x \partial y} &= \frac{\partial }{\partial x} \left(\frac{\partial u}{\partial y} \right)\\ &= \frac{-4xy}{(x^2+y^2+z^2)^2} \end{split} \end{equation*}\]

Thus,

\[\begin{equation*} \begin{split} x\frac{\partial^2 u}{\partial y \partial z} = y\frac{\partial^2 u}{\partial z \partial x} = z \frac{\partial^2 u}{\partial x \partial y} &=\\ \frac{-4xyz}{(x^2+y^2+z^2)^2} \end{split} \end{equation*}\]

  1. If \(V=f(xyz)\), show that \(xV_x = yV_y = zV_z\) and \(x^2 V_{xx} = y^2 V_{yy} = z^2 V_{zz}\)

\[\begin{equation*} \begin{split} V_x &= f'(xyz).yz\\ V_y &= f'(xyz).xz\\ V_z &= f'(xyz).xy \end{split} \end{equation*}\]

Again,

\[\begin{equation*} \begin{split} V_{xx} &= f''(xyz).yz.yz + 0\\ &= f''(xyz)y^2z^2 \end{split} \end{equation*}\]

Similarly,

\[\begin{equation*} \begin{split} V_{yy} &= f''(xyz)x^2z^2\\ V_{zz} &= f''(xyz)x^2y^2 \end{split} \end{equation*}\]

Thus,

\[\begin{equation*} \begin{split} xV_x = yV_y &= zV_z = f'(xyz)xyz \end{split} \end{equation*}\]

and,

\[\begin{equation*} \begin{split} x^2 V_{xx} = y^2 V_{yy} = z^2 V_{zz} &= \\ f''(xyz)x^2y^2z^2 \end{split} \end{equation*}\]

18.1.7 Question 7

If \(u=e^{xyz}\), prove that \(\frac{\partial^3 u}{\partial x \partial y \partial z}=(1 + 3xyz + x^2y^2z^2) e^{xyz}\)

\[\begin{equation*} \begin{split} \frac{\partial^3 u}{\partial x \partial y \partial z} &= \frac{\partial^2}{\partial x \partial y}\left(\frac{\partial u}{\partial z} \right) \\ &= \frac{\partial^2}{\partial x \partial y}\left(e^{xyz}.xy \right)\\ &= \frac{\partial }{\partial x} \frac{\partial }{\partial y} (e^{xyz}.xy)\\ &= \frac{\partial }{\partial x} (e^{xyz}.x +xye^{xyz}.xz)\\ &= \frac{\partial }{\partial x} (e^{xyz}.x) + \frac{\partial }{\partial x} (x^2yz.e^{xyz})\\ &= e^{xyz}.1 + xe^{xyz}.yz + e^{xyz}.2xyz + x^2yz.e^{xyz}.yz\\ &= e^{xyz} + 3xyze^{xyz} + x^2y^2z^2e^{xyz}\\ \frac{\partial^3 u}{\partial x \partial y \partial z} &= (1 + 3xyz + x^2y^2z^2) e^{xyz} \end{split} \end{equation*}\]

Hence, proved.

18.1.8 Question 8

If \(u=\log (x^3 + y^3 + z^3 - 3xyz)\), show that

  1. \(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} =\frac{3}{x+y+z}\)

\[\begin{equation*} \begin{split} \frac{\partial u}{\partial x} &= \frac{1}{x^3 + y^3 + z^3 -3xyz}\times (3x^2-3yz)\\ \frac{\partial u}{\partial y} &= \frac{1}{x^3 + y^3 + z^3 -3xyz}\times (3y^2 - 3xz)\\ \frac{\partial u}{\partial z} &= \frac{1}{x^3 + y^3 + z^3 -3xyz}\times (3z^2 - 3xy) \end{split} \end{equation*}\]

So,

\[\begin{equation*} \begin{split} \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} &= \\ \frac{3(x^2+y^2+z^2)-3(xy+yz+xz)}{x^3 + y^3 + z^3 -3xyz}\\ &= \frac{3(x^2+y^2+z^2-xy-yz-xz)}{x^3 + y^3 + z^3 -3xyz} \end{split} \end{equation*}\]

The expansion of \((x+y+z)(x^2+y^2+z^2-xy-yz-xz)\) yields \(x^3 + y^3 + z^3 -3xyz\). Hence, above equation becomes

\[\begin{equation*} \begin{split} &= \frac{3(x^2+y^2+z^2-xy-yz-xz)}{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} &= \frac{3}{x+y+z} \end{split} \tag{18.1} \end{equation*}\]

  1. \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} =-\frac{3}{(x+y+z)^2}\)

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} &= \frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x} \right) \\ &= \frac{\partial }{\partial x} \left(\frac{3x^2-3yz}{x^3 + y^3 + z^3 -3xyz}\right)\\ &= \frac{(x^3 + y^3 + z^3 -3xyz).6x - (3x^2 -3yz)(3x^2-3yz)}{(x^3 + y^3 + z^3 -3xyz)^2}\\ &= \frac{6x^4 + 6xy^3 + 6xz^3 -18x^2yz - 9x^4 + 18x^2yz -9y^2z^2}{(x^3 + y^3 + z^3 -3xyz)^2}\\ &= \frac{-3x^4+6xy^3+6xz^3-9y^2z^2}{(x^3 + y^3 + z^3 -3xyz)^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial y^2} &= \frac{\partial }{\partial y} \left(\frac{\partial u}{\partial y} \right)\\ &= \frac{\partial }{\partial y} \left(\frac{3y^2-3xz}{x^3 + y^3 + z^3 -3xyz}\right)\\ &= \frac{(x^3 + y^3 + z^3 -3xyz).6y - (3y^2-3xz)(3y^2-3xz)}{(x^3 + y^3 + z^3 -3xyz)^2}\\ &= \frac{6x^3y + 6y^4 + 6yz^3 - 18xy^2z-9y^4+18xy^2z-9x^2z^2}{(x^3 + y^3 + z^3 -3xyz)^2}\\ &= \frac{6x^3y - 3y^4 + 6yz^3 - 9x^2z^2}{(x^3 + y^3 + z^3 -3xyz)^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial z^2} &= \frac{\partial }{\partial z} \left(\frac{\partial u}{\partial z} \right)\\ &= \frac{(x^3 + y^3 + z^3 -3xyz).6z - (3z^2-3xy)(3z^2-3xy)}{(x^3 + y^3 + z^3 -3xyz)^2}\\ &= \frac{6x^3z + 6y^3z + 6z^4 -18xyz^2 - 9z^4 + 18xyz^2 -9x^2y^2}{(x^3 + y^3 + z^3 -3xyz)^2}\\ &= \frac{6x^3z + 6y^3z - 3z^4-9x^2y^2}{(x^3 + y^3 + z^3 -3xyz)^2} \end{split} \end{equation*}\]

Now,

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} &= \end{split} \end{equation*}\]

\[\begin{align} & \frac{-3x^4+6xy^3+6xz^3-9y^2z^2+6x^3y - 3y^4 + 6yz^3 \\ - 9x^2z^2 + 6x^3z + 6y^3z - 3z^4-9x^2y^2}{(x^3 + y^3 + z^3 -3xyz)^2}\\ &= \frac{-3x^4-3y^4-3z^4 + 6xy^3 + 6xz^3 + 6x^3y + 6yz^3 + \\ 6x^3z + 6y^3z -9y^2z^2-9x^2z^2-9x^2y^2}{(x^3 + y^3 + z^3 -3xyz)^2}\\ &= \frac{-3(x^4+y^4+z^4 - 2xy^3 - 2xz^3 - 2x^3y - 2yz^3 - 2x^3z - \\ 2y^3z + 3y^2z^2+3x^2z^2+3x^2y^2)}{(x^3 + y^3 + z^3 -3xyz)^2} \end{align}\]

The expansion of \((x+y+z)(x^2+y^2+z^2-xy-yz-xz)\) yields \(x^3 + y^3 + z^3 -3xyz\). And the expansion of \((x^2+y^2+z^2-xy-yz-xz)^2\) yields \(x^4+y^4+z^4 - 2xy^3 - 2xz^3 - 2x^3y - 2yz^3 - \\ 2x^3z - 2y^3z + 3y^2z^2+3x^2z^2+3x^2y^2\).

So, substituting these values in above equation we get,

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} &= \\ \frac{-3(x^2+y^2+z^2-xy-yz-xz)^2}{[(x+y+z)(x^2+y^2+z^2-xy-yz-xz)]^2}\\ &= -\frac{3}{(x+y+z)^2} \end{split} \end{equation*}\]

  1. \(\left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)^2 u = -\frac{9}{(x+y+z)^2}\)

We know,

\[\begin{equation*} \begin{split} \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)^2 u &= \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)\left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)u\\ &= \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z} \right)\left(\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} \right)\\ \end{split} \end{equation*}\]

From (18.1), we get

\[\begin{equation*} \begin{split} &= \left(\frac{\partial }{\partial x} + \frac{\partial }{\partial y} + \frac{\partial }{\partial z}\right)\left(\frac{3}{x+y+z}\right)\\ &= \frac{\partial }{\partial x} \left(\frac{3}{x+y+z}\right) + \frac{\partial }{\partial y} \left(\frac{3}{x+y+z}\right) + \frac{\partial }{\partial z} \left(\frac{3}{x+y+z}\right)\\ &= \frac{-3}{(x+y+z)^2} + \frac{-3}{(x+y+z)^2} + \frac{-3}{(x+y+z)^2}\\ &= -\frac{9}{(x+y+z)^2} \end{split} \end{equation*}\]

Hence, proved.

18.1.9 Question 9

Show that \(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}=0\) if

  1. \(u =\log(x^2 + y^2)\)

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} &= \frac{\partial }{\partial x} \left(\frac{\partial u}{\partial x} \right)\\ &= \frac{\partial }{\partial x} \left( \frac{1}{x^2+y^2}\times 2x\right)\\ &= \frac{(x^2+y^2).2 - 2x.2x}{(x^2+y^2)^2}\\ &= \frac{2y^2-2x^2}{(x^2+y^2)^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial y^2} &= \frac{\partial }{\partial y} \left(\frac{\partial u}{\partial y} \right)\\ &= \frac{\partial }{\partial y} \left( \frac{1}{x^2 + y^2} \times 2y\right)\\ &= \frac{(x^2 + y^2).2 - 2y.2y}{(x^2 + y^2)^2}\\ &= \frac{2x^2-2y^2}{(x^2+y^2)^2} \end{split} \end{equation*}\]

So, now

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} &= \frac{2y^2 - 2x^2 + 2x^2 -2y^2}{(x^2 + y^2)^2}\\ &= 0 \end{split} \end{equation*}\]

  1. \(u = \tan^{-1}\left(\frac{y}{x}\right)\)

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} &= \frac{\partial }{\partial x} \left(\frac{\partial u}{\partial x} \right)\\ &= \frac{\partial }{\partial x} \left[\frac{1}{1+(y/x)^2} \times \frac{-y}{x^2} \right]\\ &= \frac{\partial }{\partial x} \left(\frac{-y}{x^2 +y^2}\right)\\ &= \frac{0-(-y).2x}{(x^2+y^2)^2}\\ &= \frac{2xy}{(x^2+y^2)^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial y^2} &= \frac{\partial }{\partial y} \left(\frac{\partial u}{\partial y} \right)\\ &= \frac{\partial }{\partial y} \left[\frac{1}{1+(y/x)^2} \times \frac{1}{x} \right]\\ &= \frac{\partial }{\partial y} \left(\frac{x}{x^2+y^2}\right)\\ &= \frac{0-x.2y}{(x^2+y^2)^2}\\ &= \frac{-2xy}{(x^2+y^2)^2} \end{split} \end{equation*}\]

So, now

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} &= \frac{2xy-2xy}{(x^2 + y^2)^2}\\ &= 0 \end{split} \end{equation*}\]

  1. \(u = e^x(x \cos y - y \sin y)\)

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} &= \frac{\partial }{\partial x} \left(\frac{\partial u}{\partial x} \right)\\ &= \frac{\partial }{\partial x} \left( e^x(\cos y) + (x\cos y - y\sin y).e^x\right)\\ &= \frac{\partial }{\partial x} [e^x(\cos y + x\cos y - y\sin y)]\\ &= e^x(\cos y) + (\cos y + x\cos y -y\sin y).e^x\\ &= e^x(2\cos y + x\cos y - y\sin y) \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial y^2} &= \frac{\partial }{\partial y} \left(\frac{\partial u}{\partial y} \right)\\ &= \frac{\partial }{\partial y} \left( e^x(-x\sin y -(y\cos y + \sin y))+ 0\right)\\ &= \frac{\partial }{\partial y} \left(e^x(-x\sin y - y\cos y - \sin y)\right)\\ &= e^x(-x\cos y -(-y\sin y + \cos y)-\cos y) + 0\\ &= e^x(-x\cos y + y\sin y - 2\cos y) \end{split} \end{equation*}\]

So, now

\[\begin{equation*} \begin{split} \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} &= \\ e^x(2\cos y + x\cos y - y\sin y -x\cos y + y\sin y - 2\cos y)\\ &= 0 \end{split} \end{equation*}\]

18.1.10 Question 10

If \(x=r \cos \theta, y = r \sin \theta\), prove that

\[\begin{equation*} \begin{split} \frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} &= \frac{1}{r}\left[\left(\frac{\partial r}{\partial x} \right)^2 + \left(\frac{\partial r}{\partial y}\right)^2\right] \end{split} \end{equation*}\]

Here,

\[\begin{equation*} \begin{split} x^2 + y^2 &= r^2\cos^2 \theta + r^2 \sin^2 \theta\\ x^2 + y^2 &= r^2\\ r^2 &= x^2 + y^2 \end{split} \end{equation*}\]

Now taking partial derivative w.r.t \(x\),

\[\begin{equation*} \begin{split} 2r\frac{\partial r}{\partial x} &= 2x\\ \frac{\partial r}{\partial x} &= \frac{x}{r}\\ \frac{\partial^2 r}{\partial x^2} &= \frac{r.1-x.\frac{\partial r}{\partial x} }{r^2}\\ &= \frac{r-x.\frac{x}{r}}{r^2}\\ &= \frac{r^2-x^2}{r^3}\\ \frac{\partial^2 r}{\partial x^2} &= \frac{y^2}{r^3} \end{split} \end{equation*}\]

Again taking partial derivative of \(r^2 = x^2 + y^2\) w.r.t \(y\),

\[\begin{equation*} \begin{split} 2r\frac{\partial r}{\partial y} &= 2y\\ \frac{\partial r}{\partial y} &= \frac{y}{r}\\ \frac{\partial^2 r}{\partial y^2} &= \frac{r.1 - y.\frac{\partial r}{\partial y}}{r^2}\\ &= \frac{r-\frac{y^2}{r}}{r^2}\\ &= \frac{r^2-y^2}{r^3}\\ \frac{\partial^2 r}{\partial y^2} &= \frac{x^2}{r^3} \end{split} \end{equation*}\]

Now,

\[\begin{equation*} \begin{split} \frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} &= \frac{y^2}{r^3} + \frac{x^2}{r^3}\\ &= \frac{x^2 + y^2}{r^3} \end{split} \end{equation*}\]

Substituting \(r^2 = x^2 + y^2\), L.H.S. evaluates to

\[\begin{equation*} \begin{split} \frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} &= \frac{r^2}{r^3}\\ &= \frac{1}{r} \end{split} \end{equation*}\]

Now evaluating R.H.S,

\[\begin{equation*} \begin{split} \left(\frac{\partial r}{\partial x} \right)^2 &= \frac{x^2}{r^2}\\ \left(\frac{\partial r}{\partial y} \right)^2 &= \frac{y^2}{r^2} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{1}{r}\left[\left(\frac{\partial r}{\partial x} \right)^2 + \left(\frac{\partial r}{\partial y}\right)^2\right] &= \frac{1}{r}\left[\frac{x^2}{r^2} + \frac{y^2}{r^2}\right]\\ &= \frac{1}{r}\left[\frac{x^2+y^2}{r^2}\right]\\ &= \frac{1}{r}\frac{r^2}{r^2}\\ &= \frac{1}{r} \end{split} \end{equation*}\]

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