Functions of Several Variables: Difficult Problems with Solutions

Solution:
Let [tex]u=\frac{x+y}{1-xy}[/tex]. By the chain rule and [tex]\left(\arctan u\right)'=\frac{1}{1+u^2}[/tex],
[tex]f'_x=\frac{u'_x}{1+u^2}[/tex]
By the quotient rule ([tex]y[/tex] is a constant)
[tex]u'_x=\frac{1\times (1-xy)-(x+y)\times (-y)}{(1-xy)^2}=\frac{1+y^2}{(1-xy)^2}[/tex]
Moreover
[tex]1+u^2=\frac{(1-xy)^2+(x+y)^2}{(1-xy)^2}=\frac{1+x^2+y^2+x^2y^2}{(1-xy)^2}=\frac{(1+x^2)(1+y^2)}{(1-xy)^2}[/tex]
Therefore
[tex]f'_x=\frac{1+y^2}{(1-xy)^2}\times \frac{(1-xy)^2}{(1+x^2)(1+y^2)}=\frac{1}{1+x^2}[/tex]
By symmetry [tex]f'_y=\frac{1}{1+y^2}[/tex]. This reflects the identity [tex]f(x,y)=\arctan x+\arctan y+C[/tex], where the constant [tex]C[/tex] equals [tex]0[/tex], [tex]\pi[/tex] or [tex]-\pi[/tex] on the different parts of the domain.

Solution:
Put [tex]s=\sqrt{x^{2}-y^{2}}[/tex]. Since [tex](x+s)(x-s)=x^{2}-s^{2}=y^{2}[/tex], the argument of the logarithm equals [tex]\frac{(x+s)^{2}}{y^{2}}[/tex], so
[tex]z=2\ln|x+s|-2\ln|y|,\qquad s_x=\frac{x}{s},\quad s_y=-\frac{y}{s}[/tex]
Hence the partial derivatives are
[tex]z_x=\frac{2\left(1+\frac{x}{s}\right)}{x+s}=\frac{2(s+x)}{s(x+s)}=\frac{2}{s}[/tex]
[tex]z_y=\frac{2\times \left(-\frac{y}{s}\right)}{x+s}-\frac{2}{y}=-\frac{2\left(y^{2}+s(x+s)\right)}{y\,s\,(x+s)}[/tex]
and since [tex]s^{2}=x^{2}-y^{2}[/tex],
[tex]y^{2}+s(x+s)=y^{2}+sx+x^{2}-y^{2}=x(x+s)\Rightarrow z_y=-\frac{2x(x+s)}{y\,s\,(x+s)}=-\frac{2x}{y\,s}[/tex]
Both partial derivatives are continuous for [tex]|x|\gt|y|\gt 0[/tex], so [tex]z[/tex] is differentiable there and
[tex]dz=z_x\,dx+z_y\,dy=\frac{2}{\sqrt{x^{2}-y^{2}}}\,dx-\frac{2x}{y\sqrt{x^{2}-y^{2}}}\,dy[/tex]

Solution:
By the chain rule
[tex]z_x=f_u\,u_x+f_v\,v_x=2x\,f_u+y\,f_v[/tex]
where [tex]f_u[/tex] and [tex]f_v[/tex] are again functions of [tex]u,v[/tex] and therefore of [tex]x,y[/tex]. By the chain rule again,
[tex]\left(f_u\right)_x=f_{uu}\times 2x+f_{uv}\times y,\qquad \left(f_v\right)_x=f_{vu}\times 2x+f_{vv}\times y[/tex]
Differentiating [tex]z_x[/tex] with respect to [tex]x[/tex] by the product rule (there is no [tex]f_v[/tex] term, because [tex]\frac{\partial}{\partial x}(y)=0[/tex]):
[tex]z_{xx}=2f_u+2x\left(2x\,f_{uu}+y\,f_{uv}\right)+y\left(2x\,f_{vu}+y\,f_{vv}\right)=4x^{2}f_{uu}+2xy\,f_{uv}+2xy\,f_{vu}+y^{2}f_{vv}+2f_u[/tex]
Since the second partial derivatives of [tex]f[/tex] are continuous, Schwarz's theorem gives [tex]f_{uv}=f_{vu}[/tex], so
[tex]\frac{\partial^{2}z}{\partial x^{2}}=4x^{2}f_{uu}+4xy\,f_{uv}+y^{2}f_{vv}+2f_u[/tex]

Solution:
The function has continuous partial derivatives of all orders for [tex]x\ne 0[/tex], so near [tex]M[/tex] its second-degree Taylor polynomial is
[tex]T_2=f(M)+f'_x(M)(x-1)+f'_y(M)(y-1)+\frac{1}{2}\left[f''_{xx}(M)(x-1)^2+2f''_{xy}(M)(x-1)(y-1)+f''_{yy}(M)(y-1)^2\right][/tex]
Here [tex]f(M)=\arctan 1=\frac{\pi}{4}[/tex]. The first partial derivatives are
[tex]f'_x=\frac{1}{1+\frac{y^2}{x^2}}\times \left(-\frac{y}{x^2}\right)=-\frac{y}{x^2+y^2},\qquad f'_y=\frac{1}{1+\frac{y^2}{x^2}}\times \frac{1}{x}=\frac{x}{x^2+y^2}[/tex]
so [tex]f'_x(M)=-\frac{1}{2}[/tex] and [tex]f'_y(M)=\frac{1}{2}[/tex]. The second partial derivatives are
[tex]f''_{xx}=\frac{2xy}{(x^2+y^2)^2},\qquad f''_{xy}=\frac{y^2-x^2}{(x^2+y^2)^2},\qquad f''_{yy}=-\frac{2xy}{(x^2+y^2)^2}[/tex]
so [tex]f''_{xx}(M)=\frac{1}{2}[/tex], [tex]f''_{xy}(M)=0[/tex], [tex]f''_{yy}(M)=-\frac{1}{2}[/tex]. Substituting,
[tex]T_2(x,y)=\frac{\pi}{4}-\frac{x-1}{2}+\frac{y-1}{2}+\frac{1}{2}\left[\frac{1}{2}(x-1)^2-\frac{1}{2}(y-1)^2\right]=\frac{\pi}{4}-\frac{x-1}{2}+\frac{y-1}{2}+\frac{(x-1)^2-(y-1)^2}{4}[/tex]

Solution:
We minimise the square of the distance [tex]f=x^2+y^2+z^2[/tex] under the condition [tex]\varphi=3x-2y-6z-56=0[/tex]. A minimum exists: [tex]f[/tex] is continuous and tends to infinity as the point moves away along the plane. Since [tex]\nabla\varphi=(3, -2, -6)\ne\vec{0}[/tex], the Lagrange conditions hold at that minimum. The Lagrange function is
[tex]L=x^2+y^2+z^2-\lambda\left(3x-2y-6z-56\right)[/tex]
Setting its partial derivatives equal to zero:
[tex]L'_x=2x-3\lambda=0,\quad L'_y=2y+2\lambda=0,\quad L'_z=2z+6\lambda=0\Rightarrow x=\frac{3\lambda}{2},\ y=-\lambda,\ z=-3\lambda[/tex]
Substituting into the condition:
[tex]\frac{9\lambda}{2}+2\lambda+18\lambda=56\Rightarrow \frac{49\lambda}{2}=56\Rightarrow \lambda=\frac{16}{7}[/tex]
The only candidate is the point [tex]\left(\frac{24}{7}, -\frac{16}{7}, -\frac{48}{7}\right)[/tex], so the minimum is attained there:
[tex]f=\frac{576+256+2304}{49}=\frac{3136}{49}=64\Rightarrow d=\sqrt{64}=8[/tex]
Check with the point-to-plane distance formula: [tex]d=\frac{|-56|}{\sqrt{3^2+2^2+6^2}}=\frac{56}{7}=8[/tex].

Solution:
Stationary points:
[tex]u'_x=3x^2+12y=0,\quad u'_y=2y+12x=0,\quad u'_z=2z+2=0[/tex]
Hence [tex]z=-1[/tex], [tex]y=-6x[/tex] and [tex]3x^2-72x=0[/tex], i.e. [tex]x=0[/tex] or [tex]x=24[/tex]. The stationary points are [tex]P_1(0, 0, -1)[/tex] and [tex]P_2(24, -144, -1)[/tex]. The Hessian matrix is
[tex]H=\begin{bmatrix}6x&12&0\\12&2&0\\0&0&2\end{bmatrix}[/tex]
At [tex]P_2[/tex] it equals [tex]\begin{bmatrix}144&12&0\\12&2&0\\0&0&2\end{bmatrix}[/tex], with leading principal minors
[tex]\Delta_1=144\gt 0,\quad \Delta_2=288-144=144\gt 0,\quad \Delta_3=2\times 144=288\gt 0[/tex]
By Sylvester's criterion the second differential is positive definite, so [tex]P_2[/tex] is a point of local minimum.
At [tex]P_1[/tex] we have [tex]u(P_1)=-1[/tex], and on the line [tex]y=0[/tex], [tex]z=-1[/tex] the function equals [tex]u=x^3-1[/tex], which is less than [tex]-1[/tex] for [tex]x\lt 0[/tex] and greater than [tex]-1[/tex] for [tex]x\gt 0[/tex]. Hence there is no extremum at [tex]P_1[/tex] (there [tex]\Delta_2=-144\lt 0[/tex] as well). The value at the minimum is
[tex]u(P_2)=24^3+(-144)^2+(-1)^2+12\times 24\times (-144)+2\times (-1)=13824+20736+1-41472-2=-6913[/tex]

Solution:
The function [tex]z[/tex] is continuous and the region is closed and bounded, so by Weierstrass' theorem both values exist; they are attained at stationary points inside the region or on its boundary. Inside:
[tex]z'_x=10x=0,\quad z'_y=-8y=0\Rightarrow (x, y)=(0, 0),\quad z(0, 0)=0[/tex]
On the boundary [tex]\frac{x^2}{25}+\frac{y^2}{16}=1[/tex] we have [tex]y^2=16\left(1-\frac{x^2}{25}\right)[/tex] with [tex]-5\le x\le 5[/tex], so
[tex]z=5x^2-64+\frac{64x^2}{25}=\frac{189x^2}{25}-64[/tex]
This expression increases with [tex]x^2[/tex]: its least value is [tex]-64[/tex] at [tex]x=0[/tex] (the points [tex](0, \pm 4)[/tex]) and its greatest value is [tex]189-64=125[/tex] at [tex]x=\pm 5[/tex] (the points [tex](\pm 5, 0)[/tex]). Comparing [tex]0[/tex], [tex]-64[/tex] and [tex]125[/tex], we get
[tex]z_{\max}=125,\qquad z_{\min}=-64[/tex]

Solution:
First we compute the partial derivatives at the origin from the definition. Since [tex]\left|h\sin \frac{1}{h^{2}}\right|\le|h|[/tex],
[tex]f_x(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to 0}h\sin \frac{1}{h^{2}}=0[/tex]
and in the same way [tex]f_y(0,0)=0[/tex]. By definition, [tex]f[/tex] is differentiable at [tex](0, 0)[/tex] if, as [tex]\rho=\sqrt{h^{2}+k^{2}}\to 0[/tex],
[tex]f(h,k)-f(0,0)-f_x(0,0)\,h-f_y(0,0)\,k=o(\rho)[/tex]
Here, for [tex]\rho\gt 0[/tex],
[tex]\frac{\left|f(h,k)-0-0\times h-0\times k\right|}{\rho}=\frac{\rho^{2}\left|\sin \frac{1}{\rho^{2}}\right|}{\rho}\le\rho\to 0[/tex]
Hence [tex]f[/tex] is differentiable at [tex](0, 0)[/tex] and [tex]df(0,0)=0[/tex].
Remark: the sufficient condition (continuous partial derivatives) cannot be used here, because for [tex](x, y)\ne(0, 0)[/tex]
[tex]f_x=2x\sin \frac{1}{x^{2}+y^{2}}-\frac{2x}{x^{2}+y^{2}}\cos\frac{1}{x^{2}+y^{2}}[/tex]
and along [tex]y=0[/tex] the second term [tex]-\frac{2}{x}\cos\frac{1}{x^{2}}[/tex] is unbounded near [tex]x=0[/tex], while the first is bounded; so [tex]f_x[/tex] is not continuous at the origin, and the proof has to use the definition.

Solution:
Let [tex]g(t)=2t^{2}-3[/tex] and let [tex]G(t)=\frac{2t^{3}}{3}-3t[/tex] be an antiderivative of [tex]g[/tex]; then [tex]f(x,y)=G(4x+3y)-G(4x)[/tex] is a polynomial. Put [tex]u=4x[/tex], [tex]v=4x+3y[/tex]. Then
[tex]f_x=4g(v)-4g(u),\qquad f_y=3g(v)[/tex]
The system [tex]f_x=f_y=0[/tex] gives [tex]g(u)=g(v)=0[/tex], i.e. [tex]u,v\in\{a, -a\}[/tex], where [tex]a=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}[/tex]: four stationary points. Since [tex]g'(t)=4t[/tex],
[tex]f_{xx}=64(v-u),\quad f_{xy}=48v,\quad f_{yy}=36v,\qquad \Delta=f_{xx}f_{yy}-f_{xy}^{2}=2304v(v-u)-2304v^{2}=-2304\,uv[/tex]
If [tex]u=v[/tex] (the points [tex]\left(\pm\frac{\sqrt{6}}{8}, 0\right)[/tex]), then [tex]\Delta=-2304a^{2}\lt 0[/tex]: no extremum.
If [tex]u=a[/tex], [tex]v=-a[/tex], i.e. [tex]x=\frac{\sqrt{6}}{8}[/tex], [tex]y=\frac{v-u}{3}=-\frac{\sqrt{6}}{3}[/tex], then [tex]\Delta\gt 0[/tex] and [tex]f_{xx}=-128a\lt 0[/tex]: a local maximum. If [tex]u=-a[/tex], [tex]v=a[/tex], i.e. at the point [tex]\left(-\frac{\sqrt{6}}{8}, \frac{\sqrt{6}}{3}\right)[/tex], then [tex]\Delta\gt 0[/tex] and [tex]f_{xx}=128a\gt 0[/tex]: a local minimum.
Since [tex]a^{2}=\frac{3}{2}[/tex], we have [tex]G(a)=a\left(\frac{2a^{2}}{3}-3\right)=-2a[/tex] and [tex]G(-a)=2a[/tex]. Since [tex]f=G(v)-G(u)[/tex],
[tex]f_{\max}=G(-a)-G(a)=4a=2\sqrt{6},\qquad f_{\min}=G(a)-G(-a)=-4a=-2\sqrt{6}[/tex]

Solution:
[tex]K[/tex] is the closed square with vertices [tex](-1, 1)[/tex], [tex](1, -1)[/tex], [tex](2, 0)[/tex], [tex](0, 2)[/tex]. It is compact and [tex]F[/tex] is continuous, so by Weierstrass' theorem the least value exists; it is attained either at a stationary point inside [tex]K[/tex] or on the boundary. Stationary points:
[tex]F_x=3x^{2}-3y=0,\quad F_y=3y^{2}-3x=0\ \Rightarrow\ y=x^{2},\ x^{4}=x\ \Rightarrow\ (0, 0),\ (1, 1)[/tex]
Both points lie on the boundary (on the sides [tex]x+y=0[/tex] and [tex]x+y=2[/tex], respectively), so there are no stationary points inside [tex]K[/tex], and we examine the four sides. On the side [tex]y=-x[/tex], [tex]-1\le x\le 1[/tex]:
[tex]F=3x^{2}\in[0, 3][/tex]
On the side [tex]y=2-x[/tex], [tex]0\le x\le 2[/tex]:
[tex]F=9x^{2}-18x+8=9(x-1)^{2}-1\in[-1, 8][/tex]
and the minimum [tex]-1[/tex] is attained at [tex]x=1[/tex], i.e. at the point [tex](1, 1)[/tex]. On the side [tex]y=x-2[/tex], [tex]1\le x\le 2[/tex]:
[tex]F=2x^{3}-9x^{2}+18x-8,\quad F'=6\left(x^{2}-3x+3\right)\gt 0[/tex]
(the discriminant of the quadratic is negative), so [tex]F[/tex] increases from [tex]3[/tex] to [tex]8[/tex]. On the side [tex]y=x+2[/tex], [tex]-1\le x\le 0[/tex]:
[tex]F=2x^{3}+3x^{2}+6x+8,\quad F'=6\left(x^{2}+x+1\right)\gt 0[/tex]
so again [tex]F[/tex] increases from [tex]3[/tex] to [tex]8[/tex]. Comparing all values on the boundary, the least value of [tex]F[/tex] on [tex]K[/tex] is
[tex]\min_{K}F=F(1,1)=-1[/tex]