Wir berechnen \[ \begin{array}{l} f_x=2ax+2by,\quad f_{xx}=2a, \\[0.6ex] f_y=2bx+2cy,\quad f_{yy}=2c. \end{array} \] Das bedeutet \[ \Delta f=2(a+c)=0\quad\mbox{bzw.}\quad a=-c. \] Wir wählen also \( a=-c \) und \( b\in\mathbb R.\qquad\Box \)

Zunächst sind \[ \Delta f=\sum_{k=1}^mf_{x_kx_k}\,,\quad \Delta g=\sum_{k=1}^mg_{x_kx_k}\,,\quad \langle\nabla f,\nabla g\rangle=\sum_{k=1}^mf_{x_k}g_{x_k}\,. \] Nun berechnen wir \[ \begin{array}{lll} \Delta(fg)\!\!\! & = & \!\!\!\displaystyle \sum_{k=1}^m\partial_{x_kx_k}(fg) \,=\,\sum_{k=1}^m\partial_{x_k}(f_{x_k}g+fg_{x_k}) \\[0.6ex] & = & \!\!\!\displaystyle \sum_{k=1}^m(f_{x_kx_k}g+2f_{x_k}g_{x_k}+fg_{x_kx_k}) \\[0.6ex] & = & \!\!\!\displaystyle f\sum_{k=1}^mg_{x_kx_k}+2\sum_{k=1}^mf_{x_k}g_{x_k}+g\sum_{k=1}^mf_{x_kx_k} \end{array} \] und damit \[ \Delta(fg)=f\Delta g+2\langle\nabla f,\nabla g\rangle+g\Delta f. \] Das war zu zeigen.\( \qquad\Box \)

\[ \begin{array}{rcl} x=\sin y & \quad\text{und damit}\quad & -x=-\sin y=\sin(-y) \\[0.6ex] & \quad\text{und damit}\quad & -y=\arcsin(-x) \\[0.6ex] & \quad\text{und damit}\quad & y=-\arcsin(-x) \end{array} \]

\[ \arcsin x=-\arcsin(x). \]

\[ \begin{array}{l} \displaystyle \arccos x=y+\frac{\pi}{2} \\[1ex] \displaystyle\text{und damit}\quad x=\cos\left(x+\frac{\pi}{2}\right)=-\sin y=\sin(-y)=-\cos\left(-y+\frac{\pi}{2}\right) \\[1ex] \displaystyle\text{und damit}\quad -x=\cos\left(-y+\frac{\pi}{2}\right) \\[1ex] \displaystyle\text{und damit}\quad \arccos(-x)=-y+\frac{\pi}{2} \\[1ex] \displaystyle\text{und damit}\quad y=-\arccos(-x)+\frac{\pi}{2} \end{array} \]

\[ \begin{array}{l} \displaystyle \arccos x-\frac{\pi}{2}=-\arccos(-x)+\frac{\pi}{2} \\[1ex] \displaystyle\text{und damit}\quad \arccos x=\pi-\arccos(-x). \end{array} \]

\[ \begin{array}{l} x=\sin y=\cos\left(-y+\frac{\pi}{2}\right) \\[1ex] \displaystyle\text{und damit}\quad \arccos x=-y+\frac{\pi}{2} \\[1ex] \displaystyle\text{und damit}\quad \arccos x=-\arcsin x+\frac{\pi}{2} \end{array} \]

\[ \arccos x=\frac{\pi}{2}-\arcsin x. \]

\[ \begin{array}{l} \displaystyle \sin x=\sqrt{1-\cos^2x} \\[1ex] \displaystyle\text{und damit}\quad \sin(\arccos x)=\sqrt{1-\cos^2(\arccos x)}=\sqrt{1-x^2}\,. \end{array} \]

\[ \begin{array}{l} \displaystyle \cos x=\sqrt{1-\sin^2x} \\[1ex] \displaystyle\text{und damit}\quad \cos(\arcsin x)=\sqrt{1-\sin^2(\arcsin x)}=\sqrt{1-x^2}\,. \end{array} \]

\[ \begin{array}{l} \displaystyle \tan^x=\frac{\sin^2x}{\cos^2x} \\[1ex] \displaystyle\text{und damit}\quad 1+\tan^2x=1+\frac{\sin^2}{\cos^2x}=\frac{\cos^2x+\cos^2x}{\sin^x}=\frac{1}{\cos^2} \\[1ex] \displaystyle\text{und damit}\quad \cos^2x=\frac{1}{1+\tan^2x} \\[1ex] \displaystyle\text{und damit}\quad \cos x=\frac{1}{\sqrt{1+\tan^2x}} \end{array} \]

\[ \cos(\arctan x)=\frac{1}{\sqrt{1+\tan^2(\arctan x)}}=\frac{1}{\sqrt{1+x^2}}\,. \]

\[ \sin x=\tan x\cdot\cos x=\tan x\cdot\frac{1}{\sqrt{1+\tan^2x}} \]

\[ \sin(\arctan x)=\tan(\arctan x)\cdot\frac{1}{\sqrt{1+\tan^2(\arctan x)}}=\frac{x}{\sqrt{1+x^2}}\,. \]

\[ \frac{d}{dx}\,\arcsin x=\frac{1}{\cos(\arcsin x)}=\frac{1}{\sqrt{1-x^2}} \]

\[ \frac{d}{dx}\,\arctan x=\frac{1}{\frac{1}{\cos^2(\arctan x)}}=\cos^2(\arctan x)=\frac{1}{1+x^2}\,. \]

\[ \begin{array}{l,} f(x)=\arcsin x, & f(0)=0, \\[1ex] \displaystyle f'(x)=\frac{1}{\sqrt{1-x^2}}\,, & f'(0)=1, \\[1ex] \displaystyle f''(x)=\frac{x}{(1-x^2)^\frac{3}{2}}\,, & f''(x)=0, \\[1ex] \displaystyle f'''(x)=\frac{1}{(1-x^2)^\frac{3}{2}}+\frac{3x^2}{(1-x^2)^\frac{5}{2}}\,, & f'''(0)=1 \end{array} \]

\[ f(x)\approx T_3f(x;0)=x+\frac{1}{6}\,x^3\,. \] Damit ist alles gezeigt.\( \qquad\Box \)

Wir berechnen die partiellen Ableitungen \[ \begin{array}{rcl} f_x\!\!\! & = & \!\!\!\displaystyle ye^{xy}\sin x\cos y+e^{xy}\cos x\cos y, \\[0.6ex] f_y\!\!\! & = & \!\!\!\displaystyle xe^{xy}\sin x\cos y-e^{xy}\sin x\sin y \end{array} \] sowie (die partiellen Ableitungen zweiter Ordnung vertauschen) \[ \begin{array}{rcl} f_{xx}\!\!\! & = & \!\!\!\displaystyle y^2e^{xy}\sin x\cos y+2ye^{xy}\cos x\cos y-e^{xy}\sin x\cos y, \\[0.6ex] f_{xy}\!\!\! & = & \!\!\!\displaystyle e^{xy}\sin x\cos y+xye^{xy}\sin x\cos y-ye^{xy}\sin x\sin y+\ldots \\[0.6ex] & & \!\!\!\displaystyle \ldots+xe^{xy}\cos x\cos y-e^{xy}\cos x\sin y, \\[0.6ex] f_{yy}\!\!\! & = & \!\!\!\displaystyle x^2e^{xy}\sin x\cos y-2xe^{xy}\sin x\sin y-e^{xy}\sin x\cos y. \end{array} \] Damit ergeben sich \[ \begin{array}{l} f(0,0)=0, \\[0.6ex] f_x(0,0)=1,\quad f_y(0,0)=0, \\[0.6ex] f_{xx}(0,0)=0,\quad f_{xy}(0,0)=0,\quad f_{yy}(0,0)=0 \end{array} \] und daher als Taylorpolynom \[ T_2f(x,y;0,0)=x,\quad(x,y)\in\mathbb R^2. \] Das war gesucht.\( \qquad\Box \)

Wir berechnen \[ \begin{array}{rcl} f_x\!\!\! & = & \!\!\!\displaystyle e^{x+z}\sinh x\cosh y+e^{x+z}\cosh x\cosh y, \\[0.6ex] f_y\!\!\! & = & \!\!\!\displaystyle e^{x+z}\sinh x\sinh y-z, \\[0.6ex] f_z\!\!\! & = & \!\!\!\displaystyle e^{x+y}\sinh x\cosh y-y \end{array} \] sowie (die partiellen Ableitungen zweiter Ordnung vertauschen) \[ \begin{array}{rcl} f_{xx}\!\!\! & = & \!\!\!\displaystyle 2e^{x+z}\sinh x\cosh y+2e^{x+y}\cosh x\cosh y, \\[0.6ex] f_{xy}\!\!\! & = & \!\!\!\displaystyle e^{x+y}\sinh x\sinh y+e^{x+y}\cosh x\sinh y, \\[0.6ex] f_{xz}\!\!\! & = & \!\!\!\displaystyle e^{x+y}\sinh x\cosh y+e^{x+y}\cosh x\cosh y, \\[0.6ex] f_{yy}\!\!\! & = & \!\!\!\displaystyle e^{x+z}\sinh x\cosh y, \\[0.6ex] f_{yz}\!\!\! & = & \!\!\!\displaystyle e^{x+z}\sinh x\sinh y-1, \\[0.6ex] f_{zz}\!\!\! & = & \!\!\!\displaystyle e^{x+y}\sinh x\cosh y. \end{array} \] Damit ergeben sich \[ \begin{array}{l} f(1,0,1)=e^2\sinh 1, \\[0.6ex] f_x(1,0,1)=e^2\sinh 1+e^2\cosh 1,\quad f_y(1,0,1)=-1,\quad f_z(1,0,1)=e^2\sinh 1, \\[0.6ex] f_{xx}(1,0,1)=2e^2\sinh 1+2e^2\cosh 1,\quad f_{xy}(1,0,1)=0, \\[0.6ex] f_{xz}(1,0,1)=e^2\sinh 1+e^2\cosh 1,\quad f_{yy}(1,0,1)=e^2\sinh 1, \\[0.6ex] f_{yz}(1,0,1)=-1,\quad f_{zz}(1,0,1)=e^2\sinh 1 \end{array} \] und daher als Taylorpolynom \[ \begin{array}{l} T_2f(x,y,z;1,0,1) \\[0.6ex] \qquad\displaystyle =e^2\sinh 1+e^2(\sinh 1+\cosh 1)\cdot(x-1)-y+e^2\sinh 1\cdot(z-1)+\ldots \\[0.6ex] \qquad\quad\displaystyle \ldots+e^2(\sinh 1+\cosh 1)\cdot(x-1)^2+e^2(\sinh 1+\cosh 1)\cdot(x-1)(z-1)+\ldots \\[0.6ex] \qquad\quad\displaystyle \ldots+\frac{1}{2}\,e^2\sinh 1\cdot y^2-y(z-1)+\frac{1}{2}\,e^2\sinh 1\cdot(z-1)^2\,. \end{array} \] Das war gesucht.\( \qquad\Box \)