partielles-ableiten

Ableitung nacheindernader Variablen als Konstanten behandeln

1.6) b) $f(x,y)=x^2+2y$ $fx=2x$ $fy=2$

Mit bedingung

$f(x,y)=x^2+\frac{2x}{y},x_0=1,y_0=2$ $fx(x,y)=2x+\frac{2}{y}$ $fx(1|2)=2\ast 1+\frac{2}{2}=3$ $fy(x,y)=-\frac{2x}{y^2}$ $fy(1|2)=-\frac{2}{4}=-\frac{1}{2}$

1.10) a) $f(x,y)=x^2-3y$ $fx=2x$ $fxx=2$ $fy=-3$ $fyy=0$

1.6) g) $f(x,y)=x+\frac{1}{2y}$ $fx=1$ $fy=-\frac{1}{2y^2}$ 1.7) c) $z=f(x,y)$ $z=\sqrt{x+y}=(x+y{)}^{1/2}$ $fx=fy=\frac{1}{2\sqrt{x+y}}$ 1.8) d) $z=x\ast e^{-y}$ $z_x=e^{-y}$ $z_y=-x\ast e^{-y}$ g) $z=\frac{1}{y}+\ln (\sqrt{x})=\frac{1}{y}+\ln (x{)}^{1/2}$ $z_x=\frac{1}{2}\ast \frac{1}{x}=\frac{1}{2x}$ $z_y=-\frac{1}{y^2}$ 1.9) c) $z=x\ast \sin y$ $z_x=\sin y$ $z_y=x\ast \cos y=\cos (y)\ast x$ 1.10) c) $f(x,y)=(x\ast y{)}^2$ $fx=2\ast (x\ast y);y=2x{y}^2$ $fy=2(x\ast y)\ast x;x=2x^{2}y$ $fxy=2x\ast 2y=4xy$ $fyx=2\ast 2xy=4xy$ $fxx=2y^2$ $fyy=2x^2$ 1.10) e) $f(x,y)=\frac{x^2}{y}$ $fx=\frac{2x}{y}$ $fy=-\frac{x^2}{y^2}=$

g) $f(x,y)=x\sin (2y)$