Pochodne czastkowe

[moczul]

Wyznaczyc pochodne cząstkow funkcji

1) \(\displaystyle{ f(x,y) = \frac{xy}{x^{3}+2y}}\)

2) \(\displaystyle{ f(x,y) = \frac{\sqrt{x}}{1-y^{2}}}\)

1) \(\displaystyle{ f(x,y) = siny + arcsinxy}\)

[ariadna]

1)
\(\displaystyle{ \frac{\partial{f}}{\partial{x}}=\frac{y(x^{3}+2y)-xy(3x^{2})}{(x^{3}+2y)^{2}}}\)
\(\displaystyle{ \frac{\partial{f}}{\partial{y}}=\frac{x(x^{3}+2y)-2xy}{(x^{3}+2y)^{2}}}\)

[max]

2)
\(\displaystyle{ \frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x}(1- y^{2})}\\
\frac{\partial f}{\partial y} = \frac{2y\sqrt{x}}{(1 - y^{2})^{2}}}\)
3)
\(\displaystyle{ \frac{\partial f}{\partial x} = \frac{y}{\sqrt{1 - x^{2}y^{2}}}\\
\frac{\partial f}{\partial y} = \cos y + \frac{x}{\sqrt{1 - x^{2}y^{2}}}}\)

[flamaster2]

Czy w ostatnim przykładzie po dx nie powinno być
\(\displaystyle{ \frac{\partial f}{\partial x}=1+\frac{y}{\sqrt{1-x^2 y^2}}}\)?

[max]

Nie.
\(\displaystyle{ \frac{\partial f}{\partial x} = (\sin y + \arcsin xy)'_{x} = (\sin y)'_{x} + (\arcsin xy)'_{x} =\\
= 0 + (xy)'_{x} \frac{1}{\sqrt{1 - (xy)^{2}}}= \frac{y}{\sqrt{1- x^{2}y^{2}}}}\)