\(\displaystyle{ \ f(x,y) = arcctg(2x^{2}+7y^{5})}\)
wyznaczyć należy:
\(\displaystyle{ \ f'x(x,y)}\)
\(\displaystyle{ \ f'y(x,y)}\)
\(\displaystyle{ \ f'xx(x,y)}\)
\(\displaystyle{ \ f'yy(x,y)}\)
\(\displaystyle{ \ f'xy(x,y)}\)
\(\displaystyle{ \ f'yx(x,y)}\)
Funkcje dwóch zmiennych
-
Gotta
- Użytkownik
- Posty: 729
- Rejestracja: 19 mar 2009, o 11:18
- Płeć: Kobieta
- Podziękował: 2 razy
- Pomógł: 220 razy
Funkcje dwóch zmiennych
\(\displaystyle{ \frac{\partial f}{\partial x}f(x,y)=\frac{-4x}{1+(2x^2+7y^5)^2}}\)
\(\displaystyle{ \frac{\partial f}{\partial y}f(x,y)=\frac{-35y^4}{1+(2x^2+7y^5)^2}}\)
\(\displaystyle{ \frac{\partial^2 f}{\partial x^2}f(x,y)=\frac{-4(1+(2x^2+7y^5)^2)+4x\cdot 2(2x^2+7y^5)\cdot 4x}{(1+(2x^2+7y^5)^2)^2}}\)
\(\displaystyle{ \frac{\partial^2 f}{\partial y^2}f(x,y)=\frac{-140(1+(2x^2+7y^5)^2)+35y^4\cdot 2(2x^2+7y^5)\cdot 35y^4}{(1+(2x^2+7y^5)^2)^2}}\)
\(\displaystyle{ \frac{\partial^2 f}{\partial x \partial y}f(x,y)=\frac{4x\cdot 2(2x^2+7y^5)\cdot 35y^4}{(1+(2x^2+7y^5)^2)^2}}\)
\(\displaystyle{ \frac{\partial f}{\partial y}f(x,y)=\frac{-35y^4}{1+(2x^2+7y^5)^2}}\)
\(\displaystyle{ \frac{\partial^2 f}{\partial x^2}f(x,y)=\frac{-4(1+(2x^2+7y^5)^2)+4x\cdot 2(2x^2+7y^5)\cdot 4x}{(1+(2x^2+7y^5)^2)^2}}\)
\(\displaystyle{ \frac{\partial^2 f}{\partial y^2}f(x,y)=\frac{-140(1+(2x^2+7y^5)^2)+35y^4\cdot 2(2x^2+7y^5)\cdot 35y^4}{(1+(2x^2+7y^5)^2)^2}}\)
\(\displaystyle{ \frac{\partial^2 f}{\partial x \partial y}f(x,y)=\frac{4x\cdot 2(2x^2+7y^5)\cdot 35y^4}{(1+(2x^2+7y^5)^2)^2}}\)