Question

Oblicz pochodną kierunkową. Daje naj

Answer 1

Odpowiedź:

[tex]\displaystyle f(x,y)=\frac{x+ln(xy)}{x^{2} +y^{2} } +y\sqrt{x+3} \\P(1,1)\quad \vec{v}=[-3,2]\\f'_{\vec v_k}=grad (f(P)){\circ}\vec v_k\\f'_x=\frac{(1+\frac{1}{xy}\cdot y)(x^{2} +y^{2} ) -2x(x+ln(xy))}{(x^{2} +y^{2} )^2} +\frac{y}{2\sqrt{x+3} } \\f'_x(1,1)=\frac{(1+1)(1+1)-2\cdot1(1+0)}{(1+1)^2} +\frac{1}{2\sqrt{4} } =\frac{4-2}{4} +\frac{1}{4} =\frac{3}{4}[/tex]

[tex]\displaystyle f'_y=\frac{(\frac{1}{xy}\cdot x )(x^{2} +y^{2} )-2y(x+ln(xy))}{(x^{2} +y^{2} )^2} +\sqrt{x+3} \\f'_y(1,1)=\frac{1\cdot2-2\cdot1(1+0)}{4} +\sqrt{1+3} =\frac{2-2}{4} =2=2\\gradf(P)=\left[\frac{3}{4} ,2\right]\\|\vec v|=\sqrt{(-3)^2+2^2} =\sqrt{13} \\[/tex]

[tex]\displaystyle v_k=\frac{1}{\sqrt{13} } [-3,2]=\left[-\frac{3}{\sqrt{13} } ,\frac{2}{\sqrt{13} } \right]\\f'_{\vek v_k}=\left [\frac{3}{4} ,2\right]{\circ}\left[-\frac{3}{\sqrt{13} } ,\frac{2}{\sqrt{13} } \right]=\frac{3}{4} \cdot\left(-\frac{3}{\sqrt{13} } \right)+2\cdot\frac{2}{\sqrt{13} } =\\-\frac{9}{4\sqrt{13} } +\frac{4}{\sqrt{13} } =\frac{7}{4\sqrt{13} } =\frac{7\sqrt{13} }{52}[/tex]