Korzystając z definicji obliczyć pochodną

[mat91]

Korzystając z definicji obliczyć pochodną \(f_h^'(x_o)\)
1.\(f_{(\frac{3}{5},\frac{4}{5})}^'(1,1)\), \(f(x,y)=\frac{|x-y|}{x+y}\)
2.\(f_{(1,1,0)}^'(0,1,-1)\), \(f(x,y)=xy+z^2\)

[octahedron]

\(1)
f(x,y)=\frac{|x-y|}{x+y}
h=\[\frac{3}{5};\frac{4}{5}\]
|h|=1
f_h'(1,1)=\lim_{t\to 0}\ \frac{1}{t}\cdot\left(\frac{\|1+\frac{3}{5}t-1-\frac{4}{5}t\|}{1+\frac{3}{5}t+1+\frac{4}{5}t}-\frac{\|1-1\|}{1+1}\right)=\lim_{t\to 0}\ \frac{|t|}{t}\cdot\frac{1}{10+7t}
\lim_{t\to 0^+}\ \frac{|t|}{t}\cdot\frac{1}{10+7t}=\lim_{t\to 0^+}1\cdot\frac{1}{10+7t}=\frac{1}{10}
\lim_{t\to 0^-}\ \frac{|t|}{t}\cdot\frac{1}{10+7t}=\lim_{t\to 0^-}-1\cdot\frac{1}{10+7t}=-\frac{1}{10}\)

Pochodna nie istnieje

[octahedron]

\(2)
f(x,y,z)=xy+z^2
h=\[1;1;0\]
|h|=\sqrt{2}
f_h'(0,1,-1)=\lim_{t\to 0}\ \frac{1}{\sqrt{2}t} \left[f(0+t;1+t,-1)-f(0,1,-1)\right]=\lim_{t\to 0}\ \frac{1}{\sqrt{2}t} t(1+t)=\lim_{t\to 0}\ \frac{1+t}{\sqrt{2}}=\frac{1}{\sqrt{2}}\)