Pochodne kierunkowe funkcji
: 2 maja 2017, o 09:55
Obliczyć pochodne kierunkowe odwzorowań:
b) \(\displaystyle{ f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}}\)
\(\displaystyle{ f(x,y)=(x^{2}-y^{2},e^{x-2y})}\)
w punkcie \(\displaystyle{ (x_{0},y_{0}=(1,-1)}\)
w kierunku \(\displaystyle{ \vec{v_{1}}=(0,1)}\) i \(\displaystyle{ \vec{v_{2}}=(-1,2)}\)
\(\displaystyle{ \lim_{h \to 0} \frac{f((1,-1)+h(0,1))-f(1,-1)}{h} =
\lim_{ h \to 0 }\frac{(1-(1-2h+h^{2}),e^{1-2(-1+h)})-(0,e^{-1})}{h}=
\lim_{h \to 0 }\frac{(2h-h^{2},e^{3-2h}-e^{-1})}{h}}\)
no i teraz nie wiem jak policzyć granicę
\(\displaystyle{ \lim_{h \to 0 }\frac{e^{3-2h}-e^{-1}}{h}}\)
b) \(\displaystyle{ f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}}\)
\(\displaystyle{ f(x,y)=(x^{2}-y^{2},e^{x-2y})}\)
w punkcie \(\displaystyle{ (x_{0},y_{0}=(1,-1)}\)
w kierunku \(\displaystyle{ \vec{v_{1}}=(0,1)}\) i \(\displaystyle{ \vec{v_{2}}=(-1,2)}\)
\(\displaystyle{ \lim_{h \to 0} \frac{f((1,-1)+h(0,1))-f(1,-1)}{h} =
\lim_{ h \to 0 }\frac{(1-(1-2h+h^{2}),e^{1-2(-1+h)})-(0,e^{-1})}{h}=
\lim_{h \to 0 }\frac{(2h-h^{2},e^{3-2h}-e^{-1})}{h}}\)
no i teraz nie wiem jak policzyć granicę
\(\displaystyle{ \lim_{h \to 0 }\frac{e^{3-2h}-e^{-1}}{h}}\)