Collins Vargas
09/09/2023 · Middle School
4. (1,2 PUNTOS) Determine las derivadas parciales de 1 er y 2 do orden para cada una de las siguientes funciones y evalúe en los puntos indicados. a. \( f(x, y)=\sqrt{x^{2}-y^{2}} ; f_{x}(\sqrt{4}, \sqrt{3}) ; f_{y}(\sqrt{4}, \sqrt{3}) ; f_{x x}(\sqrt{4}, \sqrt{3}) ; f_{y y}(\sqrt{4} \) \( \sqrt{3}) ; f_{x y}(\sqrt{4}, \sqrt{3}) \) b. \( f(x, y)=\ln \left(x^{2}+y^{2}\right) ; f_{x}(1,1) ; f_{y}(1,1) ; f_{x x}(\sqrt{2}, \sqrt{3}) \) \( f_{y y}(\sqrt{2}, \sqrt{3}) ; f_{x y}(\sqrt{2}, \sqrt{3}) \)
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Step-by-step Solution
Para resolver el problema, vamos a calcular las derivadas parciales de primer y segundo orden de las funciones dadas y luego evaluarlas en los puntos indicados.
### a. \( f(x, y) = \sqrt{x^2 - y^2} \)
**Derivadas parciales de primer orden:**
1. **Derivada parcial con respecto a \( x \):**
\[
f_x(x, y) = \frac{\partial}{\partial x} \sqrt{x^2 - y^2} = \frac{1}{2\sqrt{x^2 - y^2}} \cdot 2x = \frac{x}{\sqrt{x^2 - y^2}}
\]
2. **Derivada parcial con respecto a \( y \):**
\[
f_y(x, y) = \frac{\partial}{\partial y} \sqrt{x^2 - y^2} = \frac{1}{2\sqrt{x^2 - y^2}} \cdot (-2y) = -\frac{y}{\sqrt{x^2 - y^2}}
\]
**Evaluación en \( (\sqrt{4}, \sqrt{3}) \):**
\[
f_x(\sqrt{4}, \sqrt{3}) = f_x(2, \sqrt{3}) = \frac{2}{\sqrt{4 - 3}} = \frac{2}{1} = 2
\]
\[
f_y(\sqrt{4}, \sqrt{3}) = f_y(2, \sqrt{3}) = -\frac{\sqrt{3}}{\sqrt{4 - 3}} = -\sqrt{3}
\]
**Derivadas parciales de segundo orden:**
1. **Derivada parcial de segundo orden con respecto a \( x \):**
\[
f_{xx}(x, y) = \frac{\partial}{\partial x} f_x(x, y) = \frac{\partial}{\partial x} \left( \frac{x}{\sqrt{x^2 - y^2}} \right)
\]
Usando la regla del cociente:
\[
f_{xx}(x, y) = \frac{\sqrt{x^2 - y^2} \cdot 1 - x \cdot \frac{x}{\sqrt{x^2 - y^2}}}{x^2 - y^2} = \frac{(x^2 - y^2) - x^2}{(x^2 - y^2)^{3/2}} = \frac{-y^2}{(x^2 - y^2)^{3/2}}
\]
2. **Derivada parcial de segundo orden con respecto a \( y \):**
\[
f_{yy}(x, y) = \frac{\partial}{\partial y} f_y(x, y) = \frac{\partial}{\partial y} \left( -\frac{y}{\sqrt{x^2 - y^2}} \right)
\]
Usando la regla del cociente:
\[
f_{yy}(x, y) = -\frac{\sqrt{x^2 - y^2} \cdot 1 - y \cdot \frac{-y}{\sqrt{x^2 - y^2}}}{x^2 - y^2} = -\frac{(x^2 - y^2) + y^2}{(x^2 - y^2)^{3/2}} = -\frac{x^2}{(x^2 - y^2)^{3/2}}
\]
3. **Derivada parcial mixta:**
\[
f_{xy}(x, y) = \frac{\partial}{\partial y} f_x(x, y) = \frac{\partial}{\partial y} \left( \frac{x}{\sqrt{x^2 - y^2}} \right)
\]
Usando la regla del cociente:
\[
f_{xy}(x, y) = -\frac{x \cdot \frac{-y}{\sqrt{x^2 - y^2}}}{x^2 - y^2} = \frac{xy}{(x^2 - y^2)^{3/2}}
\]
**Evaluación en \( (\sqrt{4}, \sqrt{3}) \):**
\[
f_{xx}(\sqrt{4}, \sqrt{3}) = f_{xx}(2, \sqrt{3}) = \frac{-3}{(4 - 3)^{3/2}} = -3
\
Quick Answer
a. \( f_x(2, \sqrt{3}) = 2 \), \( f_y(2, \sqrt{3}) = -\sqrt{3} \), \( f_{xx}(2, \sqrt{3}) = -3 \), \( f_{yy}(2, \sqrt{3}) = -\frac{4}{(4 - 3)^{3/2}} \), \( f_{xy}(2, \sqrt{3}) = \frac{2\sqrt{3}}{(4 - 3)^{3/2}} \)
b. \( f_x(1,1) = \frac{1}{\sqrt{2}} \), \( f_y(1,1) = -\frac{1}{\sqrt{2}} \), \( f_{xx}(\sqrt{2}, \sqrt{3}) = -\frac{3}{(2 + 3)^{3/2}} \), \( f_{yy}(\sqrt{2}, \sqrt{3}) = -\frac{2}{(2 + 3)^{3/2}} \), \( f_{xy}(\sqrt{2}, \sqrt{3}) = \frac{\sqrt{6}}{(2 + 3)^{3/2}} \)
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