Sea la función

 $F(x,y)=x^2y^2−2xy^2$  
Determine el valor de  $\frac{\partial F}{\partial x}$  como  $\frac{\partial F}{\partial y}$  en el punto  $P\left(2,1\right)$  

• 
  

   $\frac{\partial F}{\partial x}=2$   $\frac{\partial F}{\partial y}=0$  

•  $\frac{\partial F}{\partial x}=2$   $\frac{\partial F}{\partial y}=2$  

•  $\frac{\partial F}{\partial x}=0$   $\frac{\partial F}{\partial y}=0$  

•  $\frac{\partial F}{\partial x}=0$   $\frac{\partial F}{\partial y}=2$  

Sea la función

 $\theta(r,s)=\sqrt{r^2+s^2}+\frac{r}{s}$  
Determine el valor de  $\frac{\partial\theta}{\partial r}$  como  $\frac{\partial\theta}{\partial s}$  en el punto  $P\left(3,4\right)$  

•  $\frac{\partial\theta}{\partial r}=\frac{27}{20}$   $\frac{\partial\theta}{\partial s}=\frac{49}{80}$ 
  

•  $\frac{\partial\theta}{\partial r}=\frac{17}{20}$   $\frac{\partial\theta}{\partial s}=\frac{39}{80}$ 

•  $\frac{\partial\theta}{\partial r}=\frac{17}{20}$   $\frac{\partial\theta}{\partial s}=\frac{49}{80}$ 

•  $\frac{\partial\theta}{\partial r}=\frac{31}{20}$   $\frac{\partial\theta}{\partial s}=\frac{49}{80}$ 

Sea la función

 $F\left(x,y\right)=3x^3y-2x^2y^2+y^3$  
Determine el valor de  $\frac{\partial F}{\partial x}$  como  $\frac{\partial F}{\partial y}$  en el punto  $P\left(1,-2\right)$  

•  $\frac{\partial F}{\partial x}=-34$   $\frac{\partial F}{\partial y}=23$ 
  

•  $\frac{\partial F}{\partial x}=34$   $\frac{\partial F}{\partial y}=-23$ 

•  $\frac{\partial F}{\partial x}=-34$   $\frac{\partial F}{\partial y}=-23$ 

•  $\frac{\partial F}{\partial x}=34$   $\frac{\partial F}{\partial y}=23$ 

Sea la función

 $F\left(x,y\right)=\ln\left(x^2+y\right)$  
Determine el valor de  $\frac{\partial F}{\partial x}$  como  $\frac{\partial F}{\partial y}$  en el punto  $P\left(1,2\right)$  

•  $\frac{\partial F}{\partial x}=\frac{2}{3}$   $\frac{\partial F}{\partial y}=\frac{1}{3}$ 
  

•  $\frac{\partial F}{\partial x}=\frac{3}{2}$   $\frac{\partial F}{\partial y}=\frac{1}{3}$ 

•  $\frac{\partial F}{\partial x}=\frac{1}{3}$   $\frac{\partial F}{\partial y}=\frac{2}{3}$ 

•  $\frac{\partial F}{\partial x}=\frac{1}{3}$   $\frac{\partial F}{\partial y}=\frac{3}{2}$ 

Sea la función

 $F\left(x,y\right)=e^{-\frac{\left(x^2+y^2\right)}{3}}$  
Determine el valor de  $\frac{\partial F}{\partial x}$  como  $\frac{\partial F}{\partial y}$  en el punto  $P\left(1,2\right)$  

•  $\frac{\partial F}{\partial x}=-0.1259170685$   $\frac{\partial F}{\partial y}=-0.2518341371$ 
  

•  $\frac{\partial F}{\partial x}=-0.1259170685$   $\frac{\partial F}{\partial y}=0.2518341371$ 

•  $\frac{\partial F}{\partial x}=-0.2518341371$   $\frac{\partial F}{\partial y}=-0.1259170685$ 

•  $\frac{\partial F}{\partial x}=0.2518341371$   $\frac{\partial F}{\partial y}=0.1259170685$ 

Sea la función

 $F\left(x,y\right)=(3x+y)Cos(xy)$  
Determine el valor de  $\frac{\partial F}{\partial x}$  como  $\frac{\partial F}{\partial y}$  en el punto  $P\left(1,2\right)$  

•  $\frac{\partial F}{\partial x}=4.962633970$   $\frac{\partial F}{\partial y}=10.34141477$ 
  

•  $\frac{\partial F}{\partial x}=10.34141477$   $\frac{\partial F}{\partial y}=4.962633970$ 

•  $\frac{\partial F}{\partial x}=-4.962633970$   $\frac{\partial F}{\partial y}=-10.34141477$ 

•  $\frac{\partial F}{\partial x}=-10.34141477$   $\frac{\partial F}{\partial y}=-4.962633970$ 

Sea la función

 $F\left(x,y\right)=e^{-x}(2y^2−x^2)$  
Determine el valor de  $\frac{\partial F}{\partial x}$  como  $\frac{\partial F}{\partial y}$  en el punto  $P\left(1,2\right)$  

•  $\frac{\partial F}{\partial x}=1.019978389$   $\frac{\partial F}{\partial y}=-1.839397205$ 
  

•  $\frac{\partial F}{\partial x}=1.839397205$   $\frac{\partial F}{\partial y}=1.019978389$ 

•  $\frac{\partial F}{\partial x}=-1.839397205$   $\frac{\partial F}{\partial y}=1.019978389$ 

•  $\frac{\partial F}{\partial x}=1.019978389$   $\frac{\partial F}{\partial y}=1.839397205$ 

Sea la función

 $F\left(x,y\right)=\ln(1+2x^2+3y^2)$  
Determine el valor de  $\frac{\partial F}{\partial x}$  como  $\frac{\partial F}{\partial y}$  en el punto  $P\left(1,2\right)$  

•  $\frac{\partial F}{\partial x}=-\frac{4}{15}$   $\frac{\partial F}{\partial y}=-\frac{4}{5}$ 
  

•  $\frac{\partial F}{\partial x}=\frac{4}{15}$   $\frac{\partial F}{\partial y}=-\frac{4}{5}$ 

•  $\frac{\partial F}{\partial x}=\frac{4}{15}$   $\frac{\partial F}{\partial y}=\frac{4}{5}$ 

•  $\frac{\partial F}{\partial x}=\frac{4}{5}$   $\frac{\partial F}{\partial y}=\frac{4}{15}$