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Worksheets

derivadas parciales

Total questions: 8

Worksheet time: 32mins

Name
Class
Date
1.

Sea la función

 F(x,y)=x2y2−2xy2F(x,y)=x^2y^2−2xy^2  
Determine el valor de  ∂F∂x\frac{\partial F}{\partial x}  como  ∂F∂y\frac{\partial F}{\partial y}  en el punto  P(2,1)P\left(2,1\right)  

a)


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

b)

 \frac{\partial F}{\partial x}=2   ∂F∂y=2\frac{\partial F}{\partial y}=2  

c)

 ∂F∂x=0\frac{\partial F}{\partial x}=0   \frac{\partial F}{\partial y}=0  

d)

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

2.

Sea la función

 θ(r,s)=r2+s2+rs\theta(r,s)=\sqrt{r^2+s^2}+\frac{r}{s}  
Determine el valor de  ∂θ∂r\frac{\partial\theta}{\partial r}  como  ∂θ∂s\frac{\partial\theta}{\partial s}  en el punto  P(3,4)P\left(3,4\right)  

a)

 ∂θ∂r=2720\frac{\partial\theta}{\partial r}=\frac{27}{20}   \frac{\partial\theta}{\partial s}=\frac{49}{80} 

b)

 \frac{\partial\theta}{\partial r}=\frac{17}{20}   ∂θ∂s=3980\frac{\partial\theta}{\partial s}=\frac{39}{80} 

c)

 ∂θ∂r=1720\frac{\partial\theta}{\partial r}=\frac{17}{20}   ∂θ∂s=4980\frac{\partial\theta}{\partial s}=\frac{49}{80} 

d)

 ∂θ∂r=3120\frac{\partial\theta}{\partial r}=\frac{31}{20}   \frac{\partial\theta}{\partial s}=\frac{49}{80} 

3.

Sea la función

 F(x,y)=3x3y−2x2y2+y3F\left(x,y\right)=3x^3y-2x^2y^2+y^3  
Determine el valor de  ∂F∂x\frac{\partial F}{\partial x}  como  ∂F∂y\frac{\partial F}{\partial y}  en el punto  P(1,−2)P\left(1,-2\right)  

a)

 ∂F∂x=−34\frac{\partial F}{\partial x}=-34   ∂F∂y=23\frac{\partial F}{\partial y}=23 

b)

 ∂F∂x=34\frac{\partial F}{\partial x}=34   ∂F∂y=−23\frac{\partial F}{\partial y}=-23 

c)

 ∂F∂x=−34\frac{\partial F}{\partial x}=-34   ∂F∂y=−23\frac{\partial F}{\partial y}=-23 

d)

 ∂F∂x=34\frac{\partial F}{\partial x}=34   ∂F∂y=23\frac{\partial F}{\partial y}=23 

4.

Sea la función

 F(x,y)=ln⁡(x2+y)F\left(x,y\right)=\ln\left(x^2+y\right)  
Determine el valor de  ∂F∂x\frac{\partial F}{\partial x}  como  ∂F∂y\frac{\partial F}{\partial y}  en el punto  P(1,2)P\left(1,2\right)  

a)

 ∂F∂x=23\frac{\partial F}{\partial x}=\frac{2}{3}   ∂F∂y=13\frac{\partial F}{\partial y}=\frac{1}{3} 

b)

 ∂F∂x=32\frac{\partial F}{\partial x}=\frac{3}{2}   ∂F∂y=13\frac{\partial F}{\partial y}=\frac{1}{3} 

c)

 ∂F∂x=13\frac{\partial F}{\partial x}=\frac{1}{3}   ∂F∂y=23\frac{\partial F}{\partial y}=\frac{2}{3} 

d)

 ∂F∂x=13\frac{\partial F}{\partial x}=\frac{1}{3}   ∂F∂y=32\frac{\partial F}{\partial y}=\frac{3}{2} 

5.

Sea la función

 F(x,y)=e−(x2+y2)3F\left(x,y\right)=e^{-\frac{\left(x^2+y^2\right)}{3}}  
Determine el valor de  ∂F∂x\frac{\partial F}{\partial x}  como  ∂F∂y\frac{\partial F}{\partial y}  en el punto  P(1,2)P\left(1,2\right)  

a)

 ∂F∂x=−0.1259170685\frac{\partial F}{\partial x}=-0.1259170685   ∂F∂y=−0.2518341371\frac{\partial F}{\partial y}=-0.2518341371 

b)

 ∂F∂x=−0.1259170685\frac{\partial F}{\partial x}=-0.1259170685   ∂F∂y=0.2518341371\frac{\partial F}{\partial y}=0.2518341371 

c)

 ∂F∂x=−0.2518341371\frac{\partial F}{\partial x}=-0.2518341371   ∂F∂y=−0.1259170685\frac{\partial F}{\partial y}=-0.1259170685 

d)

 ∂F∂x=0.2518341371\frac{\partial F}{\partial x}=0.2518341371   ∂F∂y=0.1259170685\frac{\partial F}{\partial y}=0.1259170685 

6.

Sea la función

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

a)

 ∂F∂x=4.962633970\frac{\partial F}{\partial x}=4.962633970   ∂F∂y=10.34141477\frac{\partial F}{\partial y}=10.34141477 

b)

 ∂F∂x=10.34141477\frac{\partial F}{\partial x}=10.34141477   ∂F∂y=4.962633970\frac{\partial F}{\partial y}=4.962633970 

c)

 ∂F∂x=−4.962633970\frac{\partial F}{\partial x}=-4.962633970   ∂F∂y=−10.34141477\frac{\partial F}{\partial y}=-10.34141477 

d)

 ∂F∂x=−10.34141477\frac{\partial F}{\partial x}=-10.34141477   ∂F∂y=−4.962633970\frac{\partial F}{\partial y}=-4.962633970 

7.

Sea la función

 F(x,y)=e−x(2y2−x2)F\left(x,y\right)=e^{-x}(2y^2−x^2)  
Determine el valor de  ∂F∂x\frac{\partial F}{\partial x}  como  ∂F∂y\frac{\partial F}{\partial y}  en el punto  P(1,2)P\left(1,2\right)  

a)

 ∂F∂x=1.019978389\frac{\partial F}{\partial x}=1.019978389   ∂F∂y=−1.839397205\frac{\partial F}{\partial y}=-1.839397205 

b)

 ∂F∂x=1.839397205\frac{\partial F}{\partial x}=1.839397205   ∂F∂y=1.019978389\frac{\partial F}{\partial y}=1.019978389 

c)

 ∂F∂x=−1.839397205\frac{\partial F}{\partial x}=-1.839397205   ∂F∂y=1.019978389\frac{\partial F}{\partial y}=1.019978389 

d)

 ∂F∂x=1.019978389\frac{\partial F}{\partial x}=1.019978389   ∂F∂y=1.839397205\frac{\partial F}{\partial y}=1.839397205 

8.

Sea la función

 F(x,y)=ln⁡(1+2x2+3y2)F\left(x,y\right)=\ln(1+2x^2+3y^2)  
Determine el valor de  ∂F∂x\frac{\partial F}{\partial x}  como  ∂F∂y\frac{\partial F}{\partial y}  en el punto  P(1,2)P\left(1,2\right)  

a)

 ∂F∂x=−415\frac{\partial F}{\partial x}=-\frac{4}{15}   ∂F∂y=−45\frac{\partial F}{\partial y}=-\frac{4}{5} 

b)

 ∂F∂x=415\frac{\partial F}{\partial x}=\frac{4}{15}   ∂F∂y=−45\frac{\partial F}{\partial y}=-\frac{4}{5} 

c)

 ∂F∂x=415\frac{\partial F}{\partial x}=\frac{4}{15}   ∂F∂y=45\frac{\partial F}{\partial y}=\frac{4}{5} 

d)

 ∂F∂x=45\frac{\partial F}{\partial x}=\frac{4}{5}   ∂F∂y=415\frac{\partial F}{\partial y}=\frac{4}{15}