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

∂ F ∂ x = − 10.34141477 \frac{\partial F}{\partial x}=-10.34141477 ∂ x ∂ F = − 1 0 . 3 4 1 4 1 4 7 7 ∂ F ∂ y = − 4.962633970 \frac{\partial F}{\partial y}=-4.962633970 ∂ y ∂ F = − 4 . 9 6 2 6 3 3 9 7 0

∂ F ∂ x = 4.962633970 \frac{\partial F}{\partial x}=4.962633970 ∂ x ∂ F = 4 . 9 6 2 6 3 3 9 7 0 ∂ F ∂ y = 10.34141477 \frac{\partial F}{\partial y}=10.34141477 ∂ y ∂ F = 1 0 . 3 4 1 4 1 4 7 7

∂ F ∂ x = 10.34141477 \frac{\partial F}{\partial x}=10.34141477 ∂ x ∂ F = 1 0 . 3 4 1 4 1 4 7 7 ∂ F ∂ y = 4.962633970 \frac{\partial F}{\partial y}=4.962633970 ∂ y ∂ F = 4 . 9 6 2 6 3 3 9 7 0

∂ F ∂ x = − 4.962633970 \frac{\partial F}{\partial x}=-4.962633970 ∂ x ∂ F = − 4 . 9 6 2 6 3 3 9 7 0 ∂ F ∂ y = − 10.34141477 \frac{\partial F}{\partial y}=-10.34141477 ∂ y ∂ F = − 1 0 . 3 4 1 4 1 4 7 7

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Authored by José Torre

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MULTIPLE CHOICE QUESTION

3 mins • 1 pt

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

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

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

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

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

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 $P\left(3,4\right)$

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

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

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

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

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

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 $P\left(1,-2\right)$

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

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

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

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

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

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 $P\left(1,2\right)$

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

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

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

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

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

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 $P\left(1,2\right)$

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

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

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

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

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

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 $P\left(1,2\right)$

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

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

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

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

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

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 $P\left(1,2\right)$

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

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

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

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

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

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 $P\left(1,2\right)$

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

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

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

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

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derivadas parciales

Sea la función F(x,y)=x2y2−2xy2F(x,y)=x^2y^2−2xy^2F(x,y)=x2y2−2xy2 Determine el valor de ∂F∂x\frac{\partial F}{\partial x}∂x∂F​ como ∂F∂y\frac{\partial F}{\partial y}∂y∂F​ en el punto P(2,1)P\left(2,1\right)P(2,1)

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Sea la función F(x,y)=3x3y−2x2y2+y3F\left(x,y\right)=3x^3y-2x^2y^2+y^3F(x,y)=3x3y−2x2y2+y3 Determine el valor de ∂F∂x\frac{\partial F}{\partial x}∂x∂F​ como ∂F∂y\frac{\partial F}{\partial y}∂y∂F​ en el punto P(1,−2)P\left(1,-2\right)P(1,−2)

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

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