9-36 Find the first partial derivatives of the function. 9. f(x,y)=x4+5xy3 10. f(x,y)=x2y−3y4 11. g(x,y)=x3siny 12. g(x,t)=ext 13. z=ln(x+t2) 14. w=v2u 15. f(x,y)=yexy 16. g(x,y)=(x2+xy)3 17. g(x,y)=y(x+x2y)5 18. f(x,y)=(x+y)2x 19. f(x,y)=cx+dyax+by 20. w=u+v2ev

Answer(15)- given that function is f(x,y)= ye^(xy) differentiate partially with to 'x' and 'y'

Resuelto 9-36 Find the first partial derivatives of the

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Pregunta: 9-36 Find the first partial derivatives of the function. 9. f(x,y)=x4+5xy3 10. f(x,y)=x2y−3y4 11. g(x,y)=x3siny 12. g(x,t)=ext 13. z=ln(x+t2) 14. w=v2u 15. f(x,y)=yexy 16. g(x,y)=(x2+xy)3 17. g(x,y)=y(x+x2y)5 18. f(x,y)=(x+y)2x 19. f(x,y)=cx+dyax+by 20. w=u+v2ev
15 and 17 and 19

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Solución
Paso 1
Answer(15)- given that function is f(x,y)= $y e^{x y}$
differentiate partially with to 'x' and 'y'

$\begin{matrix} \begin{aligned} \frac{\partial f}{\partial x} & = \frac{\partial}{\partial x} y e^{x y} \\ = y e^{x y} \frac{\partial}{\partial x} x y \\ = y e^{x y} y \\ \frac{\partial f}{\partial x} & = y^{2} e^{x y} \end{aligned} \end{matrix}$

$\begin{matrix} \begin{aligned} \frac{\partial f}{\partial y} & = \frac{\partial}{\partial x} (y e^{x y}) \\ = y \frac{\partial}{\partial x} e^{x y} + e^{x y} \frac{\partial}{\partial y} y \\ = x y e^{x y} + e^{x y} .1 \\ \frac{\partial f}{\partial y} & = e^{x y} (x y + 1) \end{aligned} \end{matrix}$

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Texto de la transcripción de la imagen:

9-36 Find the first partial derivatives of the function. 9.

f (x, y) = x^{4} + 5 x y^{3}

10.

f (x, y) = x^{2} y - 3 y^{4}

11.

g (x, y) = x^{3} sin y

12.

g (x, t) = e^{x t}

13.

z = ln (x + t^{2})

14.

w = \frac{u}{v ^{2}}

15.

f (x, y) = y e^{x y}

16.

g (x, y) = (x^{2} + x y)^{3}

17.

g (x, y) = y (x + x^{2} y)^{5}

18.

f (x, y) = \frac{x}{( x + y ) ^{2}}

19.

f (x, y) = \frac{a x + b y}{c x + d y}

20.

w = \frac{e ^{v}}{u + v ^{2}}