$$\begin{array}{rcll}{\displaystyle \frac{\mathit{dz}}{\mathit{dt}}}& =& {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}\cdot {\displaystyle \frac{\mathit{dx}}{\mathit{dt}}}+{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}\cdot {\displaystyle \frac{\mathit{dy}}{\mathit{dt}}}& \\ & & & \\ & =& (y^2+1)\cdot \text{cos}<!--\; FUNCTION\; APPLICATION\; -->t+2\mathit{xy}\cdot {\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}t& \\ & & & \end{array}$$

${\displaystyle \frac{\mathit{dw}}{\mathit{dt}}}=2t+2t\text{sen}<!--\; FUNCTION\; APPLICATION\; -->2t+2t^2\text{cos}<!--\; FUNCTION\; APPLICATION\; -->2t$

$$\begin{array}{rcll}{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}& =& {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial u}}}\cdot {\displaystyle \frac{\mathit{\partial u}}{\mathit{\partial x}}}+{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial v}}}\cdot {\displaystyle \frac{\mathit{\partial v}}{\mathit{\partial x}}}& \\ & =& \left[\sqrt{u+v^2}+{\displaystyle \frac{u}{2\sqrt{u+v^2}}}\right]\cdot y+\left[{\displaystyle \frac{\mathit{uv}}{\sqrt{u+v^2}}}\right]\cdot {\displaystyle \frac{-y∕x^2}{1+{\left({\displaystyle \frac{y}{x}}\right)}^2}}& \\ & & & \end{array}$$

$$\begin{array}{rcll}{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}& =& {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial u}}}\cdot {\displaystyle \frac{\mathit{\partial u}}{\mathit{\partial y}}}+{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial v}}}\cdot {\displaystyle \frac{\mathit{\partial v}}{\mathit{\partial y}}}& \\ & =& \left[\sqrt{u+v^2}+{\displaystyle \frac{u}{2\sqrt{u+v^2}}}\right]\cdot x+\left[{\displaystyle \frac{\mathit{uv}}{\sqrt{u+v^2}}}\right]\cdot {\displaystyle \frac{1∕x}{1+{\left({\displaystyle \frac{y}{x}}\right)}^2}}& \\ & & & \end{array}$$

•

${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}={\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}\cdot y+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}$

•

Aplicamos regla del producto,
$$\begin{array}{rcll}{\displaystyle \frac{{\partial }^{2}z}{\mathit{\partial y\partial x}}}& =& {\displaystyle \frac{\partial }{\mathit{\partial y}}}\left[{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}\cdot y\right]+{\displaystyle \frac{\partial }{\mathit{\partial y}}}\left[{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}\right]& \\ & =& 1\cdot {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}+y\left[{\displaystyle \frac{\partial 2f}{\mathit{\partial u}2}}\cdot x\right]+\left[{\displaystyle \frac{\partial 2f}{\mathit{\partial u\partial v}}}\cdot x\right]& \\ & & & \end{array}$$

•

${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}=3{\text{ln}<!--\; FUNCTION\; APPLICATION\; -->}^2(\mathit{xy})\cdot {\displaystyle \frac{1}{x}}+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}\cdot y+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}$

•

Aplicamos regla del producto,

$$\begin{array}{rcll}{\displaystyle \frac{{\partial }^{2}z}{\mathit{\partial y\partial x}}}& =& 6\text{ln}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{xy})\cdot {\displaystyle \frac{1}{\mathit{xy}}}+{\displaystyle \frac{\partial }{\mathit{\partial y}}}\left[{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}\cdot y\right]+{\displaystyle \frac{\partial }{\mathit{\partial y}}}\left[{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}\right]& \\ & =& 6\text{ln}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{xy})\cdot {\displaystyle \frac{1}{\mathit{xy}}}+1\cdot {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}+y\left[{\displaystyle \frac{\partial 2f}{\mathit{\partial u}2}}\cdot x\right]+\left[{\displaystyle \frac{\partial 2f}{\mathit{\partial u\partial v}}}\cdot x\right]& \\ & & & \end{array}$$

Primero hacemos un cambio de variable: $z=f(U,V)$ con $U=x\mathit{sen}(y)$ $V=g(x).$

$\begin{array}{ccc}\hfill {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial U}}}\cdot U_y+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial V}}}\cdot V_y\hfill \\ \hfill \\ \hfill & \hfill =\hfill & {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial U}}}\cdot x\text{cos}<!--\; FUNCTION\; APPLICATION\; -->y+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial V}}}\cdot 0\hfill \end{array}y\begin{array}{ccc}\hfill {\displaystyle \frac{{\partial }^{2}z}{\mathit{\partial x\partial y}}}& \hfill =\hfill & {\displaystyle \frac{\partial }{\mathit{\partial x}}}\left({\displaystyle \frac{\mathit{\partial f}(U,V)}{\mathit{\partial U}}}\cdot x\text{cos}<!--\; FUNCTION\; APPLICATION\; -->y\right)\hfill \\ \hfill \\ \hfill & \hfill =\hfill & \left({\displaystyle \frac{{\partial }^{2}f}{\partial U^2}}\cdot U_x+{\displaystyle \frac{{\partial }^{2}f}{\mathit{\partial V}\mathit{\partial U}}}\cdot V_x\right)x\text{cos}<!--\; FUNCTION\; APPLICATION\; -->y+\text{cos}<!--\; FUNCTION\; APPLICATION\; -->y{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial U}}}\hfill \\ \hfill \\ \hfill & \hfill =\hfill & \left({\displaystyle \frac{{\partial }^{2}f}{\partial U^2}}\cdot \text{sen}<!--\; FUNCTION\; APPLICATION\; -->y+{\displaystyle \frac{{\partial }^{2}f}{\mathit{\partial V}\mathit{\partial U}}}\cdot g^{\prime }(x)\right)x\text{cos}<!--\; FUNCTION\; APPLICATION\; -->y+\text{cos}<!--\; FUNCTION\; APPLICATION\; -->y{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial U}}}\hfill \\ \hfill \end{array}$

${\displaystyle \frac{\mathit{\partial T}}{\mathit{\partial t}}}=2e^{-3x}\text{sen}<!--\; FUNCTION\; APPLICATION\; -->(2t-3x)$

${\displaystyle \frac{{\partial }^{2}T}{\mathit{\partial x\partial t}}}=-6e^{-3x}\text{cos}<!--\; FUNCTION\; APPLICATION\; -->(2t-3x)+6e^{-3x}\text{sen}<!--\; FUNCTION\; APPLICATION\; -->(2t-3x)$

${\displaystyle \frac{\mathit{\partial T}}{\mathit{\partial x}}}=-3e^{-3x}\text{sen}<!--\; FUNCTION\; APPLICATION\; -->(2t-3x)-3e^{-3x}\text{cos}<!--\; FUNCTION\; APPLICATION\; -->(2t-3x)$

$K=24$.

$$\bullet {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}=f(y)+y{g}^{\prime }(x)\bullet {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}=x{f}^{\prime }(y)+g(x)\bullet {\displaystyle \frac{{\partial }^{2}z}{\mathit{\partial x\partial y}}}=f^{\prime }(y)+g^{\prime }(x)$$

Si $K=2,$ entonces

$$\begin{array}{ccc}2\mathit{xy}(f^{\prime }(y)+g^{\prime }(x))-2x(f(y)+y{g}^{\prime }(x))-2y(x{f}^{\prime }(y)+g(x))+2z\hfill & \hfill \hfill & \hfill \\ \hfill \\ =2\mathit{xy}{f}^{\prime }(y)+2\mathit{xy}{g}^{\prime }(x)-2\mathit{xf}(y)-2\mathit{xy}{g}^{\prime }(x)-2\mathit{yx}{f}^{\prime }(y)-2\mathit{yg}(x)+2z\hfill \\ \hfill & \hfill \hfill & \hfill \\ =-2\mathit{xf}(y)-2\mathit{yg}(x)+2z=0✓\text{ pues }2z=2\mathit{xf}(y)+2\mathit{yg}(x)\hfill \end{array}$$

$$\begin{array}{rcll}{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial u}}}& =& {\displaystyle \frac{\partial 2f}{\mathit{\partial x}2}}\cdot {\displaystyle \frac{\mathit{\partial x}}{\mathit{\partial u}}}+{\displaystyle \frac{\partial 2f}{\mathit{\partial y\partial x}}}\cdot {\displaystyle \frac{\mathit{\partial y}}{\mathit{\partial u}}}+{\displaystyle \frac{\partial 2f}{\mathit{\partial x\partial y}}}\cdot {\displaystyle \frac{\mathit{\partial x}}{\mathit{\partial u}}}+{\displaystyle \frac{\partial 2f}{\mathit{\partial y}2}}\cdot {\displaystyle \frac{\mathit{\partial y}}{\mathit{\partial u}}}& \\ & =& {\displaystyle \frac{\partial 2f}{\mathit{\partial x}2}}\cdot 2u+{\displaystyle \frac{\partial 2f}{\mathit{\partial y\partial x}}}\cdot 1+{\displaystyle \frac{\partial 2f}{\mathit{\partial x\partial y}}}\cdot 2u+{\displaystyle \frac{\partial 2f}{\mathit{\partial y}2}}\cdot 1& \end{array}$$

$$\begin{array}{rcll}{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial v}}}& =& {\displaystyle \frac{\partial 2f}{\mathit{\partial x}2}}\cdot {\displaystyle \frac{\mathit{\partial x}}{\mathit{\partial v}}}+{\displaystyle \frac{\partial 2f}{\mathit{\partial y\partial x}}}\cdot {\displaystyle \frac{\mathit{\partial y}}{\mathit{\partial v}}}+{\displaystyle \frac{\partial 2f}{\mathit{\partial x\partial y}}}\cdot {\displaystyle \frac{\mathit{\partial x}}{\mathit{\partial u}}}+{\displaystyle \frac{\partial 2f}{\mathit{\partial y}2}}\cdot {\displaystyle \frac{\mathit{\partial y}}{\mathit{\partial u}}}& \\ & =& {\displaystyle \frac{\partial 2f}{\mathit{\partial x}2}}\cdot 1+{\displaystyle \frac{\partial 2f}{\mathit{\partial y\partial x}}}\cdot 2v+{\displaystyle \frac{\partial 2f}{\mathit{\partial x\partial y}}}\cdot 1+{\displaystyle \frac{\partial 2f}{\mathit{\partial y}2}}\cdot 2v& \end{array}$$

${\displaystyle \frac{\mathit{\partial w}}{\mathit{\partial 𝜃}}}=-g^{\prime }(x)\cdot r\text{sen}<!--\; FUNCTION\; APPLICATION\; -->𝜃-{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial x}}}\cdot r\text{sen}<!--\; FUNCTION\; APPLICATION\; -->𝜃+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial y}}}\cdot r\text{cos}<!--\; FUNCTION\; APPLICATION\; -->𝜃$

•

${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}=2x{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}+y{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}$

•

Aplicamos la regla del producto,
$$\begin{array}{rcll}{\displaystyle \frac{{\partial }^{2}z}{\partial x^2}}& =& {\displaystyle \frac{\partial }{\mathit{\partial x}}}\left[2x{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}\right]+{\displaystyle \frac{\partial }{\mathit{\partial x}}}\left[y{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}\right]& \\ & =& 2{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}+2x\left[2x{\displaystyle \frac{\partial 2f}{\mathit{\partial u}2}}+y{\displaystyle \frac{\partial 2f}{\mathit{\partial v\partial u}}}\right]+y\left[2x{\displaystyle \frac{\partial 2f}{\mathit{\partial u\partial v}}}+y{\displaystyle \frac{\partial 2f}{\mathit{\partial v}2}}\right]& \\ & & & \end{array}$$

•

Simplificando se obtiene el resultado.

Primero debemos hacer un cambio de variable para poder aplicar la regla de la cadena. Sea $z=f(u,v)-g(w).$ Entonces,
$z_y=f_u\cdot -\text{sen}<!--\; FUNCTION\; APPLICATION\; -->y-g(w)\cdot 6\mathit{xy}.$

$z_{\mathit{xy}}=-\text{sen}<!--\; FUNCTION\; APPLICATION\; -->y(f_{\mathit{uu}}\cdot 2x+f_{\mathit{uv}}\cdot 2x)-g^{″<!--\; FUNCTION\; APPLICATION\; -->}(w)\cdot 3y^2\cdot 6\mathit{xy}+6y\cdot g^{\prime }(w)$

Primero debemos hacer un cambio de variable para poder aplicar la regla de la cadena. Sea $z=x^2{f}^4(u,v).$ Entonces,

$z_x=2x{f}^4(u,v)+x^2(4f^3(u,v)\cdot (f_u\cdot y+f_v\cdot 0))$

$z_y=x^2(4f^3(u,v)\cdot (f_u\cdot x+f_v\cdot 2y))$

Sea $u=\frac{y}{x}$, $v=e^{3x}+3x$ y $w=x^2{y}^2$. Entonces $z=x\cdot f\left(u,v\right)+h\left(w\right)$

$$\begin{array}{rcll}\frac{\mathit{\partial z}}{\mathit{\partial y}}& =& x\cdot \frac{\mathit{\partial f}}{\mathit{\partial y}}+\frac{\mathit{\partial h}}{\mathit{\partial y}}& \\ & =& x\cdot \frac{\mathit{\partial f}}{\mathit{\partial y}}+\frac{\mathit{\partial h}}{\mathit{\partial y}}& \\ & =& x\cdot \left(\frac{\mathit{\partial f}}{\mathit{\partial u}}\cdot \frac{\mathit{\partial u}}{\mathit{\partial y}}+\frac{\mathit{\partial f}}{\mathit{\partial v}}\cdot \frac{\mathit{\partial v}}{\mathit{\partial y}}\right)+h^{\prime }\left(w\right)\cdot \frac{\mathit{\partial w}}{\mathit{\partial y}}& \\ & =& x\cdot \left(\frac{\mathit{\partial f}}{\mathit{\partial u}}\cdot \frac{1}{x}+\frac{\mathit{\partial f}}{\mathit{\partial v}}\cdot 0\right)+h^{\prime }\left(w\right)\cdot 2x^{2}y& \\ & =& \frac{\mathit{\partial f}}{\mathit{\partial u}}+h^{\prime }\left(w\right)\cdot 2x^{2}y& \end{array}$$

Ahora

$$\begin{array}{rcll}\frac{{\partial }^{2}z}{\mathit{\partial x\partial y}}& =& \frac{\partial \left(\frac{\mathit{\partial z}}{\mathit{\partial y}}\right)}{\mathit{\partial x}}& \\ & =& \frac{\partial \left(\frac{\mathit{\partial f}}{\mathit{\partial u}}+h^{\prime }\left(w\right)\cdot 2x^{2}y\right)}{\mathit{\partial x}}& \\ & =& \frac{{\partial }^{2}f}{\partial u^2}\cdot \frac{\mathit{\partial u}}{\mathit{\partial x}}+\frac{{\partial }^{2}f}{\mathit{\partial v\partial u}}\cdot \frac{\mathit{\partial v}}{\mathit{\partial x}}+h^{\mathit{\prime \prime }}\left(w\right)\frac{\mathit{\partial w}}{\mathit{\partial x}}\cdot 2x^{2}y+h^{\prime }\left(w\right)\cdot 4\mathit{xy}& \\ & =& \frac{{\partial }^{2}f}{\partial u^2}\cdot \frac{-y}{x^2}+\frac{{\partial }^{2}f}{\mathit{\partial v\partial u}}\cdot \left(3e^{3x}+3\right)+h^{\mathit{\prime \prime }}\left(w\right)\cdot 2x{y}^2\cdot 2x^{2}y+h^{\prime }\left(w\right)\cdot 4\mathit{xy}& \\ & =& \frac{{\partial }^{2}f}{\partial u^2}\cdot \frac{-y}{x^2}+\frac{{\partial }^{2}f}{\mathit{\partial v\partial u}}\cdot \left(3e^{3x}+3\right)+h^{\mathit{\prime \prime }}\left(w\right)\cdot 4x^3{y}^3+h^{\prime }\left(w\right)\cdot 4\mathit{xy}& \end{array}$$

Cambio de variable: $z=f(u,v)+g(w)$
${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}={\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}\cdot (2x^2-1)-x\cdot {\displaystyle \frac{\mathit{dg}}{\mathit{dw}}}$

${\displaystyle \frac{{\partial }^{2}z}{\mathit{\partial x\partial y}}}=4x{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}+(2x^2-1)\left({\displaystyle \frac{{\partial }^{2}f}{\partial u^2}}4\mathit{xy}+{\displaystyle \frac{{\partial }^{2}f}{\mathit{\partial v\partial u}}}2x\right)-g^{\prime }(w)-x\cdot g^{″}(w)\cdot (2x-y)$

Las derivadas parciales que se piden aparecen cuando calculamos ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}$ y ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}.$ La idea es despejar a partir de estos dos cálculos.

$\{\begin{array}{ccccc}\hfill {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial s}}}\cdot 2x+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial t}}}\cdot y\hfill & \hfill \hfill & \hfill \\ \hfill & \hfill \hfill & \hfill & \hfill ⟹\hfill & {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial t}}}={\displaystyle \frac{1}{y+2x^2}}\left[{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}+2x\cdot {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}\right]\hfill \\ \hfill {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial s}}}\cdot -1+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial t}}}\cdot x\hfill & \hfill \hfill & \hfill \end{array}$

Sea $u=\mathit{xy}.$ Con un cálculo directo obtenemos ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}=\mathit{yf}(u)$ y ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}=x{f}^{\prime }(u)$ Sustituyendo en la ecuación diferencial se obtiene el resultado.

Observe que

${\displaystyle \frac{\partial }{\mathit{\partial x}}}\left(r^2\right)={\displaystyle \frac{\partial }{\mathit{\partial x}}}\left(x^2+y^2+z^2\right)⟹2r{\displaystyle \frac{\mathit{\partial r}}{\mathit{\partial x}}}=2x$, es decir, ${\displaystyle \frac{\mathit{\partial r}}{\mathit{\partial x}}}={\displaystyle \frac{x}{r}}.$ De manera análoga,

${\displaystyle \frac{\mathit{\partial r}}{\mathit{\partial y}}}={\displaystyle \frac{y}{r}}$

${\displaystyle \frac{\mathit{\partial r}}{\mathit{\partial z}}}={\displaystyle \frac{z}{r}}$

Entonces

$$x{\displaystyle \frac{\mathit{\partial u}}{\mathit{\partial x}}}+y{\displaystyle \frac{\mathit{\partial u}}{\mathit{\partial y}}}+z{\displaystyle \frac{\mathit{\partial u}}{\mathit{\partial z}}}=x\cdot f^{\prime }(r)\cdot {\displaystyle \frac{x}{r}}+y\cdot f^{\prime }(r)\cdot {\displaystyle \frac{y}{r}}+z\cdot f^{\prime }(r){\displaystyle \frac{z}{r}}=f^{\prime }(r)\cdot {\displaystyle \frac{x^2+y^2+z^2}{r}}=r{f}^{\prime }(r)$$

$${\displaystyle \frac{{\partial }^{2}z}{\mathit{\partial x\partial y}}}=-{\displaystyle \frac{3}{x^2}}\cdot f_v+3x{f}_{\mathit{vu}}$$

Al aplicar el cambio de variable, $z$ es una función de $u$ y $v,$ es decir, $z=z(u,v).$ Por tanto

${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}={\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial u}}}{u}_x+{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial v}}}{v}_y=y{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial u}}}+{\displaystyle \frac{1}{y}}{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial v}}}$ y ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}=x{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial u}}}-{\displaystyle \frac{x}{y^2}}{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial v}}}$

Sustituyendo ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}},$ ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}$, $x=\sqrt{\mathit{uv}}$ y $y=\sqrt{u∕v}$en $\text{(∗),}$ nos queda

$${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial u}}}=v\text{sen}<!--\; FUNCTION\; APPLICATION\; -->u$$