Sea $z(x,y)=\sqrt{\text{arctan}<!--\; FUNCTION\; APPLICATION\; -->(y∕x)+\text{tan}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{xy})}.$ Podemos hacer un cambio de variable y calcular ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}$ usando la regla de la cadena. Sea $u(x,y)=\text{arctan}<!--\; FUNCTION\; APPLICATION\; -->(y∕x)$ y $v(x,y)=\text{tan}<!--\; FUNCTION\; APPLICATION\; -->(\mathit{xy}),$ entonces $z=\sqrt{u+v}.$

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\mathit{\partial z}}{\partial x}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial u}}}{\displaystyle \frac{\mathit{\partial u}}{\partial x}}+{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial v}}}{\displaystyle \frac{\mathit{\partial v}}{\partial x}}\hfill \\ \hfill & \hfill \hfill & \hfill \\ \hfill & \hfill =\hfill & {\displaystyle \frac{1}{2\sqrt{u+v}}}{\displaystyle \frac{1}{1+{(y∕x)}^2}}\cdot {\displaystyle \frac{-y\cdot 1}{x^2}}+{\displaystyle \frac{1}{2\sqrt{u+v}}}y{\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^2(xy)\hfill \end{array}$$

Al sustituir $u$ y $v$ obtenemos el resultado completo, si fuera necesario.

Sea $z(x,y)=x^2+3y^2,$ donde $x=e^t$ y $y=\text{cos}<!--\; FUNCTION\; APPLICATION\; -->(t)$ entonces

$\begin{array}{ccc}\hfill {\displaystyle \frac{\mathit{dz}}{\mathit{dt}}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial z}}{\partial x}}{\displaystyle \frac{dx}{\mathit{dt}}}+{\displaystyle \frac{\mathit{\partial z}}{\partial y}}{\displaystyle \frac{dy}{\mathit{dt}}}\hfill \\ \hfill & \hfill \hfill & \hfill \\ \hfill & \hfill =\hfill & 2x{e}^t-6y\text{sen}<!--\; FUNCTION\; APPLICATION\; -->(t)=2e^{2t}-6\text{cos}<!--\; FUNCTION\; APPLICATION\; -->(t)\text{sen}<!--\; FUNCTION\; APPLICATION\; -->(t)\hfill \end{array}$

Sea $z(u,v)=x^2{e}^{y^3},$ donde $x=\mathit{uv}$ y $y=u^2-v^3$ entonces

$\begin{array}{ccc}\hfill {\displaystyle \frac{\mathit{\partial z}}{\partial u}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial z}}{\partial x}}{\displaystyle \frac{\partial x}{\partial u}}+{\displaystyle \frac{\mathit{\partial z}}{\partial y}}{\displaystyle \frac{\partial y}{\partial u}}\hfill \\ \hfill & \hfill \hfill & \hfill \\ \hfill & \hfill =\hfill & 2x{e}^{y^3}{\displaystyle \frac{\mathit{\partial x}}{\partial u}}+3x^2{y}^2{e}^{y^3}{\displaystyle \frac{\mathit{\partial y}}{\partial u}}=2x{e}^{y^3}v+3x^2{y}^2{e}^{y^3}2u\hfill \end{array}$

$\begin{array}{ccc}\hfill {\displaystyle \frac{\mathit{\partial z}}{\partial v}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial z}}{\partial x}}{\displaystyle \frac{\partial x}{\partial v}}+{\displaystyle \frac{\mathit{\partial z}}{\partial y}}{\displaystyle \frac{\partial y}{\partial v}}\hfill \\ \hfill & \hfill \hfill & \hfill \\ \hfill & \hfill =\hfill & 2x{e}^{y^3}{\displaystyle \frac{\mathit{\partial x}}{\partial v}}+3x^2{y}^2{e}^{y^3}{\displaystyle \frac{\mathit{\partial y}}{\partial v}}=2x{e}^{y^3}u+3x^2{y}^2{e}^{y^3}\cdot -3v^2\hfill \end{array}$

Sea $f$ una función diferenciable y $z(x,y)=f(x^2,x{y}^2).$ Para derivar usando la regla de la cadena usamos el cambio de variable $u=x^2$ y $v=x{y}^2,$ entonces $z(x,y)=f(u,v)$ y

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

$\begin{array}{ccc}\hfill {\displaystyle \frac{\mathit{\partial z}}{\partial y}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}\cdot {\displaystyle \frac{\mathit{\partial u}}{\partial y}}+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}\cdot {\displaystyle \frac{\mathit{\partial v}}{\partial y}}={\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial u}}}\cdot 0+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}\cdot 2\mathit{xy}\hfill \end{array}$

Sea $f$ una función derivable y $z=f(x,y)$ con $x=r\text{cos}<!--\; FUNCTION\; APPLICATION\; -->𝜃,y=r\text{sen}<!--\; FUNCTION\; APPLICATION\; -->𝜃,$ entonces

$\begin{array}{ccc}\hfill {\displaystyle \frac{\mathit{\partial z}}{\partial r}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial x}}}{\displaystyle \frac{\mathit{\partial x}}{\partial r}}+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial y}}}{\displaystyle \frac{\mathit{\partial y}}{\partial r}}={\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial x}}}\text{cos}<!--\; FUNCTION\; APPLICATION\; -->𝜃+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial y}}}\text{sen}<!--\; FUNCTION\; APPLICATION\; -->𝜃\hfill \end{array}$

$\begin{array}{ccc}\hfill {\displaystyle \frac{\mathit{\partial z}}{\partial 𝜃}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial x}}}{\displaystyle \frac{\mathit{\partial x}}{\partial 𝜃}}+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial y}}}{\displaystyle \frac{\mathit{\partial y}}{\partial 𝜃}}={\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial x}}}\cdot -r\text{sen}<!--\; FUNCTION\; APPLICATION\; -->𝜃+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial y}}}r\text{cos}<!--\; FUNCTION\; APPLICATION\; -->𝜃\hfill \end{array}$

Sea $V=V(P,T).$ Si $P(V-b)e^{\mathit{RV}}=\mathit{RT},$ con $b,R$ constantes, calcule ${\displaystyle \frac{\mathit{\partial V}}{\mathit{\partial T}}}.$

Solución. $V$ es función de $P$ y $T.$ Derivamos a ambos lados respecto a $T,$

$$\begin{array}{rcll}{\displaystyle \frac{\partial }{\mathit{\partial T}}}\left[P(V-b)e^{\mathit{RV}}\right]& =& {\displaystyle \frac{\partial }{\mathit{\partial T}}}\left[\mathit{RT}\right]& \\ P\left[V_T{e}^{\mathit{RV}}+(V-b)e^{\mathit{RV}}R{V}_T\right]& =& R& \\ \therefore V_T& =& {\displaystyle \frac{R}{P{e}^{\mathit{RV}}(1+(V-b)R).}}& \\ & & & \end{array}$$

Si $z=g(x^2)+f(\text{arctan}<!--\; FUNCTION\; APPLICATION\; -->x,\text{tan}<!--\; FUNCTION\; APPLICATION\; -->y),$ calcule ${\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial x}}}$, ${\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial y}}},$ ${\displaystyle \frac{\partial 2z}{\mathit{\partial x\partial y}}}$ y ${\displaystyle \frac{\partial 2z}{\mathit{\partial x}2}}$

Solución. : Primero un cambio de variable $z=g(u)+f(v,w)$ con $\{\begin{array}{ccc}\hfill u& \hfill =\hfill & x^2\hfill \\ \hfill v& \hfill =\hfill & \text{arctan}<!--\; FUNCTION\; APPLICATION\; -->x\hfill \end{array}$

1.

${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}=g^{\prime }(u)\cdot 2x+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}\cdot {\displaystyle \frac{1}{1+x^2}}+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial w}}}\cdot 0$

2.

$\begin{array}{ccc}{\displaystyle \frac{\partial 2z}{\mathit{\partial x}2}}\hfill & \hfill =\hfill & {\displaystyle \frac{\partial }{\mathit{\partial x}}}\left(2x\cdot g^{\prime }(u)+{\displaystyle \frac{1}{1+x^2}}\cdot {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}(v,w)\right)\hfill \\ \end{array}$

3.

${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}=0+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial v}}}\cdot 0+{\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial w}}}\cdot {\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}y$

4.

$\begin{array}{ccc}{\displaystyle \frac{\partial 2z}{\mathit{\partial x\partial y}}}\hfill & \hfill =\hfill & {\displaystyle \frac{\partial }{\mathit{\partial x}}}\left({\text{sec}<!--\; FUNCTION\; APPLICATION\; -->}^{2}y\cdot {\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial w}}}(v,w)\right)\hfill \\ \end{array}$

Si $z(x,y)=g(y)\cdot f(x-2y,y^3).$ Calcule $z_x$ y $z_{\mathit{xy}}.$

Solución. Cambio de variable: Sea $u=x-2y,$ $v=y^3.$ Entonces $z(x,y)=g(y)f(u,v).$

$z_x=g(y)\left[f_u\cdot 1+f_v\cdot 0\right]=g(y)f_u(u,v)$

$z_{xy}=g^{\prime }(y)\cdot 1\cdot f_u+g(y)\left[-2f_{uu}+3y^2{f}_{uv}\right]$

Cambio de variable: Sea $z(x,y)=g(u,v)$ con $u=x^2{y}^2$ y $v=\mathit{xy}.$ Calcule ${\displaystyle \frac{\partial 2z}{\mathit{\partial y\partial x}}}.$

Solución.

$\begin{array}{ccccc}\hfill {\displaystyle \frac{\mathit{\partial z}}{\partial x}}& \hfill =\hfill & {\displaystyle \frac{\mathit{\partial g}}{\mathit{\partial u}}}{\displaystyle \frac{\mathit{\partial u}}{\partial x}}+{\displaystyle \frac{\mathit{\partial g}}{\mathit{\partial v}}}{\displaystyle \frac{\mathit{\partial v}}{\partial x}}\hfill & \hfill =\hfill & {\displaystyle \frac{\mathit{\partial g}}{\mathit{\partial u}}}\cdot 2x{y}^2+{\displaystyle \frac{\mathit{\partial g}}{\mathit{\partial v}}}\cdot y\hfill \\ \hfill \end{array}$

$\begin{array}{ccccc}\hfill {\displaystyle \frac{\partial 2z}{\partial y\mathit{\partial x}}}& \hfill =\hfill & {\displaystyle \frac{\partial }{\partial y}}\left[{\displaystyle \frac{\mathit{\partial g}(u,v)}{\mathit{\partial u}}}\right]\cdot 2x{y}^2+{\displaystyle \frac{\partial }{\partial y}}\left[2x{y}^2\right]\cdot {\displaystyle \frac{\mathit{\partial g}}{\mathit{\partial u}}}+{\displaystyle \frac{\mathit{\partial g}(u,v)}{\mathit{\partial v}}}+y\cdot {\displaystyle \frac{\partial }{\partial y}}\left[{\displaystyle \frac{\mathit{\partial g}(u,v)}{\mathit{\partial v}}}\right]\hfill & \hfill \hfill & \hfill \\ \hfill \\ \hfill & \hfill \hfill & \hfill & \hfill \hfill & \hfill \\ \hfill & \hfill =\hfill & \left[{\displaystyle \frac{{\partial }^{2}g}{\partial u^2}}\cdot u_y+{\displaystyle \frac{{\partial }^{2}g}{\mathit{\partial v}\mathit{\partial u}}}\cdot v_y\right]\cdot 2x{y}^2+4\mathit{xy}\cdot {\displaystyle \frac{\mathit{\partial g}}{\mathit{\partial u}}}+{\displaystyle \frac{\mathit{\partial g}(u,v)}{\mathit{\partial v}}}+y\cdot \left[{\displaystyle \frac{{\partial }^{2}g}{\mathit{\partial u}\mathit{\partial v}}}\cdot u_y+{\displaystyle \frac{{\partial }^{2}g}{\partial v^2}}\cdot v_y\right]\hfill & \hfill \hfill & \hfill \\ \hfill \\ \hfill & \hfill \hfill & \hfill & \hfill \hfill & \hfill \\ \hfill \\ \hfill & \hfill =\hfill & \left[{\displaystyle \frac{{\partial }^{2}g}{\partial u^2}}\cdot 2y{x}^2+{\displaystyle \frac{{\partial }^{2}g}{\mathit{\partial v}\mathit{\partial u}}}\cdot x\right]\cdot 2x{y}^2+4\mathit{xy}\cdot {\displaystyle \frac{\mathit{\partial g}}{\mathit{\partial u}}}+{\displaystyle \frac{\mathit{\partial g}(u,v)}{\mathit{\partial v}}}+y\cdot \left[{\displaystyle \frac{{\partial }^{2}g}{\mathit{\partial u}\mathit{\partial v}}}\cdot 2y{x}^2+{\displaystyle \frac{{\partial }^{2}g}{\partial v^2}}\cdot x\right]\hfill & \hfill \hfill & \hfill \end{array}$

Sea $F(u,v)=-u-v$ con $u^2=x-y$ y $v^2=x+y.$ Si $u\ne 0$ y $v\ne 0,$ verifique

a.)

$F_x=-{\displaystyle \frac{u+v}{2\mathit{uv}}}.$

b.)

$F_y={\displaystyle \frac{v-u}{2\mathit{uv}}}.$

Solución. Primero veamos que $2u{u}_x=1,$ $2v{v}_x=1,$ $2u{u}_y=-1$ y $2v{v}_y=1.$ Por lo tanto

a.)

$F_x=F_u{u}_x+F_v{v}_x=-1\cdot {\displaystyle \frac{1}{2u}}-1\cdot {\displaystyle \frac{1}{2v}}=-{\displaystyle \frac{u+v}{2\mathit{uv}}}.$

b.)

$F_y=F_u{u}_y+F_v{v}_y=-1\cdot \left(-{\displaystyle \frac{1}{2u}}\right)-1\cdot {\displaystyle \frac{1}{2v}}={\displaystyle \frac{v-u}{2\mathit{uv}}}.$