Recordemos que en una variable, si $k$ es una constante, $\{\begin{array}{ccc}{\displaystyle \frac{d}{\mathit{du}}}\left(k\cdot f(u)\right)\hfill & \hfill =\hfill & k\cdot {\displaystyle \frac{\mathit{df}}{\mathit{du}}}\hfill \\ \end{array}$

1.

Si $z=x^2{y}^2+y,$ calcular ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}$ y ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}.$

Solución.

• $z=x^2{y}^2+y⟹{\displaystyle \frac{\mathit{\partial z}}{\partial x}}={\displaystyle \frac{d}{\mathit{dx}}}\left(x^2\right)y^2+{\displaystyle \frac{d}{\mathit{dx}}}\left(y\right)=2x{y}^2+0$

• $z=x^2{y}^2+y⟹{\displaystyle \frac{\mathit{\partial z}}{\partial y}}=x^2{\displaystyle \frac{d}{\mathit{dy}}}\left(y^2\right)+{\displaystyle \frac{d}{\mathit{dy}}}\left(y\right)=x^{2}2y+1$

2.

Si $z={\displaystyle \frac{x^3}{y^5}},$ calcular ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}$ y ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}.$

Solución.

• $z={\displaystyle \frac{x^3}{y^5}}⟹{\displaystyle \frac{\mathit{\partial z}}{\partial x}}={\displaystyle \frac{\partial }{\partial x}}\left({\displaystyle \frac{1}{y^5}}\cdot x^3\right)={\displaystyle \frac{1}{y^5}}\cdot {\displaystyle \frac{d}{\mathit{dx}}}\left(x^3\right)={\displaystyle \frac{1}{y^5}}\cdot 3x^2$

• $z={\displaystyle \frac{x^3}{y^5}}⟹{\displaystyle \frac{\mathit{\partial z}}{\partial y}}={\displaystyle \frac{\partial }{\partial y}}\left({\displaystyle \frac{x^3}{y^5}}\right)={\displaystyle \frac{-x^3\cdot {\displaystyle \frac{d}{\mathit{dy}}}\left(y^5\right)}{y^{10}}}={\displaystyle \frac{-x^3\cdot 5y^4}{y^{10}}}$

Recordemos que, $\{\begin{array}{ccc}{\displaystyle \frac{d}{\mathit{dx}}}\left(f(u)\right)\hfill & \hfill =\hfill & {\displaystyle \frac{\mathit{df}}{du}}\cdot {\displaystyle \frac{du}{\mathit{dx}}},\text{ en particular }{\displaystyle \frac{d}{\mathit{dx}}}\left(f^n(u)\right)=n{f}^{n-1}(u)\cdot {\displaystyle \frac{\mathit{df}}{du}}\cdot {\displaystyle \frac{du}{\mathit{dx}}}\hfill \\ \\ {\displaystyle \frac{d}{\mathit{dx}}}\left(f(x)g(x)\right)\hfill & \hfill =\hfill & {\displaystyle \frac{\mathit{df}}{\mathit{dx}}}\cdot g(x)+f(x)\cdot {\displaystyle \frac{\mathit{dg}}{\mathit{dx}}}\hfill \\ \\ {\displaystyle \frac{d}{\mathit{dx}}}\left({\displaystyle \frac{f(x)}{g(x)}}\right)\hfill & \hfill =\hfill & {\displaystyle \frac{{\displaystyle \frac{\mathit{df}}{\mathit{dx}}}\cdot g(x)-f(x)\cdot {\displaystyle \frac{\mathit{dg}}{\mathit{dx}}}}{g^2(x)}}\hfill \\ \end{array}$

Si $w={\displaystyle \frac{y+z^2{\text{cos}<!--\; FUNCTION\; APPLICATION\; -->}^4(z{x}^3)}{1+y^2}},$ calcule ${\displaystyle \frac{\mathit{\partial w}}{\mathit{\partial x}}},$ ${\displaystyle \frac{\mathit{\partial w}}{\mathit{\partial y}}}$ y ${\displaystyle \frac{\mathit{\partial w}}{\mathit{\partial z}}}$

Solución.

• ${\displaystyle \frac{\mathit{\partial w}}{\mathit{\partial x}}}={\displaystyle \frac{1}{1+y^2}}\cdot {\displaystyle \frac{\partial }{\mathit{\partial x}}}\left(y+z^2{\text{cos}<!--\; FUNCTION\; APPLICATION\; -->}^4(z{x}^3)\right)={\displaystyle \frac{0+4z^2{\text{cos}<!--\; FUNCTION\; APPLICATION\; -->}^3(z{x}^3)\cdot -\text{sen}<!--\; FUNCTION\; APPLICATION\; -->(z{x}^3)\cdot 3x^{2}z}{1+y^2}}$

$\begin{array}{ccc}{\displaystyle \frac{\mathit{\partial w}}{\mathit{\partial y}}}\hfill & \hfill =\hfill & {\displaystyle \frac{{\displaystyle \frac{\partial }{\mathit{\partial y}}}\left(y+z^2{\text{cos}<!--\; FUNCTION\; APPLICATION\; -->}^4(z{x}^3)\right)\cdot (1+y^2)-{\displaystyle \frac{\partial }{\mathit{\partial y}}}\left(1+y^2\right)\cdot (y+z^2{\text{cos}<!--\; FUNCTION\; APPLICATION\; -->}^4(z{x}^3))}{{(1+y^2)}^2}}\hfill \\ \end{array}$

$\begin{array}{ccc}{\displaystyle \frac{\mathit{\partial w}}{\mathit{\partial z}}}\hfill & \hfill =\hfill & {\displaystyle \frac{1}{1+y^2}}\cdot {\displaystyle \frac{\partial }{\mathit{\partial z}}}\left(y+z^2{\text{cos}<!--\; FUNCTION\; APPLICATION\; -->}^4(z{x}^3)\right)\hfill \\ \\ \hfill & \hfill =\hfill & {\displaystyle \frac{0+{\displaystyle \frac{\partial }{\mathit{\partial z}}}\left(z^2\right)\cdot {\text{cos}<!--\; FUNCTION\; APPLICATION\; -->}^4(z{x}^3)+z^2\cdot {\displaystyle \frac{\partial }{\mathit{\partial z}}}\left({\text{cos}<!--\; FUNCTION\; APPLICATION\; -->}^4(z{x}^3)\right)}{1+y^2}}\hfill \\ \end{array}$

El volumen de un cono es $V={\displaystyle \frac{\pi r^{2}h}{3}},$ calcule ${{\displaystyle \frac{\mathit{\partial V}}{\mathit{\partial r}}}|}_{r=2,h=4}$ y ${{\displaystyle \frac{\mathit{\partial V}}{\mathit{\partial h}}}|}_{r=2,h=4}$

Solución.

• ${{\displaystyle \frac{\mathit{\partial V}}{\mathit{\partial r}}}|}_{r=2,h=4}={{\displaystyle \frac{2\mathit{\pi rh}}{3}}|}_{r=2,h=4}={\displaystyle \frac{16\pi }{3}}$

• ${{\displaystyle \frac{\mathit{\partial V}}{\mathit{\partial h}}}|}_{r=2,h=4}={{\displaystyle \frac{\pi r^2}{3}}|}_{r=2,h=4}={\displaystyle \frac{4\pi }{3}}$

Verifique que si $z=\text{arctan}<!--\; FUNCTION\; APPLICATION\; -->(y∕x),$ entonces $x{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}+y{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}=0.$

Solución.

Si $w=z^2\text{ln}<!--\; FUNCTION\; APPLICATION\; -->(x^2)\text{cos}<!--\; FUNCTION\; APPLICATION\; -->(y^2),$ determine $g(z)$ tal que $x\text{ln}<!--\; FUNCTION\; APPLICATION\; -->(x^2){\displaystyle \frac{\mathit{\partial w}}{\mathit{\partial x}}}+g(z){\displaystyle \frac{\mathit{\partial w}}{\mathit{\partial z}}}=4w$

Solución.

Ahora,

Por lo tanto, si $g(z)=z$ tendríamos lo que se pide:

Si $f(t,𝜃)=e^{2𝜃}\varphi (t,𝜃),$ calcule ${\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial t}}}$ y ${\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial 𝜃}}}$

Solución. : En este caso, como $\varphi$ no es conocida, sus derivadas parciales solo se dejan indicadas.

• ${\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial t}}}=e^{2𝜃}{\displaystyle \frac{\mathit{\partial \varphi }}{\mathit{\partial t}}}$

• ${\displaystyle \frac{\mathit{\partial f}}{\mathit{\partial 𝜃}}}={\displaystyle \frac{\partial }{\mathit{\partial 𝜃}}}\left(e^{2𝜃}\right)\cdot \varphi (t,𝜃)+e^{2𝜃}{\displaystyle \frac{\mathit{\partial \varphi }}{\mathit{\partial 𝜃}}}=2e^{2𝜃}\cdot \varphi (t,𝜃)+e^{2𝜃}{\displaystyle \frac{\mathit{\partial \varphi }}{\mathit{\partial 𝜃}}}$

Sean $f,g:ℝ\to ℝ$ funciones derivables y $u=x^5+y^3.$ Si $z=x^{2}g(u)+f^4(u),$ calcule ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}$ y ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}.$

Solución. : Como $f$ y $g$ no son conocidas, sus derivadas solo se dejan indicadas.

$\begin{array}{ccc}{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}\hfill & \hfill =\hfill & 2x\cdot g(u)+x^2{\displaystyle \frac{\mathit{dg}}{\mathit{du}}}\cdot {\displaystyle \frac{\mathit{du}}{\mathit{dx}}}+4f^3(u)\cdot {\displaystyle \frac{\mathit{df}}{\mathit{du}}}\cdot {\displaystyle \frac{\mathit{du}}{\mathit{dx}}}\hfill \\ \end{array}$

$\begin{array}{ccc}{\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}\hfill & \hfill =\hfill & x^2{\displaystyle \frac{\mathit{dg}}{\mathit{du}}}\cdot {\displaystyle \frac{\mathit{du}}{\mathit{dy}}}+4f^3(u)\cdot {\displaystyle \frac{\mathit{df}}{\mathit{du}}}\cdot {\displaystyle \frac{\mathit{du}}{\mathit{dy}}}\hfill \\ \end{array}$

Recordemos que en una variable, si $a>0$ y, $f$ y $g$ son derivables, entonces aplicando regla de la cadena,

$\{\begin{array}{ccc}{\displaystyle \frac{d}{\mathit{dx}}}\left(a^{g(u)}\right)\hfill & \hfill =\hfill & a^{g(u)}\cdot \text{ln}<!--\; FUNCTION\; APPLICATION\; -->a\cdot {\displaystyle \frac{\mathit{dg}}{\mathit{du}}}\cdot {\displaystyle \frac{\mathit{du}}{\mathit{dx}}}\hfill \\ \end{array}$

Si $z={(\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x)}^{y^2},$ calcular ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial x}}}$ y ${\displaystyle \frac{\mathit{\partial z}}{\mathit{\partial y}}}.$

Solución.

• ${\displaystyle \frac{\mathit{\partial z}}{\partial x}}={\displaystyle \frac{\partial }{\partial x}}\left({[\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x]}^{y^2}\right)=y^2\cdot {[\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x]}^{y^2-1}\cdot {\displaystyle \frac{d}{\mathit{dx}}}\left(\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x\right)=y^2\cdot {[\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x]}^{y^2-1}\cdot \text{cos}<!--\; FUNCTION\; APPLICATION\; -->x$

• ${\displaystyle \frac{\mathit{\partial z}}{\partial y}}={\displaystyle \frac{\partial }{\partial y}}\left({[\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x]}^{y^2}\right)={[\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x]}^{y^2}\cdot \text{ln}<!--\; FUNCTION\; APPLICATION\; -->(\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x)\cdot {\displaystyle \frac{d}{\mathit{dy}}}\left(y^2\right)={[\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x]}^{y^2}\cdot \text{ln}<!--\; FUNCTION\; APPLICATION\; -->(\text{sen}<!--\; FUNCTION\; APPLICATION\; -->x)\cdot 2y$

Sea $f(x,y)=\sqrt[3]{x}\sqrt[3]{y}.$ Cuando derivamos respecto a $x,$ el factor $\sqrt[3]{y}$ lo podemos ver como una "constante" y cuando derivamos respecto a $y,$ el factor $\sqrt[3]{x}$ lo podemos ver como una "constante".

Aplicando la regla: ${\displaystyle \frac{d}{\mathit{du}}}\left(k\cdot f(u)\right)=k\cdot {\displaystyle \frac{\mathit{df}}{\mathit{du}}}$, con $k$ constante, tenemos

${\displaystyle \frac{\mathit{\partial f}}{\partial x}}={\displaystyle \frac{\partial }{\partial x}}\genfrac{}{}{0ex}{}{\left(\sqrt[3]{x}\sqrt[3]{y}\right)}{}={\displaystyle \frac{d}{\mathit{dx}}}\left(\sqrt[3]{x}\right)\sqrt[3]{y}=\frac{1}{3}{x}^{2∕3}\sqrt[3]{y}$

${\displaystyle \frac{\mathit{\partial f}}{\partial y}}={\displaystyle \frac{\partial }{\partial y}}\genfrac{}{}{0ex}{}{\left(\sqrt[3]{x}\sqrt[3]{y}\right)}{}=\sqrt[3]{x}{\displaystyle \frac{d}{\mathit{dy}}}\left(\sqrt[3]{y}\right)=\sqrt[3]{x}\frac{1}{3}{y}^{2∕3}$

Esta es la manera de derivar $f$ respecto a $x$ y respecto a $y$ usando teoremas de derivadas. Sin embargo esto no decide si la función es derivable o no en $(0,0).$ Para saber si estas derivadas parciales existen en $(0,0),$ se debe calcular usando la definición,