다변수 연쇄법칙

By definition,

$$\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) =\frac{\partial }{\partial x}\left[\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right) \right]$$

We'll compute the derivatives from the inside out. Let's call $F(x,y)=f(2x+3y,xy)$ so that the innermost derivative is $G(x,y)=\frac{\partial }{\partial y}f(2x+3y,xy)=\frac{\partial }{\partial y}F(x,y)$. By the chain rule

$$\begin{aligned} G(x,y)=\frac{\partial }{\partial y}F(x,y) &=f_1(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_2(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_1(2x+3y,xy)+xf_2(2x+3y,xy) \end{aligned}$$

Here the subscript $1$ means take the partial derivative of $f$ with respect to the first argument while holding the second argument fixed, and the subscript $2$ means take the partial derivative of $f$ with respect to the second argument while holding the first argument fixed. Next call the middle derivative $H(x,y)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right)$ so that

$$\begin{aligned} H(x,y) &= \frac{\partial }{\partial y} G(x,y) \\ &=\frac{\partial }{\partial y}\Big(3f_1(2x+3y,xy)+xf_2(2x+3y,xy)\Big) \\ &=3\frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) +x\frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) \end{aligned}$$

By the chain rule (twice),

$$\begin{aligned} \frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) &=f_{11}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{12}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \\ \frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) &=f_{21}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{22}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy) \end{aligned}$$

so that

$$\begin{aligned} H(x,y)&=3\Big(3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \Big) \\[-0.05in] & +x\Big(3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy)\Big) \\ &=9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy) \end{aligned}$$

In the last equality we used that $f_{21}(2x+3y,xy)=f_{12}(2x+3y,xy)$. The notation $f_{21}$ means first differentiate with respect to the second argument and then differentiate with respect to the first argument. For example, if $f(x,y)=e^{2y}\sin x$, then

$$\begin{aligned} f_{21}(x,y) =\frac{\partial }{\partial x} \Big[\frac{\partial }{\partial y}\big(e^{2y}\sin x\big)\Big] =\frac{\partial }{\partial x} \Big[2e^{2y}\sin x\Big] =2e^y\cos x \end{aligned}$$

Finally, we get to

$$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=\frac{\partial }{\partial x} H(x,y) \\ &=\frac{\partial }{\partial x}\Big(9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy)\Big)\\ &=9\frac{\partial }{\partial x}\Big(f_{11}(2x+3y,xy)\Big)\\& +6f_{12}(2x+3y,xy) +6x\frac{\partial }{\partial x}\Big(f_{12}(2x+3y,xy)\Big) \\& +2xf_{22}(2x+3y,xy) +x^2\frac{\partial }{\partial x}\Big(f_{22}(2x+3y,xy)\Big) \end{aligned}$$

By three applications of the chain rule

$$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=9\Big(2f_{111}+yf_{112}\Big)\\& +6f_{12}+6x\Big(2f_{121}+yf_{122}\Big)\\& +2xf_{22}+x^2\Big(2f_{221}+yf_{222}\Big)\\ &=6\,f_{12}+2x\,f_{22}+18\,f_{111}+(9y+12x)\,f_{112} +(6xy+2x^2)\,f_{122}+x^2y\,f_{222} \end{aligned}$$

All functions on the right hand side have arguments $(2x+3y,xy)$.