# 14.6 | The Chain Rule

The Chain Rule (Case 1): Suppose that $$z = f(x,y)$$ is a differentiable function of $$x$$ and $$y$$, where $$x = g(t)$$ and $$y = h(t)$$ are both differentiable functions of $$t$$. Then $$z$$ is a differentiable function of $$t$$ and
$$\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$$
The Chain Rule (Case 2): Suppose that $$z = f(x,y)$$ is a differentiable function of $$x$$ and $$y$$, where $$x = g(s,t)$$ and $$y = h(s,t)$$ are differentiable functions of $$s$$ and $$t$$. Then
$$\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{ \partial y}\frac{\partial y}{\partial s}$$
and
$$\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{ \partial y}\frac{\partial y}{\partial t}$$
The Chain Rule (General Version): Suppose that $$u$$ is a differentiable function of $$n$$ variables $$x_1, x_2, \ldots, x_n$$ and each $$x_j$$ is a differentiable function of the $$m$$ variables $$t_1, t_2, \ldots, t_m$$. Then $$u$$ is a function of $$t_1, t_2, \ldots, t_m$$ and

$$\frac{\partial u}{\partial t_i} = \frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial t_i} + \frac{\partial u}{\partial x_2}\frac{\partial x_2}{\partial t_i} + \cdots + \frac{\partial u}{\partial x_n}\frac{\partial x_n}{\partial t_i}$$
\begin{aligned} \frac{\partial u}{\partial t_i} = & \frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial t_i} + \frac{\partial u}{\partial x_2}\frac{\partial x_2}{\partial t_i} \\ &+ \cdots + \frac{\partial u}{\partial x_n}\frac{\partial x_n}{\partial t_i} \end{aligned}

for each $$i = 1,2, \ldots, m$$.

Implicit Function Theorem: If $$F$$ is defined on a disk containing $$(a,b)$$, where $$F(a,b)=0$$, $$F_y(a,b) \not = 0$$, and $$F_x$$ and $$F_y$$ are continuous on the disk, then the equation $$F(x,y) = 0$$ defines $$y$$ as a function of $$x$$ near the point $$(a,b)$$
Suppose $$F(x,y)=0$$ defines $$y$$ implicitly as a function of $$x$$, that is, $$y = f(x)$$. If $$F$$ is differentiable then
$$\frac{dy}{dx} = – \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{F_x}{F_y}$$
Suppose $$F(x,y,z)=0$$ defines $$z$$ implicitly as a function of $$(x,y)$$, that is, $$z = f(x,y)$$. If $$F$$ is differentiable then
$$\frac{\partial z}{\partial x} = – \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} = -\frac{F_x}{F_z}$$
and
$$\frac{\partial z}{\partial y} = – \frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}} = -\frac{F_y}{F_z}$$

Figure 1

Let
$$R = \ln(u^2 +v^2 + w^2),$$
where $$u = x+2y$$, $$v=2x-y$$, and $$w = 2xy$$. Find $$\frac{\partial R}{\partial x}$$ and $$\frac{\partial R}{\partial y}$$ when $$x = y = 1$$.

Diagrams, like the one in figure 1, can be used to find formulas for partial derivatives. To construct the diagram as in figure 1, first illustrate the dependencies between variables. For example $$R$$ depends on $$u,v,$$ and $$w$$ and hence there are directed arrows connecting $$R$$ to $$u,v,$$ and $$w$$. In turn, $$u,v,$$ and $$w$$ are dependent on $$x$$ and $$y$$ and hence there are directed edges connecting each of $$u,v,$$ and $$w$$ to $$x$$ and $$y$$. Although $$R$$ is dependent on $$x$$ and $$y$$, a separate edge between them is not introduced as these dependencies are already captured by the paths through $$u,v,$$ and $$w$$.

Edges are then labeled:

• If there is more than one edge emanating from $$R$$, label each edge with the partial derivative of $$R$$ with respect to the terminal vertex of the respective edge. For example, the edge connecting $$R$$ to $$u$$ would be labeled $$\partial R/ \partial u$$ while the edge connecting $$R$$ to $$v$$ would be labeled $$\partial R/ \partial v$$.
• If there is only one edge emanating from $$R$$, label that edge with the derivative of $$R$$ with respect to the terminal vertex of that edge. For example, if $$R$$ had only one edge emanating from it with terminal vertex $$u$$, then the label along that edge would be $$dR/du$$.

Finally, the diagram is used to find the formula for say $$\partial R/ \partial x$$:

• locate all paths from $$R$$ to $$x$$
• the number of paths determines the number of terms that will be present in the formula
• each term is obtained by traversing a path from $$R$$ to $$x$$, multiplying the labels along the edges.

There are three paths leading from $$R$$ to $$x$$ from which there will be three terms in the formula for $$\partial R/ \partial x$$. The path from $$R$$ to $$x$$ passing through $$u$$ corresponds to
$$\frac{\partial R}{\partial u}\frac{\partial u}{\partial x}$$
and the other two paths correspond to
$$\frac{\partial R}{\partial v}\frac{\partial v}{\partial x} \text{ and } \frac{\partial R}{\partial w}\frac{\partial w}{\partial x}$$
So, the formula for the partial derivative of $$R$$ with respect to $$x$$ is given by
$$\frac{\partial R}{\partial x} = \frac{\partial R}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial R}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial R}{\partial w}\frac{\partial w}{\partial x}$$
Apply the formula:
\begin{aligned} \frac{\partial R}{\partial x} & = \frac{\partial R}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial R}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial R}{\partial w}\frac{\partial w}{\partial x} \\ & = \frac{\partial}{\partial u}[\ln(u^2 +v^2 + w^2)] \frac{\partial }{\partial x}[x+2y] \\ & \; \; + \frac{\partial }{\partial v}[\ln(u^2 +v^2 + w^2)]\frac{\partial }{\partial x}[2x-y] \\ & \; \; + \frac{\partial }{\partial w}[\ln(u^2 +v^2 + w^2)]\frac{\partial }{\partial x}[2xy] \\ & = \frac{1}{u^2 + v^2 + w^2} \left(2u + 4v + 4yw \right) \end{aligned}

Note that when $$x = y = 1$$, $$u = 3, v=1,$$ and $$w = 2$$. Hence
$$\frac{\partial R}{\partial x} \Big|_{\substack{(3,1,2) \\ x=y=1}} = \frac{18}{14}.$$

The formula for the partial derivative of $$R$$ with respect to $$y$$ is given by
$$\frac{\partial R}{\partial y} = \frac{\partial R}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial R}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial R}{\partial w}\frac{\partial w}{\partial y}$$
Apply the formula:
\begin{aligned} \frac{\partial R}{\partial y} &= \frac{\partial R}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial R}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial R}{\partial w}\frac{\partial w}{\partial y}\\ & = \frac{\partial}{\partial u}[\ln(u^2 +v^2 + w^2)] \frac{\partial }{\partial y}[x+2y] \\ & \; \; + \frac{\partial }{\partial v}[\ln(u^2 +v^2 + w^2)]\frac{\partial }{\partial y}[2x-y] \\ & \; \; + \frac{\partial }{\partial w}[\ln(u^2 +v^2 + w^2)]\frac{\partial }{\partial y}[2xy] \\ & = \frac{1}{u^2 + v^2 + w^2} \left(4u -2v + 4xw \right) \end{aligned}

Hence
$$\frac{\partial R}{\partial y} \Big|_{\substack{(3,1,2) \\ x=y=1}} = \frac{18}{14}.$$