Multivariate chain rules

Next, we consider differentiating a composite function such as $f(\phi (t),\psi (t))$ or $f(\phi (u,v),\psi (u,v))$. In the case of univariate functions, we have the chain rule. That is,

$$\frac{dg(f(x))}{dx}=g^{\prime }(f(x))\cdot f^{\prime }(x).$$

We have corresponding multivariate versions.

Theorem [Chain rule (1)]

Let $f(x,y)$ be a totally differentiable function on an open region $U(\subset {\mathbb{R}}^2)$. Let $x=\phi (t),y=\psi (t)$ be differentiable functions on an open interval $I$. Suppose that $(\phi (t),\psi (t))\in U$ for all $t\in I$. Then the function $z=f(\phi (t),\psi (t))$ of $t$ on $I$ is differentiable on $I$, and its derivative is given by

$$\begin{array}{}\text{(Eq:ChainRule1)}& \frac{d}{dt}f(\phi (t),\psi (t))=\frac{\partial f}{\partial x}(\phi (t),\psi (t))\frac{d\phi }{dt}(t)+\frac{\partial f}{\partial y}(\phi (t),\psi (t))\frac{d\psi }{dt}(t).\end{array}$$

Remark.  Eq. (Eq:ChainRule1) may be a little too cluttered and difficult to read. We could simplify the notation by using dependent variables:

$$\frac{dz}{dt}(t)=\frac{\partial z}{\partial x}(x,y)\cdot \frac{dx}{dt}(t)+\frac{\partial z}{\partial y}(x,y)\cdot \frac{dy}{dt}(t).$$

Or, even more simply,

$$\frac{dz}{dt}=\frac{\partial z}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial z}{\partial y}\cdot \frac{dy}{dt}.$$

But, Eq. (Eq:ChainRule1) is the most accurate expression. □

Proof. For an arbitrary $t_0\in I$, let $x_0=\phi (t_0)$ and $y_0=\psi (t_0)$. For an arbitrary real number $\delta$ such that $t_0+\delta \in I$, let $h(\delta )=\phi (t_0+\delta )-\phi (t_0)$ and $k(\delta )=\psi (t_0+\delta )-\psi (t_0)$. We have

$$\begin{array}{rcl}\underset{\delta \to 0}{lim}\frac{h(\delta )}{\delta }& =& \frac{d\phi }{dt}(t_0),\\ \underset{\delta \to 0}{lim}\frac{k(\delta )}{\delta }& =& \frac{d\psi }{dt}(t_0).\end{array}$$

Since $f(x,y)$ is totally differentiable,

$$\begin{array}{rcl}f(\phi (t_0+\delta ),\psi (t_0+\delta ))-f(\phi (t_0),\psi (t_0))& =& f(x_0+h,y_0+k)-f(x_0,y_0)\\ & =& \frac{\partial f(x_0,y_0)}{\partial x}\cdot h+\frac{\partial f(x_0,y_0)}{\partial y}\cdot k\\ & & +o(\sqrt{h^2+k^2}).\end{array}$$

As $\delta \to 0$,

$$\frac{o(\sqrt{h^2+k^2})}{\delta }=\frac{h}{\delta }\cdot \frac{o(\sqrt{h^2+k^2})}{h}\to \frac{d\phi (t_0)}{dt}\cdot 0=0.$$

Therefore,

$$\begin{array}{rcl}& & \underset{\delta \to 0}{lim}\frac{f(\phi (t_0+\delta ),\psi (t_0+\delta ))-f(\phi (t_0),\psi (t_0))}{\delta }\\ \text{(Eq:Dt0)}& & & =\frac{\partial f}{\partial x}(x_0,y_0)\cdot \frac{d\phi }{dt}(t_0)+\frac{\partial f}{\partial y}(x_0,y_0)\cdot \frac{d\psi }{dt}(t_0).\end{array}$$

This shows that the function $z=f(\phi (t),\psi (t))$ is differentiable at $t=t_0$ and its differential coefficient is given by the right-hand side of Eq. (Eq:Dt0). ■

Example. If $z=f(x,y),x=at+b,y=ct+d$, then

$$\begin{array}{rcl}\frac{dz}{dt}& =& \frac{\partial z}{\partial x}\cdot \frac{dx}{dt}+\frac{\partial z}{\partial y}\cdot \frac{dy}{dt}\\ & =& a{f}_x(at+b,ct+d)+c{f}_y(at+b,ct+d).\end{array}$$ □

Theorem [Chain rule (2)]

Let $z=f(x,y)$ be a totally differentiable function on an open region $U\subset {\mathbb{R}}^2$. Let $x=\phi (u,v),y=\psi (u,v)$ be functions on an open region $V\subset {\mathbb{R}}^2$ such that $(\phi (u,v),\psi (u,v))\in U$ for all $(u,v)\in V$. Suppose that the partial derivatives ${\phi }_u(u,v),{\phi }_v(u,v)$, ${\psi }_u(u,v),{\psi }_v(u,v)$ exist on $V$. Then, the function $z=f(\phi (u,v),\psi (u,v))$ on $V$ has the partial derivatives with respect to $u$ and $v$ given by

$$\begin{array}{rcl}\frac{\partial }{\partial u}f(\phi (u,v),\psi (u,v))& =& \frac{\partial f}{\partial x}(\phi (u,v),\psi (u,v))\frac{\partial \phi }{\partial u}(u,v)\\ & & +\frac{\partial f}{\partial y}(\phi (u,v),\psi (u,v))\frac{\partial \psi }{\partial u}(u,v),\\ \frac{\partial }{\partial v}f(\phi (u,v),\psi (u,v))& =& \frac{\partial f}{\partial x}(\phi (u,v),\psi (u,v))\frac{\partial \phi }{\partial v}(u,v)\\ & & +\frac{\partial f}{\partial y}(\phi (u,v),\psi (u,v))\frac{\partial \psi }{\partial v}(u,v).\end{array}$$

Remark. In a more simplified notation, the above partial derivatives may be written as

$$\begin{array}{rcl}\frac{\partial z}{\partial u}& =& \frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial u},\\ \frac{\partial z}{\partial v}& =& \frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial v}.\end{array}$$

Proof. The partial differential coefficient

$$\frac{\partial z(u_0,v_0)}{\partial u}$$

at $(u,v)=(u_0,v_0)$ is the same as the differential coefficient of the "one-variable" function $z=f(\phi (u,v_0),\psi (u,v_0))$ at $u=u_0$ where $v=v_0$ is fixed. Then, we can apply the above Theorem [Chain Rule (1)]. Noting that the differential coefficients of the one-variable functions $\phi (u,v_0)$ and $\psi (u,v_0)$ are calculated as $\frac{\partial \phi (u_0,v_0)}{\partial u}$ and $\frac{\partial \psi (u_0,v_0)}{\partial u}$, respectively, we have the claimed result. The same argument applies to the partial derivative with respect to $v$. ■

Example. Let $f(x,y)=e^{x^2+y^2}$ and $\phi (u,v)=u\cos v,\psi (u,v)=u\sin v$. Let $g(u,v)=f(\phi (u,v),\psi (u,v))$. Find $g_u(u,v)$ and $g_v(u,v)$. Let $x=u\cos v$ and $y=u\sin v$. Then, $x^2+y^2=u^2$. Thus,

$$\begin{array}{rcl}{g}_u(u,v)& =& \frac{\partial }{\partial u}f(\phi (u,v),\psi (u,v))\\ & =& \frac{\partial f}{\partial x}(\phi (u,v),\psi (u,v))\frac{\partial \phi }{\partial u}(u,v)+\frac{\partial f}{\partial y}(\phi (u,v),\psi (u,v))\frac{\partial \psi }{\partial u}(u,v)\\ & =& [e^{u^2}\cdot 2u\cos v]\cos v+[e^{u^2}\cdot 2u\sin v]\sin v=2u{e}^{u^2},\\ g_v(u,v)& =& \frac{\partial }{\partial v}f(\phi (u,v),\psi (u,v))\\ & =& \frac{\partial f}{\partial x}(\phi (u,v),\psi (u,v))\frac{\partial \phi }{\partial v}(u,v)+\frac{\partial f}{\partial y}(\phi (u,v),\psi (u,v))\frac{\partial \psi }{\partial v}(u,v)\\ & =& [e^{u^2}\cdot 2u\cos v]\cdot (-u\sin v)+[e^{u^2}\cdot 2u\sin v]\cdot (u\cos v)=0.\end{array}$$ □