Multivariate chain rules

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

\[\frac{dg(f(x))}{dx} = g'(f(x))\cdot f'(x).\]

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 \(x = \varphi(t), y = \psi(t)\) be differentiable functions on an open interval \(I\). Suppose that \((\varphi(t), \psi(t))\in U\) for all \(t \in I\). Then the function \(z = f(\varphi(t), \psi(t))\) of \(t\) on \(I\) is differentiable on \(I\), and its derivative is given by

\[\frac{d}{dt}f(\varphi(t),\psi(t)) = \frac{\partial f}{\partial x}(\varphi(t),\psi(t))\frac{d\varphi}{dt}(t) + \frac{\partial f}{\partial y}(\varphi(t),\psi(t))\frac{d\psi}{dt}(t).\tag{Eq:ChainRule1}\]

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

\[\begin{eqnarray} \lim_{\delta \to 0}\frac{h(\delta)}{\delta} &=& \frac{d\varphi}{dt}(t_0),\\ \lim_{\delta \to 0}\frac{k(\delta)}{\delta} &=& \frac{d\psi}{dt}(t_0). \end{eqnarray}\]

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

\[\begin{eqnarray*} f(\varphi(t_0+\delta),\psi(t_0+\delta)) -f(\varphi(t_0),\psi(t_0)) &=& f(x_0 + h, y_0 + k) - f(x_0, y_0)\\ &=&\frac{\partial f(x_0,y_0)}{\partial x}\cdot h + \frac{\partial f(x_0,y_0)}{\partial y}\cdot k\\ && + o(\sqrt{h^2 + k^2}). \end{eqnarray*}\]

As \(\delta \to 0\),

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

Therefore,

\[\begin{eqnarray} &&\lim_{\delta\to 0}\frac{f(\varphi(t_0+\delta),\psi(t_0+\delta)) - f(\varphi(t_0),\psi(t_0))}{\delta}\nonumber\\ && = \frac{\partial f}{\partial x}(x_0,y_0)\cdot \frac{d\varphi}{dt}(t_0) + \frac{\partial f}{\partial y}(x_0,y_0)\cdot \frac{d\psi}{dt}(t_0).\tag{Eq:Dt0} \end{eqnarray}\]

This shows that the function \(z = f(\varphi(t), \psi(t))\) 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

\[\begin{eqnarray} \frac{dz}{dt} &=& \frac{\partial z}{\partial x}\cdot\frac{d x}{dt} + \frac{\partial z}{\partial y}\cdot\frac{d y}{dt}\\ &=& af_x(at+b, ct+d) + cf_y(at + b, ct + d). \end{eqnarray}\] □

Theorem [Chain rule (2)]

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

\[\begin{eqnarray*} \frac{\partial}{\partial u}f(\varphi(u,v), \psi(u,v)) &=&\frac{\partial f}{\partial x}(\varphi(u,v), \psi(u,v))\frac{\partial \varphi}{\partial u}(u,v)\\ && +\frac{\partial f}{\partial y}(\varphi(u,v), \psi(u,v))\frac{\partial \psi}{\partial u}(u,v) \ ,\\ \frac{\partial}{\partial v}f(\varphi(u,v), \psi(u,v)) &=&\frac{\partial f}{\partial x}(\varphi(u,v), \psi(u,v))\frac{\partial \varphi}{\partial v}(u,v)\\ &&+\frac{\partial f}{\partial y}(\varphi(u,v), \psi(u,v))\frac{\partial \psi}{\partial v}(u,v) \ . \end{eqnarray*}\]

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

\[\begin{eqnarray*} \frac{\partial z}{\partial u} &=& \frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial u},\\ \frac{\partial z}{\partial v} &=& \frac{\partial z}{\partial x}\cdot\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\cdot\frac{\partial y}{\partial v}. \end{eqnarray*}\]

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 \(z = f(\varphi(u, v_0), \psi(u, v_0))\) 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 \(\varphi(u,v_0)\) and \(\psi(u, v_0)\) are calculated as \(\frac{\partial \varphi(u_0,v_0)}{\partial u}\) 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 \(\varphi(u,v) = u\cos v, \psi(u,v) = u\sin v\). Let \(g(u,v) = f(\varphi(u,v), \psi(u,v))\). 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,

\[\begin{eqnarray*} g_u(u,v) &=& \frac{\partial}{\partial u}f(\varphi(u,v),\psi(u,v))\\ &=& \frac{\partial f}{\partial x}(\varphi(u,v), \psi(u,v))\frac{\partial \varphi}{\partial u}(u,v) + \frac{\partial f}{\partial y}(\varphi(u,v), \psi(u,v))\frac{\partial \psi}{\partial u}(u,v)\\ &=& \left[e^{u^2}\cdot 2u\cos v\right] \cos v +\left[e^{u^2}\cdot 2u\sin v\right] \sin v = 2ue^{u^2},\\ g_v(u,v) &=& \frac{\partial}{\partial v}f(\varphi(u,v),\psi(u,v))\\ &=& \frac{\partial f}{\partial x}(\varphi(u,v), \psi(u,v))\frac{\partial \varphi}{\partial v}(u,v) + \frac{\partial f}{\partial y}(\varphi(u,v), \psi(u,v))\frac{\partial \psi}{\partial v}(u,v)\\ &=& \left[e^{u^2}\cdot 2u\cos v\right]\cdot (-u \sin v) + \left[e^{u^2}\cdot 2u\sin v\right]\cdot(u\cos v) = 0. \end{eqnarray*}\] □





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