Theorem 10 Sums, Products, Quotients

• (i) Constant Multiple Rule. Let \(f\colon\, U \subset {\mathbb R}^n \rightarrow {\mathbb R}^m\) be differentiable at \({\bf x}_0\) and let \(c\) be a real number. Then \(h ({\bf x}) = cf ( {\bf x})\) is differentiable at \({\bf x}_0\) and \[ {\bf D} h ( {\bf x}_0) = c {\bf D}\! f ( {\bf x}_0) \qquad \hbox{(equality of matrices)}. \]
• (ii) Sum Rule. Let \(f\colon\, U \subset {\mathbb R}^n \rightarrow {\mathbb R}^m\) and \(g\colon\, U \subset {\mathbb R}^n \rightarrow {\mathbb R}^m\) be differentiable at \({\bf x}_0\). Then \(h ( {\bf x}) = f ( {\bf x})+ g ( {\bf x})\) is differentiable at \({\bf x}_0\) and \[ {\bf D} h ( {\bf x}_0) = {\bf D}\! f ( {\bf x}_0) + {\bf D} g ( {\bf x}_0) \qquad \hbox{(sum of matrices).} \]
• (iii) Product Rule. Let \(f\colon\, U \subset {\mathbb R}^n \rightarrow {\mathbb R}\) and \(g\colon\, U \subset {\mathbb R}^n \rightarrow {\mathbb R}\) be differentiable at \({\bf x}_0\) and let \(h ( {\bf x}) = g ({\bf x}) f ({\bf x})\). Then \(h\colon\, U \subset {\mathbb R}^n \rightarrow {\mathbb R}\) is differentiable at \({\bf x}_0\) and \[ {\bf D} h ( {\bf x}_0 ) = g ( {\bf x}_0) {\bf D}\! f ( {\bf x}_0)+ f ( {\bf x}_0) {\bf D} g ( {\bf x}_0). \] (Note that each side of this equation is a \(1\times n\) matrix; a more general product rule is presented in Exercise \(31\) at the end of this section.)
• (iv) Quotient Rule. With the same hypotheses as in rule (iii), let \(h ( {\bf x}) = f ({\bf x}) / g ( {\bf x})\) and suppose \(g\) is never zero on \(U\). Then \(h\) is differentiable at \({\bf x}_0\) and \[ {\bf D} h ( {\bf x}_0) = \frac{ g ( {\bf x}_0) {\bf D}\! f ( {\bf x}_0) - f ( {\bf x}_0) {\bf D} g ( {\bf x}_0)}{ [ g ( {\bf x}_0)]^2} . \]

proof

The proofs of rules (i) through (iv) proceed almost exactly as in the one-variable case, with a slight difference in notation. We shall prove rules (i) and (ii), leaving the proofs of rules (iii) and (iv) as Exercise 27.

• (i) To show that \({\bf D} h ( {\bf x}_0) = c {\bf D}\! f ({\bf x}_0)\), we must show that \[ {\mathop {{\rm limit} }_{{\bf x} \to {\bf x}_0}} \ \frac{ \| h ( {\bf x} ) - h ( {\bf x}_0) - c {\bf D}\! f ( {\bf x}_0) ( {\bf x}- {\bf x}_0) \| }{ \| {\bf x} - {\bf x}_0 \| } =0, \] that is, that \[ {\mathop {{\rm limit} }_{{\bf x} \to {\bf x}_0}} \ \frac{ \| cf( {\bf x}) - cf( {\bf x}_0) - c {\bf D}\! f ( {\bf x}_0) ( {\bf x}- {\bf x}_0) \| }{ \| {\bf x} - {\bf x}_0 \| } =0, \] [see equation (4) of Section 2.3]. This is certainly true, since \(f\) is differentiable and the constant \(c\) can be factored out [see Theorem 3(i), Section 2.2].
• (ii) By the triangle inequality, we may write \begin{eqnarray*} &&\frac{ \| h ({\bf x}) - h ( {\bf x}_0) - [{\bf D}\! f ( {\bf x}_0) + {\bf D} g ({\bf x}_0) ] ( {\bf x}- {\bf x}_0) \| }{ \| {\bf x} - {\bf x}_0 \|} \\[3pt] && =\frac{ \| f({\bf x})-f({\bf x}_0)-[{\bf D}\! f({\bf x}_0)]({\bf x}-{\bf x}_0)+g({\bf x})-g({\bf x}_0)-[{\bf D}g({\bf x}_0)]({\bf x}-{\bf x}_0) \|}{\| {\bf x}-{\bf x}_0 \| } \\[3pt] && \leq \frac{ \| f({\bf x})-f({\bf x}_0)-[{\bf D}\! f({\bf x}_0)]({\bf x}-{\bf x}_0) \| }{ \| {\bf x}-{\bf x}_0 \| }+ \frac{ \| g({\bf x})-g({\bf x}_0)-[{\bf D}g({\bf x}_0)] ({\bf x}-{\bf x}_0) \| }{ \| {\bf x}-{\bf x}_0 \| }, \end{eqnarray*} and each term approaches 0 as \({\bf x}\rightarrow {\bf x}_0\). Hence, rule (ii) holds.

example 1

Verify the formula for \({\bf D}h\) in rule (iv) of Theorem 10 with \[ f(x,y,z)=x^2+y^2+z^2 \hbox{ and } g(x,y,z)=x^2+1. \]

solution Here \[ h(x,y,z)=\frac{x^2+y^2+z^2}{x^2+1}, \] so that by direct differentiation \begin{eqnarray*} {\bf D}h(x,y,z) &\!=\!& \bigg[\frac{\partial h}{\partial x},\frac{\partial h}{\partial y}, \frac{\partial h}{\partial z}\bigg] \!=\!\bigg[\frac{(x^2+1)2x-(x^2+y^2+z^2)2x}{(x^2+1)^2}, \frac{2y}{x^2+1},\frac{2z}{x^2+1}\bigg]\\[6pt] &=&\!\bigg[\frac{2x(1-y^2-z^2)}{(x^2+1)^2},\frac{2y}{x^2+1},\frac{2z}{x^2+1}\bigg]. \end{eqnarray*}

By rule (iv), we get \[ {\bf D}h=\frac{g{\bf D}\! f-f{\bf D}g}{g^2}= \frac{(x^2+1)[2x,2y,2z]-(x^2+y^2+z^2)[2x,0,0]}{(x^2+1)^2}, \] which is the same as what we obtained directly.

Theorem 11 Chain Rule

Le} \(U\subset {\mathbb R}^n\) and \(V \subset {\mathbb R}^m\) be open sets. Let \(g\colon\, U\subset {\mathbb R}^n\rightarrow {\mathbb R}^m\) and \(f\colon\, V \subset {\mathbb R}^m\rightarrow {\mathbb R}^p\) be given functions such that \(g\) maps \(U\) into \(V\), so that \(f \circ g\) is defined. Suppose \(g\) is differentiable at \({\bf x}_0\) and \(f\) is differentiable at \({\bf y}_0=g({\bf x}_0)\). Then \(f \circ g\) is differentiable at \({\bf x}_0\) and \begin{equation*} {\bf D}(f\circ g)({\bf x}_0)={\bf D}\! f({\bf y}_0){\bf D}g({\bf x}_0).\tag{1} \end{equation*}

The right-hand side is the matrix product of \({\bf D} f({\bf y}_0)\) with \({\bf D} g({\bf x}_0)\).

proof of equation (2)

By definition, \[ \frac{dh}{dt}(t_0)=\mathop {{\rm limit}\ }_{t\rightarrow t_0}\frac{h(t)-h(t_0)}{t-t_0}. \]

Adding and subtracting two terms, we write \begin{eqnarray*} \frac{h(t)-h(t_0)}{t-t_0}&=&\frac{f(x(t),y(t),z(t))-f(x(t_0),y(t_0),z(t_0))}{t-t_0}\\[5pt] &=&\frac{f(x(t),y(t),z(t))-f(x(t_0),y(t),z(t))}{t-t_0}\\[5pt] & & +\frac{f(x(t_0),y(t),z(t))-f(x(t_0),y(t_0),z(t))}{t-t_0}\\[5pt] & &+\frac{f(x(t_0),y(t_0),z(t))-f(x(t_0),y(t_0),z(t_0))}{t-t_0}.\\[-15.8pt] \end{eqnarray*}

Now we invoke the mean-value theorem from one-variable calculus, which states: If \(g\colon\, [a,b]\rightarrow {\mathbb R}\) is continuous and is differentiable on the open interval \((a,b)\), then there is a point \(c\) in \((a,b)\) such that \(g(b)-g(a)=g'(c)(b-a)\). Applying this to \(f\) as a function of \(x\), we can assert that for some \(c\) between \(x\) and \(x_0\), \[ f(x,y,z)-f(x_0,y,z)=\bigg[\frac{\partial f}{\partial x}(c,y,z)\bigg] (x-x_0). \]

In this way, we find that \begin{eqnarray*} \frac{h(t)-h(t_0)}{t-t_0}&=&\bigg[\frac{\partial f}{\partial x}(c,y(t),z(t))\bigg] \frac{x(t)-x(t_0)}{t-t_0}+ \bigg[\frac{\partial f}{\partial y}(x(t_0),d,z(t))\bigg] \frac{y(t)-y(t_0)}{t-t_0} \\[6pt] && + \bigg[\frac{\partial f}{\partial z}(x(t_0),y(t_0),e)\bigg] \frac{z(t)-z(t_0)}{t-t_0}, \end{eqnarray*} where \(c,d\), and \(e\) lie between \(x(t)\) and \(x(t_0)\), between \(y(t)\) and \(y(t_0)\), and between \(z(t)\) and \(z(t_0)\), respectively. Taking the limit \(t\rightarrow t_0\), using the continuity of the partials \(\partial f/\partial x, \partial f/\partial y, \partial f/\partial z\), and the fact that \(c,d\), and \(e\) converge to \(x(t_0),y(t_0)\), and \(z(t_0)\), respectively, we obtain formula (2).

proof of the second special case of the chain rule

By definition, \(\partial h/ \partial x\) is obtained by differentiating \(h\) with respect to \(x\), holding \(y\) and \(z\) fixed. But then \((u(x,y,z),v(x,y,z),w(x,y,z))\) may be regarded as a vector function of the single variable \(x\). The first special case applies to this situation and, after the variables are renamed, gives \begin{equation*} \frac{\partial h}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial x}.\tag{3'} \end{equation*}

Similarly, \begin{equation*} \frac{\partial h}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial y}\tag{3''} \end{equation*} and \begin{equation*} \frac{\partial h}{\partial z}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial z}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial z}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial z}.\tag{3'''} \end{equation*}

In actual calculations, at the last step, we usually express \(\frac{\partial f}{\partial u}\), \(\frac{\partial f}{\partial v}\), and \(\frac{\partial f}{\partial w}\) in terms of \(x,y,z\). These equations are exactly what would be obtained by multiplying out the matrices in equation (3).

proof of theorem 11

The general case in equation (1) may be proved in two steps. First, equation (2) is generalized to \(m\) variables; that is, for \(f(x_1,\ldots, x_m)\) and \({\bf c}(t)=(x_1(t),\ldots ,x_m(t))\), we have \[ \frac{{\it dh}}{{\it dt}}= \sum^m_{i=1}\frac{\partial f}{\partial x_i}\frac{dx_i}{{\it dt}}, \] where \(h(t)=f(x_1(t),\ldots ,x_m(t))\). Second, the result obtained in the first step is used to obtain the formula \[ \frac{\partial h_j}{\partial x_i}=\sum^m_{k=1}\frac{\partial f_j}{\partial y_k}\frac{\partial y_k}{\partial x_i}, \] where \(f=(f_1,\ldots, f_p)\) is a vector function of arguments \(y_1,\ldots,y_m; g(x_1,\ldots ,x_n) = (y_1(x_1,\) \(\ldots,\) \(x_n),\ldots ,y_m(x_1,\ldots, x_n));\) and \(h_j(x_1,\ldots ,x_n)=f_j(y_1(x_1,\ldots , x_n),\ldots , y_m (x_1,\ldots, x_n))\). (Using the letter \(y\) for both functions and arguments is an abuse of notation, but it can help us remember the formula.) This formula is equivalent to formula (1) after the matrices are multiplied out.

example 2

Verify the chain rule in the form of formula \((3')\) for \[ f(u,v,w)=u^2+v^2-w, \] where \[ u(x,y,z)=x^2y, v(x,y,z)=y^2, w(x,y,z) =e^{-xz}. \]

solution Here \begin{eqnarray*} h(x,y,z)&=&f(u(x,y,z), v(x,y,z),w(x,y,z))\\[3pt] &=&(x^2y)^2+y^4-e^{-xz}=x^4y^2+y^4-e^{-xz}. \end{eqnarray*}

Thus, differentiating directly, \[ \frac{\partial h}{\partial x}=4x^3y^2+ze^{-xz}. \]

On the other hand, using the chain rule, \begin{eqnarray*} \frac{\partial h}{\partial x}&=&\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial f}{\partial w}\frac{\partial w}{\partial x}=2u(2xy)+2v \,{\cdot}\, 0+(-1)(-ze^{-xz})\\[3pt] &=& (2x^2y)(2xy)+ze^{-xz}, \end{eqnarray*} which is the same as the preceding equation.

example 3

Given \(g(x,y)=(x^2+1,y^2)\) and \(f(u,v)= (u+v,u,v^2),\) compute the derivative of \(f\circ g\) at the point \((x,y)= (1,1)\) using the chain rule.

solution The matrices of partial derivatives are \[ {\bf D}\! f(u,v)=\left[ \begin{array}{c@{\quad}c} \\[-10.5pt] \displaystyle \frac{\partial f_1}{\partial u} &\displaystyle \frac{\partial f_1}{\partial v}\\[10pt] \displaystyle \frac{\partial f_2}{\partial u} &\displaystyle \frac{\partial f_2}{\partial v}\\[10pt] \displaystyle \frac{\partial f_3}{\partial u}&\displaystyle \frac{\partial f_3}{\partial v} \end{array} \right] =\left[ \begin{array}{c@{\quad}c} 1&1\\ 1&0\\ 0&2v \end{array} \right]\!\!\quad\hbox{and}\quad {\bf D}g(x,y)=\bigg[ \begin{array}{c@{\quad}c} 2x&0\\ 0&2y \end{array} \bigg]. \]

When \((x,y)=(1,1)\), note that \(g(x,y)=(u,v)=(2,1)\). Hence, \[ {\bf D} (f\circ g)(1,1)={\bf D}\! f(2,1){\bf D}g(1,1)=\left[ \begin{array}{c@{\quad}c} 1&1\\ 1&0\\ 0&2 \end{array} \right] \left[ \begin{array}{c@{\quad}c} 2&0\\ 0&2 \end{array} \right] =\left[ \begin{array}{c@{\quad}c} 2&2\\ 2&0\\ 0&4 \end{array} \right] \] is the required derivative.

example 4

Let \(f(x,y)\) be given and make the substitution \(x=r\cos \theta, y=r\sin \theta\) (polar coordinates). Write a formula for \(\partial f/\partial \theta\).

solution By the chain rule, \[ \frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}, \] that is, \[ \frac{\partial f}{\partial \theta}=-r \sin \theta \frac{\partial f}{\partial x}+ r\cos \theta \frac{\partial f}{\partial y}. \]

example 5

Let \(f(x,y) = (\cos y +x^2,e^{x+y})\) and \(g(u,v)=(e^{u^2},u-\sin v)\). (a) Write a formula for \(f\circ g\). (b) Calculate \({\bf D} (f\circ g)(0,0)\) using the chain rule.

solution (a) We have \begin{eqnarray*} (f \circ g)(u,v)&=&f(e^{u^2},u-\sin v)\\[2pt] &=& \big(\cos\, (u-\sin v)+e^{2u^2},e^{e^{u^{2}}\,+\,u-\sin v} \big). \end{eqnarray*}

(b) By the chain rule, \[ {\bf D}(f\circ g)(0,0)=[{\bf D}\! f(g(0,0))][{\bf D}g(0,0)]=[{\bf D}\! f(1,0)][{\bf D}g(0,0)]. \]

Now \[ {\bf D}g(0,0)=\left[ \begin{array}{c@{\quad}c} 2ue^{u^2}&0\\ 1&{-}{\cos v} \end{array} \right]_{(u,v)=(0,0)}=\left[\begin{array}{c@{\quad}c} 0&0\\ 1&-1 \end{array} \right] \] and \[ {\bf D}\! f(1,0)=\left[ \begin{array}{c@{\quad}c} 2x&{-}{\sin y}\\ e^{x+y}&e^{x+y} \end{array} \right]_{(x,y)=(1,0)}=\left[\begin{array}{c@{\quad}c} 2&0\\ e&e \end{array} \right]\!. \] [Remember that \({\bf D}\! f\) is evaluated at \(g(0,0)\), not at \((0,0)\)!]. Thus, \[ {\bf D}(f\circ g)(0,0)=\left[ \begin{array}{c@{\quad}c} 2&0\\ e&e \end{array} \right]\left[ \begin{array}{c@{\quad}c} 0&0\\ 1&-1 \end{array} \right]=\left[ \begin{array}{c@{\quad}c} 0&0\\ e&-e \end{array} \right]\!. \]

example 6

Let \(f\colon\, U\subset {\mathbb R}^{n}\rightarrow {\mathbb R}^{m}\) be differentiable, with \(f=(f_1,\ldots ,f_m)\), and let \(g({\bf x})=\sin\,[f({\bf x})\,{ \cdot}\, f{{\bf (x)}}]\). Compute \({\bf D}g({\bf x})\).

solution By the chain rule, \({\bf D}g({\bf x})=\cos\, [f({\bf x})\,{ \cdot}\, f({\bf x})]{\bf D}h({\bf x})\), where \(h({\bf x})= [f({\bf x})\,{\cdot}\, f({\bf x})]=f^2_1({\bf x})+ \cdots +f^2_m({\bf x})\). Then \begin{eqnarray*} {\bf D}h(x)&=&\bigg[ \frac{\partial h}{\partial x_1} \quad\cdots\quad \frac{\partial h}{\partial x_n}\bigg]\\[4pt] &=&\bigg[2\! f_1\frac{\partial f_1}{\partial x_1}+\cdots +2\! f_m \frac{\partial f_m}{\partial x_1} \quad\cdots\quad 2\! f_1\frac{\partial f_1}{\partial x_n}+ \cdots + 2\! f_m \frac{\partial f_m}{\partial x_n}\bigg], \end{eqnarray*} which can be written \(2f({\bf x}){\bf D}\! f({\bf x})\), where we regard \(f\) as a row matrix, \[ f=[f_1\quad\cdots\quad f_m] \hbox{and} {\bf D}\! f=\left[ \begin{array}{c@{\quad}c@{\quad}c}\\[-10pt] \displaystyle \frac{\partial f_1}{\partial x_1}&\cdots & \displaystyle \frac{\partial f_1}{\partial x_n}\\ \vdots & &\vdots\\ \displaystyle \frac{\partial f_m}{\partial x_1} &\cdots & \displaystyle \frac{\partial f_m}{\partial x_n} \end{array} \right]\!. \]

Thus, \({\bf D}g({\bf x})=2[\cos\,(f({\bf x}) \,{ \cdot}\, f({\bf x}))]f({\bf x}){\bf D}\! f({\bf x})\).