THEOREM Chain Rule

If a function \(g\) is differentiable at \(x_0\), and a function \(f\) is differentiable at \(g(x_0)\), then the composite function \(f\circ g\) is differentiable at \(x_0\) and \[ (f\circ g)'(x_0)=f'(g(x_0))\cdot g'(x_0). \]

Using Leibniz notation, if \(y=f(u)\) and \(u=g(x)\), then \[ \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}, \]

where \(\displaystyle \frac{dy}{du}\) is evaluated at \(u_0=g(x_0)\) and \(\displaystyle \frac{du}{dx}\) is evaluated at \(x_0\).

Proof

For the fixed number \(x_0\), let \(\Delta u=g(x_0+\Delta x)-g(x_0)\). Since the function \(u=g(x)\) is differentiable at \(x_0\), it is also continuous at \(x_0\), and therefore \(\Delta u\to 0\) as \(\Delta x\to 0\).

For the fixed number \(u_0=g(x_0)\), let \(\Delta y=f(u_0+\Delta u)-f(u_0)\). Since the function \(y = f(u)\) is differentiable at the number \(u_0\), we can write \(\displaystyle \lim_{\Delta u\to 0}\frac{\Delta y}{\Delta u}=\frac{dy}{du}(u_0)\). This implies that for any \(\Delta u\ne 0\): \begin{equation}\label{Dy-over-Du} \frac{\Delta y}{\Delta u}=\frac{dy}{du}(u_0)+\alpha,\tag{1} \end{equation}

where \(\alpha=\alpha(\Delta u)\) is a function of \(\Delta u\) such that \(\alpha\to 0\) as \(\Delta u\to 0\). Now we define \(\alpha=0\) when \(\Delta u=0\), so that \(\alpha=\alpha(\Delta u)\) is continuous at \(0\). Multiplying (\ref{Dy-over-Du}) by \(\Delta u\ne 0\), we obtain \begin{equation}\label{Dy-itself} \Delta y=\frac{dy}{du}(u_0)\cdot\Delta u+\alpha(\Delta u)\cdot\Delta u. \tag{2} \end{equation}

Notice that the equation (\ref{Dy-itself}) is true for all \(\Delta u\):

• If \(\Delta u\) is not equal to 0, then (\ref{Dy-itself}) is a consequence of (\ref{Dy-over-Du}).
• If \(\Delta u\) equals 0, then the left hand side of (\ref{Dy-itself}) is \[ \Delta y=f(u_0+\Delta u)-f(u_0)=f(u_0)-f(u_0)=0 \]

  and the right-hand side of 2 is also 0.

  Now divide (\ref{Dy-itself}) by \(\Delta x\ne 0\): \begin{equation}\label{Dy-over-Dx} \frac{\Delta y}{\Delta x}=\frac{dy}{du}(u_0)\cdot\frac{\Delta u}{\Delta x}+\alpha(\Delta u)\cdot\frac{\Delta u}{\Delta x}. \tag{3} \end{equation}

Since the function \(u=g(x)\) is differentiable at \(x_0\), \(\displaystyle \lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x}=\frac{du}{dx}(x_0)\). Also, since \(\Delta u\to 0\) when \(\Delta x\to 0\) and \(\alpha(\Delta u)\) is continuous at \(\Delta u=0\), we conclude that \(\alpha(\Delta u)\to 0\) as \(\Delta x\to 0\). So we can take the limit of (\ref{Dy-over-Dx}) as \(\Delta x\to 0\), which proves that the derivative \(\displaystyle \frac{dy}{dx}(x_0)\) exists and is equal to \begin{eqnarray*} \frac{dy}{dx}(x_0)&=&\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}=\lim_{\Delta x\to 0} \left[\frac{dy}{du}(u_0)\cdot\frac{\Delta u}{\Delta x}+\alpha(\Delta u)\cdot\frac{\Delta u}{\Delta x}\right] \\[4pt] &=&\frac{dy}{du}(u_0)\cdot\left[\lim_{\Delta x\to 0} \frac{\Delta u}{\Delta x}\right]+\left[\lim_{\Delta x\to 0} \alpha(\Delta u)\right]\cdot\left[\lim_{\Delta x\to 0} \frac{\Delta u}{\Delta x}\right] \\[4pt] &=&\frac{dy}{du}(u_0)\cdot\frac{du}{dx}(x_0)+0\cdot\frac{du}{dx}(x_0)=\frac{dy}{du}(u_0)\cdot\frac{du}{dx}(x_0). \end{eqnarray*}

THEOREM Cauchy's Mean Value Theorem

If the functions \(f\) and \(g\) are continuous on the closed interval \( [a,b] \) and differentiable on the open interval \((a,b) ,\) and if \(g^{\prime} (x) \neq 0\) on \((a,b) \), then there is a number \(c\) in \((a,b)\) for which \begin{equation*} \dfrac{f^{\prime} (c) }{g^{\prime} (c) } = \dfrac{f(b) -f(a) }{g(b) -g(a) } \end{equation*}

Origins

The theorem was named after the French mathematician Augustin Cauchy (1789–1857).

Proof

Define the function \(h\) as \begin{equation*} h( x) = [ g( b) -g( a)] [ f( x) -f(a)] - [ g( x) -g (a)] [ f( b) -f(a)]\qquad a\leq x\leq b \end{equation*}

Then \(h\) is continuous on \([a,b]\) and differentiable on \(( a,b) \) and \(h(a) = h( b) =0\). So by Rolle's Theorem, there is a number \(c\) in the interval \((a,b) \) for which \(h^{\prime} (c) = 0\). That is, \begin{eqnarray*} h^{\prime} ( c) &=& [ g( b) -g(a) ] f^{\prime} (c) -g^{\prime} (c) [ f( b) -f(a) ] = 0\\[4pt] {[}g(b) - g(a) {]} f^{\prime} (c) &=& g^{\prime} (c) [f( b) - f(a)]\\[4pt] \dfrac{f' (c)}{g^{\prime} (c)} &=& \dfrac{f( b) -f(a)}{g( b) -g(a) } \end{eqnarray*}

THEOREM L'Hôpital's Rule

Suppose the functions \(f\) and \(g\) are differentiable on an open interval \(I\) containing the number \(c\), except possibly at \(c\), and \(g^{\prime} (x) \neq 0\) for all \(x\neq c\) in \(I\). Let \(L\) denote either a real number or \(\pm \infty \), and suppose \(\dfrac{f(x) }{g(x) }\) is an indeterminate form at \(c\) of the type \(\dfrac{0}{0}\) or \( \dfrac{\infty }{\infty}\). If \(\lim\limits_{x\rightarrow c}\dfrac{f^{\prime}(x) }{g^{\prime} ( x) }=L\), then \(\lim\limits_{x \rightarrow c}\dfrac{f(x) }{g(x) } = L\).

Partial Proof

Suppose \(\dfrac{f( x) }{g( x) }\) is an indeterminate form at \(c\) of the type \( \dfrac{0}{0}\), and suppose \(\lim\limits_{x\rightarrow c}\dfrac{f^{\prime} (x) }{g^{\prime} ( x) } = L\), where \(L\) is a real number. We need to prove \(\lim\limits_{x\rightarrow c}\dfrac{f( x) }{ g( x) } = L\). First define the functions \(F\) and \(G\) as follows: \begin{equation*} F( x) =\left\{ \begin{array}{c@{\quad}c@{\quad}c} f( x) &\quad \hbox{if} & x \neq c \\ 0 & \quad\hbox{if} & x = c \end{array} \right.,\qquad G( x) =\left\{ \begin{array}{c@{\quad}c@{\quad}c} g( x) & \quad\hbox{if} & x \neq c \\ 0 & \quad\hbox{if} & x = c \end{array} \right. \end{equation*}

Both \(F\) and \(G\) are continuous at \(c\), since \(\lim\limits_{x\rightarrow c}F( x) =\lim\limits_{x\rightarrow c}f( x) =0=F( c) \) and \(\lim\limits_{x\rightarrow c}G( x) =\lim\limits_{x\rightarrow c}g( x) =0=G( c) \). Also, \[ F^{\prime} ( x) =f^{\prime} ( x)\qquad \hbox{and}\qquad G^{\prime} (x) = g^{\prime} (x) \]

for all \(x\) in the interval \(I\), except possibly at \(c\). Since the conditions for Cauchy's Mean Value Theorem are met by \(F\) and \(G\) in either \( [x,c] \) or \([c,x] \), there is a number \(u\) between \(c \) and \(x\) for which \[ \dfrac{F( x) -F( c) }{G( x) -G( c) } = \dfrac{F^{\prime} (u) }{G^{\prime} (u)} = \dfrac{ f^{\prime} (u) } {g^{\prime} (u) } \]

Since \(F( c) = 0\) and \(G( c) = 0\), this simplifies to \( \dfrac{f( x) }{g( x) } = \dfrac{f^{\prime} (u) }{g^{\prime} (u)}\).

Since \(u\) is between \(c\) and \(x\), it follows that \begin{equation*} \lim\limits_{x\rightarrow c}\dfrac{f( x) }{g( x) } =\lim\limits_{u\rightarrow c}\dfrac{f^{\prime} (u) }{g^{\prime} (u) } = L \end{equation*}

A similar argument is used if \(L\) is infinite. The proof when \(\dfrac{ f( x) }{g( x) }\) is an indeterminate form at \(\infty \) of the type \(\dfrac{\infty }{\infty }\) is omitted here, but it may be found in books on advanced calculus.

THEOREM

Let \(z=f(x,y)\) be a function of two variables whose domain is \(D\). Let \((x_{0},y_{0})\) be an interior point of \( D\). If the partial derivatives \(f_{x}\) and \(f_{y}\) exist at each point of some disk centered at \((x_{0},y_{0})\), and if \(f_{x}\) and \(f_{y}\) are each continuous at \((x_{0},y_{0})\), then \(f\) is differentiable at \((x_{0},y_{0})\).

Proof

The proof depends on the Mean Value Theorem for derivatives. Let \(\Delta x\) and \(\Delta y\) be changes, not both \(0\), in \(x\) and in \(y\), respectively, so that the point \((x_{0}+\Delta x, y_{0}+\Delta y) \) lies in some disk centered at \((x_{0},y_{0})\). The change in \(z\) is \[ \Delta z = f( x_{0} + \Delta x,\,y_{0} + \Delta y) -f ( x_{0},y_{0}) \]

Adding and subtracting \(f( x_{0},y_{0}+\Delta y) \) on the right-hand side, we obtain \begin{equation} \Delta z=f( x_0+\Delta x,\,y_{0}+\Delta y) -f( x_{0},y_{0}+\Delta y) + f( x_{0},y_{0}+\Delta y) -f ( x_{0},y_{0})\tag{4} \end{equation}

The expression \(f(x, y_{0}+\Delta y)\) is a function of \(x\) alone, and its partial derivative \(f_{x}(x, y_{0}+\Delta y)\) exists in the disk centered at \((x_{0}, y_{0})\). Then by the Mean Value Theorem, there is a real number \( u \) between \(x_{0}\) and \(x_{0}+\Delta x\) for which \begin{equation} f(x_{0}+\Delta x, y_{0}+\Delta y)-f(x_{0}, y_{0}+\Delta y)=f_{x}(u, y_{0}+\Delta y)\Delta x\tag{5} \end{equation}

Similarly, the expression \(f(x_{0},y)\) is a function of \(y\) alone, and the partial derivative \(f_{y}(x_{0},y)\) exists in the disk centered at \( (x_{0},y_{0})\). Again, by the Mean Value Theorem, there is a real number \(v\) between \(y_{0}\) and \(y_{0}+\Delta y\) for which \begin{equation} f(x_{0}, y_{0}+\Delta y)-f(x_{0},y_{0})=f_{y}(x_{0},v)\Delta y\tag{6} \end{equation}

Substitute (5) and (6) back into equation (4) for \(\Delta z\) to obtain \begin{equation*} \Delta z=f_{x}(u, y_{0}+\Delta y)\Delta x+f_{y}(x_{0}, v)\Delta y \end{equation*}

Now introduce the functions \(\eta _{1}\) and \(\eta _{2}\) defined by \[ \begin{equation*} \eta _{1}=f_{x}(u, y_{0}+\Delta y)-f_{x}(x_{0},y_{0}) \quad and \quad \eta _{2}=f_{y}(x_{0},v)-f_{y}(x_{0},y_{0}) \end{equation*} \]

As \((\Delta x, \Delta y)\rightarrow (0, 0)\), then \(u\rightarrow x_0\) and \(v\rightarrow y_0\). Since \(f_{x}\) and \(f_{y}\) are continuous at \((x_{0},y_{0})\), \(\eta _{1}\) and \(\eta _{2}\) have the desired property that \begin{eqnarray*} \lim_{(\Delta x, \Delta y)\rightarrow (0, 0)}\eta _{1}&=&\lim_{(\Delta x, \Delta y)\rightarrow (0, 0)} [ f_{x}(u, y_{0}+\Delta y)-f_{x}(x_{0}, y_{0})]\\[4pt] &=&f_{x} ( {x_{0},y_{0}}) -f_{x}({x_{0},y_{0}}) =0 \end{eqnarray*} and \begin{eqnarray*} \lim_{(\Delta x, \Delta y)\rightarrow (0, 0)}\eta _{2}&=&\lim_{(\Delta x, \Delta y)\rightarrow (0, 0)}\left[ f_{y}(x_{0},v)-f_{y}(x_{0},y_{0}) \right] \\[4pt] &=&f_{y}(x_{0}, y_{0})-f_{y}(x_{0}, y_{0})=0 \end{eqnarray*}

As a result, \(\Delta z\) can be written as \begin{eqnarray*} \Delta z &=&f_{x}(u, y_{0}+\Delta y)\Delta x+f_{y}(x_{0}, v)\Delta y \\[3pt] &=&\left[ \eta _{1}+f_{x}(x_{0},y_{0}) \right] \Delta x+\left[ \eta _{2}+f_{y}( x_{0},y_{0}) \right] \Delta y \\[3pt] &=&f_{x}(x_{0},y_{0})\Delta x+f_{y}(x_{0},y_{0})\Delta y+\eta _{1}\Delta x+\eta _{2}\Delta y \end{eqnarray*} proving that \(f\) is differentiable at \((x_{0}, y_{0})\).