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 $(x_{0}, y_{0})$.