The algebra of differentiable functions.

Theorem 6.5.1   If $ f$ and $ g$ are two differentiable functions at $ x_0$ , then $ f+g$ is differentiable at $ x_0$ and

$\displaystyle (f+g)'(x_0)=f'(x_0)+g'(x_0)$    

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
Be careful! The converse is not true!}
\end{figure}
For example, take $ f$ and $ g$ as follows:

\begin{displaymath}\begin{cases}f(x)=-1 \text{ if } x \leq 0  f(x)=1 \text{ if...
...1 \text{ if } x \leq 0  g(x)=-1 \text{ if } x > 0 \end{cases}\end{displaymath}    

The functions $ f$ and $ g$ are not differentiable at 0 (as they are not continous at 0, see theorem 7.1), but $ f+g$ is differentiable at 0 ($ f+g$ is constant on $ \mathbb{R}$ ).

Theorem 6.5.2   If $ f$ and $ g$ are two differentiable functions at $ x_0$ , then $ fg$ is differentiable at $ x_0$ and

$\displaystyle (fg)'(x_0)=f'(x_0)g(x_0)+f(x_0)g'(x_0)$    

If $ u$ and $ v$ are two functions differentiable on the same domain $ D$ , then $ uv$ is differentiable on $ D$ and we have:

$\displaystyle (uv)'=u'v+uv'$    

This is the so-called Leibniz's formula.

Example 6.5.3   Let $ f(x)=\underbrace{x}_{u(x)} \; \underbrace{\sin x}_{v(x)}$ . Then we have:

$\displaystyle \forall x \in \mathbb{R} \; f'(x)=1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x.$    

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
Be careful! The converse is not true!}
\end{figure}
For example, take $ f$ and $ g$ as follows:

\begin{displaymath}\begin{cases}f(x)=-1 \text{ if } x \leq 0  f(x)=1 \text{ if...
...1 \text{ if } x \leq 0  g(x)=-1 \text{ if } x > 0 \end{cases}\end{displaymath}    

The functions $ f$ and $ g$ are not differentiable at 0, but $ fg$ is differentiable at 0 ($ fg$ is constant on $ \mathbb{R}$ ).

Using the fact that a constant function is differentiable on its domain, Theorems 5.1 and  5.2 imply that the set of all the differentiable functions on the interval $ I$ is a real vector space.

Theorem 6.5.4   If $ g$ ia a differentiable function at $ x_0$ , and if $ g(x_0) \neq 0$ , then $ \frac 1g$ is differentiable at $ x_0$ and

$\displaystyle \left( \frac 1g \right)'(x_0)= - \frac {g'(x_0)}{(g(x_0)^2}$    

Example 6.5.5   Let $ f(x)= \frac {1}{\cos x}$ and $ x_0=0$ . The cosine function is differentiable at 0 and $ \cos 0 =1$ . Thus, $ f$ is differentiable at 0 and $ f'(0)= - \frac {\sin 0}{\cos ^2 0}=0$ .

Theorem 6.5.6   If $ f$ and $ g$ are two differentiable functions at $ x_0$ , and if $ g(x_0) \neq 0$ , then $ \frac fg$ is differentiable at $ x_0$ and

$\displaystyle \left( \frac fg \right)'(x_0) = \frac {f'(x_0)g(x_0)-f(x_0)g'(x_0)}{g(x_0)^2}.$    

Example 6.5.7   Let $ f(x)=x/ \sin x$ . The we have:

$\displaystyle \forall x \in \mathbb{R}-\{ k \pi, \; k \in \mathbb{Z} \}, \; f'(...
...\sin x - x \cdot \cos x }{\sin^2 x} =\frac {\sin x - x \cdot \cos x }{\sin^2 x}$    

Example 6.5.8   Let $ f(x)=\tan x = \frac {\sin x}{\cos x}$ . As the cosine and sine function are differentiable at 0 and $ \cos 0 \neq 0$ , the tangent function is differentiable at 0; we have:

$\displaystyle (\tan )' (0) = \frac { \cos 0 \cdot \cos 0 - \sin 0 \cdot (- \sin 0)}{\cos ^ 2 0} = \frac {1 + 0}{1} = 1$    

If $ u$ and $ v$ are two functions differentiable on the same domain $ D$ , then $ uv$ is differentiable on $ D$ and we have:

$\displaystyle (\frac uv)'=\frac {u'v-uv'}{v^2}$    

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
Be careful! The converse is not true!}
\end{figure}

Theorem 6.5.9   If $ f$ is differentiable at $ x_0$ and if $ g$ is differentiable at $ y_0=f(x_0)$ , then $ gof$ is differentiable at $ x_0$ , and

$\displaystyle (gof)'(x_0)=\underbrace{g'(f(x_0)}_{\text{outer derivation}} \cdot \underbrace{f'(x_0)}_{\text{inner derivation}}.$    

Example 6.5.10   Let $ f(x)=\sin (x^2+1)$ . According to the notations of Theorem 5.9, $ f$ is a polynomial function and $ g$ is the sine function. Both are differentiable over the whole of $ \mathbb{R}$ . Let us differentiate the function $ f$ ;

$\displaystyle \forall x \in \mathbb{R}, \; f'(x)=\cos (x^2+1) \cdot 2x = 2x \cos (x^2+1).$    

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
Be careful! The converse is not true!}
\end{figure}
For example, take the function $ f$ defined as follows:

\begin{displaymath}\begin{cases}f(x)=-1, \; x\leq 2  f(x)=1, \; x > 2 \end{cases}\end{displaymath}    

At the point 2, $ f$ is not differentiable. Now take a function constant over $ \mathbb{R}$ , say $ g: \; x \mapsto 4$ . Then $ g \circ f$ is constant over $ \mathbb{R}$ , thus is differentiable over $ \mathbb{R}$ .

Remark 6.5.11 (Chain Rule)   We will use the notations of 1.4 with the following setting:

$\displaystyle x \; \overset{f}{\rightarrow} \; y \; \overset{g}{\rightarrow} \; z$    

Then:
$ \frac{dz}{dx}=\frac {dz}{dy} \cdot \frac {dy}{dx}.$

Example 6.5.12   Let $ f(x)= x^2+3$ and $ g(x)=\cos x$ . Then for every $ x \in \mathbb{R}$ , we have:

$\displaystyle (gof)'(x)= \underbrace{- \sin (x^2+3)}_{=g'(f(x))} \cdot \underbrace{2x}_{=f'(x)} = -2x \; \sin (x^2+3).$    

Example 6.5.13   Let $ f(x)= x^2+3$ and $ g(x)= \sqrt{x}$ . Then for every $ x \in \mathbb{R}$ , $ f(x) > 0$ , hence the function $ gof$ is defined by $ (gof)(x)=\sqrt{x^2+3}$ and is differentiable at every point in $ \mathbb{R}$ . We have:

$\displaystyle \forall x \in \mathbb{R}, \; (gof)'(x)= \underbrace{\frac {1}{2\s...
... \underbrace{2x}_{=f'(x)} = \frac {2x}{2\sqrt{x^2+3}}= \frac {x}{\sqrt{x^2+3}}.$    

This last example is an illustration of the following result:

Corollary 6.5.14   Let $ u$ be a function, differentiable on the domain $ D$ . Suppose that for any $ x \in D, \; u(x)>0$ . Then the function $ \sqrt{u}$ is differentiable on $ D$ and we have:

$ (\sqrt{u} )'= \frac {u'}{2 \sqrt{u}}$

Example 6.5.15   Let $ f(x)=\sqrt{x^2+1}$ . As the radicand is positive for any real number $ x$ , the function is defined over $ \mathbb{R}$ and is differentiable over $ \mathbb{R}$ . Now let us differentiate the function:

$\displaystyle \forall x \in \mathbb{R}, \; f'(x)= \frac {2x}{2\sqrt{x^2+1}} = \frac {x}{\sqrt{x^2+1}}.$    

Another very useful result is the following:

Corollary 6.5.16   Let $ u$ be a function, differentiable on the domain $ D$ . Then $ \ln \vert u\vert$ is differentiable on $ D$ and we have:

$ (\ln \vert u\vert )'= \frac {u'}{u}$

Example 6.5.17   Let $ f(x)=\ln (x^2+1)$ . This function is defined, and differentiable, on $ \mathbb{R}$ , as the polynomial $ x^2+1$ has only positive values.

We have:

$\displaystyle \forall x \in \mathbb{R}, \; f'(x)= \frac {2x}{x^2+1}.$    

Of course we can iterate the process.

Example 6.5.18   Let $ f: \; x \mapsto \ln (\ln ( \ln x))$ . The function is defined whenever $ x>0$ and $ \ln x > 0 $ and $ \ln (\ln x))>0$ , i.e. $ D_f=(e, +\infty)$ . As we compose three differentiable functions, the function $ f$ is differentiable over $ (e, +\infty)$ , and we have:

$\displaystyle \forall x \in (e, +\infty), \; f'(x)= \frac {1}{\ln ( \ln x)} \cdot \frac {1}{\ln x} \cdot \frac 1x = \frac {1}{x \ln x \ln (\ln x)}.$    

Example 6.5.19   Let $ f(x)=\ln \vert \sin x \vert$ . This function is defined, and differentiable, on $ \mathbb{R}-\{ k \pi, k \in \mathbb{Z} \}$ .

We have:

$\displaystyle \forall x \in \mathbb{R}- \{ k \pi , k \in \mathbb{Z} \}, \; f'(x)= \frac {\cos x}{\sin x} = \cot x$    

Example 6.5.20   Let $ f(x)=\frac 14 \ln \left( 2x+\sqrt{1+4x^2} \right) + \frac 12 x \sqrt{1+4x^2}$ . This function is defined, and differentiable on $ \mathbb{R}$ (please, check this!).

We have:

$\displaystyle \forall x \in \mathbb{R}, \; f'(x)= \sqrt{1+4x^2}.$    

An important corollary of Theorem 5.9 is the following; it enables us to differentiate the inverse function (under certain conditions) of a known function:

Corollary 6.5.21   Let $ f$ be a function defined over the interval I; denote $ J=f(I)$ . Suppose that the following properties are verified:
(i)
$ f$ is invertible;
(ii)
$ f$ is differentiable at $ x_0 \in I$ ;
(iii)
$ f'(x_0) \neq 0$ .
If $ y_0=f(x_0)$ , then $ f^{-1}$ is differentiable at $ y_0$ and

$\displaystyle (f^{-1})'(y_0)=\frac {1}{f'(x_0)} = \frac {1}{f'(f^{-1}(y0)}.$    

Example 6.5.22   The function $ f$ such that $ f(x)=x^2$ for $ x \geq 0$ is invertible and $ f^{-1}(x)=\sqrt{x}$ .

As $ f'(0)=0$ , Corollary 5.21 does not apply for $ f^{-1}$ at ). At every other (positive) point, the corollary applies, and we have:

$\displaystyle (f^{-1})'(x)=\frac {1}{2 f'(f^{-1}(x))} = \frac {1}{2\sqrt{x}}.$    

We obtained already this derivative by applying Definition 1.1.

For example, Figure 4 displays the tangents at $ A(1,1)$ to the graphs of $ f$ and of $ f^{-1}$ :

Figure 4:
\begin{figure}\centering
\mbox{\epsfig{file=TangentsInverseGraphs.eps, height=5cm}}\end{figure}

Example 6.5.23 (Inverse trigonometric functions)       

(i)
Let $ f(x)=\arcsin x$ , for $ -1 \leq x \leq 1$ . As the derivative of the sine function is equal to 0 at the points $ -\pi /2$ and $ \pi /2$ , we cannot apply Corollary 5.21 for $ \arcsin$ at $ -1$ and at $ 1$ .

At any other point in the open interval $ (-1,1)$ , Corollary 5.21 applies, and we have:

$\displaystyle \forall x \in (-1,1), \; (\arcsin)'(x)$ $\displaystyle = \frac {1}{(\sin)'(\arcsin x))}$    
$\displaystyle \quad$ $\displaystyle =\frac {1}{\cos (\arcsin x)}$    
$\displaystyle \quad$ $\displaystyle = \frac {1}{\sqrt{1-x^2}}$   by example identities with inverse trigonometric functions$\displaystyle .$    

(ii)
Let $ f(x)=\arccos x$ , for $ -1 \leq x \leq 1$ . By an argument similar to the previous one, we conclude that this function is differentiable over $ (-1,1)$ and that

$\displaystyle \forall x \in (-1,1), \; (\arccos)'(x)=- \; \frac {1}{\sqrt{1-x^2}}.$    

(iii)
Let $ f(x)=\tan x$ , for $ -\pi /2 < x < \pi /2$ . The tangent function is differentiable over the interval $ (-\pi /2 , \pi /2)$ and its first derivative never vanishes. Therefore Corollary 5.21 applies at any real point and we have:

$\displaystyle \forall x \in \left(-\frac{\pi}{2} , \frac{\pi}{2} \right), \; (\arctan)'(x)$ $\displaystyle =\frac {1}{(\tan)'(\arctan x)}$    
$\displaystyle \quad$ $\displaystyle = \frac {1}{1+\tan^2 (\arctan x)}$    
$\displaystyle \quad$ $\displaystyle = \frac {1}{1+x^2},$   by example identities with inverse trigonometric functions$\displaystyle .$    

The formulas that we showed here should be added to the table of derivatives of elementary functions.

Noah Dana-Picard 2007-12-28