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3.2 Basic properties of differentiable functions

Let us start with a very basic fact that however is not immediately obvious from the definition:

Proposition 3.2.1.

If $f:I\rightarrow\mathbb{R}$ is differentiable at some $x_{0}\in I$ then $f$ is also continuous at $x_{0}$ . If $f$ is differentiable then it is continuous.

Proof.

The differentiability of $f$ in $x_{0}$ implies the convergence of the difference quotient

\frac{f(x)-f(x_{0})}{x-x_{0}}

to a number $f^{\prime}(x_{0})$ , as $x\rightarrow x_{0}$ , and hence

	$\displaystyle f(x)-f(x_{0})=$	$\displaystyle\Big{(}\frac{f(x)-f(x_{0})}{x-x_{0}}-f^{\prime}(x_{0})\Big{)}(x-x% _{0})+f^{\prime}(x_{0})(x-x_{0})$
		$\displaystyle\rightarrow 0\cdot 0+f^{\prime}(x_{0})\cdot 0=0\mbox{ as }x% \rightarrow x_{0},$

so $f$ is continuous at $x_{0}$ by definition of continuity. ∎

Key idea of proof.

write $f(x)-f(x_{0})$ in terms of difference quotient and $f^{\prime}(x_{0})$

Example 3.2.2.

Example 3.1.2(1) provides an example of a differentiable function. According to Proposition 3.2.1, it is therefore continuous – a fact familiar to us from previous modules. Example 3.1.2(2) instead gives us a function which is not differentiable. However, we know from MATH113 that it is continuous. This proves that the converse of Proposition 3.2.1 is not true in general, namely:

Comment: A differentiable function is always continuous. But a continuous function need not be differentiable.

Exercise W3.3.

Construct three substantially different functions which are continuous but not differentiable. Explain but you do not have to prove all details.

Definition 3.2.3.

A local extremum of a function $f:I\rightarrow\mathbb{R}$ is a point $x_{0}\in I$ such that there is $\delta>0$ and either $f(x)\geq f(x_{0})$ at all $x\in I\cap(x_{0}-\delta,x_{0}+\delta)$ (local minimum) or $f(x)\leq f(x_{0})$ at all $x\in I\cap(x_{0}-\delta,x_{0}+\delta)$ (local maximum).

Proposition 3.2.4.

Let $f:I\rightarrow\mathbb{R}$ be a continuous function which has a local extremum at some $x_{0}\in I$ . If $f$ is differentiable at $x_{0}$ then $f^{\prime}(x_{0})=0$ .

Proof.

Exercise A3.1. ∎

Proposition 3.2.5 (Sum, Product and Quotient Rules).

Let $f,g:I\rightarrow\mathbb{R}$ be differentiable at some $x_{0}\in I$ , and let $\lambda\in\mathbb{R}$ . Then:

(i)

$f+\lambda g$ is differentiable at $x_{0}$ with

$(f+\lambda g)^{\prime}(x_{0})=f^{\prime}(x_{0})+\lambda g^{\prime}(x_{0}).$
(ii)

$f\cdot g$ is differentiable at $x_{0}$ with

$(f\cdot g)^{\prime}(x_{0})=f^{\prime}(x_{0})g(x_{0})+f(x_{0})g^{\prime}(x_{0}).$
(iii)

If $g(x_{0})\not=0$ then $\frac{f}{g}$ is differentiable at $x_{0}$ with

$\Big{(}\frac{f}{g}\Big{)}^{\prime}(x_{0})=\frac{f^{\prime}(x_{0})g(x_{0})-f(x_% {0})g^{\prime}(x_{0})}{g(x_{0})^{2}}.$

Remark 3.2.6.

Before we can prove this, we have to recall a few simple facts from MATH113:

•

If $f,g:I\rightarrow\mathbb{R}$ are continuous at some $x_{0}\in I$ then $f+g$ and $f\cdot g$ are continuous at $x_{0}$ .
•

If additionally $g(x_{0})\not=0$ then $g(x)\not=0$ for $x$ in some neighbourhood of $x_{0}$ , and $\frac{f}{g}$ is defined in that neighbourhood and continuous at $x_{0}$ . More precisely, continuity of $g$ at $x_{0}$ implies that there is $\delta>0$ such that, for $x\in I\cap(x_{0}-\delta,x_{0}+\delta)$ , we have $g(x)\not=0$ , so on $I\cap(x_{0}-\delta,x_{0}+\delta)$ we can define $\frac{f}{g}$ and it is continuous at $x_{0}$ .

Proof.

(i) Exercise X (easy).

(ii) We already know that $f$ and $g$ are differentiable at $x_{0}$ , so their difference quotients at $x_{0}$ converge to $f^{\prime}(x_{0})$ and $g^{\prime}(x_{0})$ ; continuity at $x_{0}$ implies $g(x)\rightarrow g(x_{0})$ and $f(x)\rightarrow f(x_{0})$ as $x\rightarrow x_{0}$ . Combining these facts, we can calculate the difference quotient of $f g$ :

Comment: A standard trick we have already encountered in Proposition 3.2.1: add and subtract auxiliary terms in order to split into more familiar subexpressions.

	$\displaystyle\frac{f(x)g(x)-f(x_{0})g(x_{0})}{x-x_{0}}=$	$\displaystyle\frac{f(x)g(x)-f(x_{0})g(x)+f(x_{0})g(x)-f(x_{0})g(x_{0})}{x-x_{0}}$
	$\displaystyle=$	$\displaystyle\frac{f(x)-f(x_{0})}{x-x_{0}}g(x)+f(x_{0})\frac{g(x)-g(x_{0})}{x-% x_{0}}$
	$\displaystyle\rightarrow$	$\displaystyle f^{\prime}(x_{0})g(x_{0})+f(x_{0})g^{\prime}(x_{0}),\quad x% \rightarrow x_{0},$

which completes the proof.

(iii) Since $g$ is continuous at $x_{0}$ and $g(x_{0})\not=0$ , we have that $\frac{1}{g}$ is well-defined in a neighbourhood of $x_{0}$ and continuous at $x_{0}$ ; thus in particular $\frac{1}{g(x)}\rightarrow\frac{1}{g(x_{0})}$ , as $x\rightarrow x_{0}$ . We can now compute the difference quotient:

	$\displaystyle\frac{f(x)/g(x)-f(x_{0})/g(x_{0})}{x-x_{0}}=$	$\displaystyle\frac{f(x)/g(x)-f(x_{0})/g(x)+f(x_{0})/g(x)-f(x_{0})/g(x_{0})}{x-% x_{0}}$
	$\displaystyle=$	$\displaystyle\frac{1}{g(x)}\frac{f(x)-f(x_{0})}{x-x_{0}}-\frac{f(x_{0})}{g(x)g% (x_{0})}\frac{g(x)-g(x_{0})}{x-x_{0}}$
	$\displaystyle\rightarrow$	$\displaystyle\frac{1}{g(x_{0})}f^{\prime}(x_{0})-\frac{f(x_{0})}{g(x_{0})^{2}}% g^{\prime}(x_{0}),\quad x\rightarrow x_{0},$
	$\displaystyle=$	$\displaystyle\frac{f^{\prime}(x_{0})g(x_{0})-f(x_{0})g^{\prime}(x_{0})}{g(x_{0% })^{2}},$

which concludes the proof. ∎

Key idea of proof.

add and subtract auxiliary terms in difference quotients

Exercise L.

Study differentiability of $f:\mathbb{R}\rightarrow\mathbb{R}$ defined by $f(x)=x^{n}$ .

Example 3.2.7.

Let us study

f:(-\pi/2,\pi/2)\rightarrow\mathbb{R},\quad f(x)=\tan(x)^{2}=\Big{(}\frac{\sin% (x)}{\cos(x)}\Big{)}^{2}.

This is well-defined since $\cos$ is strictly positive on $(-\pi/2,\pi/2)$ . Moreover, $\sin$ and $\cos$ are differentiable on this domain and hence the quotient rule implies that $\tan$ is differentiable, and the product rule in turn implies that $\tan^{2}$ is differentiable on this domain. We can calculate it explicitly using the formulas of the product and quotient rule, namely

f^{\prime}(x)=2\tan(x)\tan^{\prime}(x)=2\tan(x)\frac{\cos(x)^{2}+\sin(x)^{2}}{% \cos(x)^{2}}=2\frac{\sin(x)}{\cos(x)^{3}}.

In contrast, the function

g:(0,\pi)\rightarrow\mathbb{R},\quad g(x)=\Big{(}\frac{\sin(x)}{\cos(x)}\Big{)% }^{2}

is not well-defined since $\cos(\pi/2)=0$ , so it does not make sense to speak about continuity or differentiability at all.

Example 3.2.8.

We would like to determine the derivative of

f:[0,\infty)\rightarrow[0,\infty),\quad f(x)=x^{p/q},

with $p,q\in\mathbb{N}$ , if it exists.

(1) Consider first $p=1$ and $q\in\mathbb{N}$ arbitrary. Then we have $f^{q}(x)=x$ , and $f^{q}$ is differentiable with derivative equal $1$ and constant difference quotient. The product rule then implies

1=\frac{f^{q}(x)-f^{q}(x_{0})}{x-x_{0}}=(f^{q-1}(x)+f^{q-2}(x)f(x_{0})+\ldots+% f^{q-1}(x_{0}))\frac{f(x)-f(x_{0})}{x-x_{0}}.

Suppose $x_{0}>0$ . Then $f(x)$ and all its powers are strictly positive, so we can write

\frac{f(x)-f(x_{0})}{x-x_{0}}=\frac{1}{f^{q-1}(x)+f^{q-2}(x)f(x_{0})+\ldots+f^% {q-1}(x_{0})}\rightarrow\frac{1}{qf^{q-1}(x_{0})}=\frac{1}{q}x_{0}^{-\frac{q-1% }{q}},

as $x\rightarrow x_{0}$ . So $f$ is differentiable on $(0,\infty)$ with $f^{\prime}(x_{0})=\frac{1}{q}x_{0}^{-\frac{q-1}{q}}$ . Instead for $x_{0}=0$ , we find

\frac{f(x)-f(x_{0})}{x-x_{0}}=\frac{x^{1/q}}{x}=x^{-\frac{q-1}{q}},

which converges to $1$ if $q=1$ and diverges otherwise if $q>1$ . Thus $f$ is not differentiable at $0$ except when $q=1$ .

(2) Now let both $p,q\in\mathbb{N}$ be arbitrary and $g(x)=x^{1/q}$ , so that $f=g^{p}$ ; we know from part (1) that the function $g$ is differentiable on $(0,\infty)$ for all $q$ and at $0$ if and only if $q=1$ . Then the product rule applied $p-1$ times together with (1) implies that $f$ is differentiable on $(0,\infty)$ with

f^{\prime}(x_{0})=(g^{p})^{\prime}(x_{0})=pg^{p-1}(x_{0})g^{\prime}(x_{0})=px_% {0}^{\frac{p-1}{q}}\frac{1}{q}x_{0}^{-\frac{q-1}{q}}=\frac{p}{q}x_{0}^{\frac{p% -q}{q}}=\frac{p}{q}x_{0}^{\frac{p}{q}-1}.

It can be shown to be differentiable at $0$ if and only if $p/q\in\mathbb{N}$ , and in that case

f^{\prime}(0)=\left\{\begin{array}[]{l@{\: : \:}l}1\hfil\>:\>&p/q=1\\ 0\hfil\>:\>&p/q\not=1.\end{array}\right.

The formula actually holds for all positive real exponents, not only the rational ones, but that is a bit more involved to show, see MATH210/215.

Theorem 3.2.9 (Chain rule).

Let $I,J\subset\mathbb{R}$ be intervals, $x_{0}\in I$ , and $f:I\rightarrow J$ and $g:J\rightarrow\mathbb{R}$ functions such that $f$ is differentiable at $x_{0}$ and $g$ is differentiable at $f(x_{0})$ . Then the composition $g\circ f:I\rightarrow\mathbb{R}$ is differentiable at $x_{0}$ with

(g\circ f)^{\prime}(x_{0})=g^{\prime}(f(x_{0}))f^{\prime}(x_{0}).

Proof.

Exercise X. ∎

Example 3.2.10.

Consider the functions

f:\mathbb{R}\rightarrow\mathbb{R},\quad f(x)=\sin(x)

and

g:\mathbb{R}\rightarrow\mathbb{R},\quad g(x)=|x|

and the point $x_{0}=0$ . We notice that almost all assumptions of the chain rule are fulfiled except that $g$ is not differentiable at $\sin(x_{0})=\sin(0)=0$ , according to Example 3.1.2(2). Thus we cannot apply the chain rule to $g\circ f$ . It is actually possible to show that $g\circ f$ is not differentiable at $x_{0}=0$ , in the same way as in Example 3.1.2(2): we compute the difference quotient

\frac{g\circ f(x)-g\circ f(x_{0})}{x-x_{0}}=\frac{|\sin(x)|-|\sin(x_{0})|}{x-x% _{0}}=\left\{\begin{array}[]{l@{\; : \:}l}\frac{\sin(x)}{x}\rightarrow 1\hfil% \;:\>&x>0\\ \frac{-\sin(x)}{x}\rightarrow-1\hfil\;:\>&x<0,\end{array}\right.

\frac{g\circ f(x)-g\circ f(x_{0})}{x-x_{0}}\rightarrow\left\{\begin{array}[]{l% @{\; : \:}l}1\hfil\;:\>&x\downarrow 0\\ -1\hfil\;:\>&x\uparrow 0.\end{array}\right.

Since the left and right-sided limits do not coincide, $g\circ f$ is not differentiable at $0$ . Now let us look at a different point, say $x_{0}=2$ . Then $f$ is differentiable at $2$ and $g$ is differentiable at $f(2)=\sin(2)>0$ , with $(g\circ f)^{\prime}(2)=g^{\prime}(\sin(2))\cos(2)=\cos(2)$ , according to the chain rule.

What about generic points? One finds that $g\circ f$ is differentiable at all points except at $\pi\mathbb{Z}$ , multiples of $\pi$ .

If instead we take $f(x)=\sin(x)+2$ then $f(\mathbb{R})\subset(0,\infty)$ , so $g\circ f$ is differentiable.

Comment: This example shows how crucial it is to verify that all assumptions of the theorem are fulfiled. Always remember the full theorem, not only the calculation formula!

Exercise W4.1.

Let $\alpha>0$ be a rational number and

f:[0,\infty)\rightarrow\mathbb{R},\quad f(x)=x^{\alpha}\sin\Big{(}\frac{1}{x}% \Big{)}.

Determine for which $\alpha$ values the function $f$ is differentiable. (Hint: split into the two cases $x_{0}=0$ and $x_{0}\not=0$ and use Example 3.2.8.)

Theorem 3.2.11 (Inverse function theorem).

Let $f:I\rightarrow\mathbb{R}$ be a strictly monotone continuous function, with inverse $f^{-1}:f(I)\rightarrow I$ . If $f$ is differentiable at $x_{0}\in I$ and $f^{\prime}(x_{0})\not=0$ then $f^{-1}$ is differentiable at $y_{0}=f(x_{0})$ with

(f^{-1})^{\prime}(y_{0})=\frac{1}{f^{\prime}(x_{0})}=\frac{1}{f^{\prime}(f^{-1% }(y_{0}))}.

In order to prove this theorem we need a general tool:

Lemma 3.2.12.

Let $f:I\rightarrow\mathbb{R}$ be a continuous function. If $f$ is differentiable at $x_{0}\in I$ then there is a continuous function $\phi:I\rightarrow\mathbb{R}$ such that

f(x)-f(x_{0})=(x-x_{0})\phi(x),\quad x\in I.{}

Proof.

Let

\phi(x)=\left\{\begin{array}[]{l@{\; : \;}l}\frac{f(x)-f(x_{0})}{x-x_{0}}\hfil% \;:\;&x\in I,\;x\not=x_{0}\\ f^{\prime}(x_{0})\hfil\;:\;&x=x_{0}.\end{array}\right.

Then $\phi$ is continuous at $x\not=x_{0}$ by the quotient rule for continuity (Remark 3.2.6 above, or MATH113 for details) because numerator and denominator are continuous and the denominator is nonzero at $x\not=x_{0}$ and $(*)$ is clearly fulfiled. Furthermore, differentiability of $f$ at $x_{0}$ implies that

\frac{f(x)-f(x_{0})}{x-x_{0}}\rightarrow f^{\prime}(x_{0}),\quad x\rightarrow x% _{0},

thus $\phi(x)\rightarrow\phi(x_{0})$ as $x\rightarrow x_{0}$ , so $\phi$ is continuous at $x=x_{0}$ as well. Equation $(*)$ in this case becomes $0=0$ which is trivially true. ∎

Exercise W4.2.

Prove the following statement:

Let $f:I\rightarrow\mathbb{R}$ be strictly monotone and continuous with $x_{0}\in I$ and write $y_{0}:=f(x_{0})$ and $y=f(x)$ . Then

y\rightarrow y_{0}\mbox{ if and only if }x\rightarrow x_{0}.

Proof of Theorem 3.2.11.

We would like to compute the difference quotient

\frac{f^{-1}(y)-f^{-1}(y_{0})}{y-y_{0}}

and show that it converges as $y\rightarrow y_{0}$ . Let $\phi:I\rightarrow\mathbb{R}$ be as in Lemma 3.2.12. Then $\phi(x)\not=0$ for all $x\in I$ because $f$ is strictly monotone and $\phi(x_{0})=f^{\prime}(x_{0})\not=0$ by assumption. Write $y=f(x)$ so that $x=f^{-1}(y)$ as in Exercise W4.2. Then

\frac{f^{-1}(y)-f^{-1}(y_{0})}{y-y_{0}}=\frac{x-x_{0}}{f(x)-f(x_{0})}=\frac{1}% {\phi(x)}=\frac{1}{\phi(f^{-1}(y))}

which is well-defined because $\phi(x)\not=0$ . From Exercise W4.2 we know that $y\rightarrow y_{0}$ if and only if $x\rightarrow x_{0}$ . Furthermore, $\phi$ and $f^{-1}$ are continuous, so the composition $\phi\circ f^{-1}$ is continuous as well and so $\phi(f^{-1}(y))\rightarrow\phi(f^{-1}(y_{0}))=f^{\prime}(f^{-1}(y_{0}))\not=0$ , as $y\rightarrow y_{0}$ . This implies that

\frac{f^{-1}(y)-f^{-1}(y_{0})}{y-y_{0}}=\frac{1}{\phi(f^{-1}(y))}\rightarrow% \frac{1}{f^{\prime}(f^{-1}(y_{0}))},\quad y\rightarrow y_{0},

which proves the theorem. ∎

Key idea of proof.

express difference quotient of $f^{-1}$ as inverse of difference quotient of $f$

Example 3.2.13.

Consider the function

f:[-\pi/2,\pi/2]\rightarrow\mathbb{R},\quad f(x)=\sin(x).

Then $f$ is continuous and strictly monotone. Notice that if we changed the domain to e.g. $[0,\pi]$ then $f$ would no longer be monotone.

Let $x_{0}\in[-\pi/2,\pi/2]$ . Then $f$ is differentiable at $x_{0}$ with $f^{\prime}(x_{0})=\cos(x_{0})\geq 0$ , which is nonzero if and only if $x_{0}\in(-\pi/2,\pi/2)$ , so let us make this additional assumption, $x_{0}\in(-\pi/2,\pi/2)$ . Then we can apply the inverse function rule to find that $f^{-1}=\arcsin:[-1,1]\rightarrow[-\pi/2,\pi/2]$ is differentiable at $y_{0}\in(-1,1)$ with

	$\displaystyle\arcsin^{\prime}(y_{0})=$	$\displaystyle(f^{-1})^{\prime}(y_{0})=\frac{1}{f^{\prime}(f^{-1}(y_{0}))}=% \frac{1}{\cos(\arcsin(y_{0}))}$
	$\displaystyle=$	$\displaystyle\frac{1}{\sqrt{1-\big{(}\sin(\arcsin(y_{0}))\big{)}^{2}}}=\frac{1% }{\sqrt{1-y_{0}^{2}}}.$

Consider now the case $x_{0}=\pi/2$ . Then

\frac{f^{-1}(y)-f^{-1}(y_{0})}{y-y_{0}}=\frac{1}{\frac{f(x)-f(x_{0})}{x-x_{0}}}

and the denominator converges to $f^{\prime}(x_{0})=\cos(\pi/2)=0$ , so the fraction does not converge, so $f^{-1}$ is not differentiable at $x_{0}=\pi/2$ . Analogously for $x_{0}=-\pi/2$ .