MATH 409 Lecture 15

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Lecture Slides

Review

Derivative is defined as $f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h}$ or equivalently $f^{\prime }(a)=\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}$.

Notations:

• Lagrange: $f^{\prime }$
• Newton: $\dot{f}$
• Leibniz: $\frac{\mathrm{d}f}{\mathrm{d}x}$
• Euler: $D_{x}f$
• $f^{(1)}$ (higher order derivatives)

Derivative at a point denoted ${(f(x){)}^{\prime }|}_{x=a}$.

Differentiability Theorems

• Sum and Difference Rules
• Product Rule
• Quotient Rule

Derivatives of Elementary Functions

Higher-Order Derivatives

Defined inductively:

For $n\ge 2$ for any $a\in \mathbb{ℝ}$, the $n$-th derivative of $f$ at a point $a$, denoted $f^{(n)}(a)$, is defined by $f^{(n)}(a)=(f^{(n-1)}{)}^{\prime }(a)$.

Derivative Spaces

Let $I$ be an interval of the real line $\mathbb{ℝ}$. We denote $C(I)$ or $C^0(I)$ as the set of all continuous functions on $I$.

For any $n\in \mathbb{\mathbb{N}}$, we denote $C^n(I)$ as the set of all functions that are $n$ times continuously differentiable on $I$. (that is, the $n$-th derivative is continuous)

$C^{\mathrm{\infty }}(I)$ represents the set of all infinitely continuously differentiable on $I$.

Examples

$f(0)=0$, and $f(x)=x\sin \frac{1}{x}$ for $x\ne 0$.

Product and Chain rules show that $f$ is differentiable on $\mathbb{ℝ}\setminus \left\{0\right\}$.

For $x\ne 0$.

$f^{\prime }(x)=(x\sin \frac{1}{x})=\sin \frac{1}{x}+x(\sin \frac{1}{x})=\sin \frac{1}{x}-\frac{1}{x}\cos \frac{1}{x}$

The function $f$ is continuous at $0$, but the derivative is not continuous at $0$, so $f$ is not differentiable at $0$. Indeed $\frac{f(h)-f(0)}{h}=\sin \frac{1}{h}$ has no limit as $h\to 0$.

$g(0)=0$, and $g(x)=x^2\sin \frac{1}{x}$ for $x\ne 0$.

As above, the Product and Chain rules show that $f$ is differentiable on $\mathbb{ℝ}\setminus \left\{0\right\}$.

For $x\ne 0$,

$\begin{array}{rl}{g}^{\prime }(x)& =(x\cdot x\sin \frac{1}{x})\\ & =2x\sin \frac{1}{x}-\cos \frac{1}{x}\end{array}$

The function is differentiable at 0. Indeed $\frac{g(h)-g(0)}{h}=h\sin \frac{1}{h}\to 0$ as $h\to 0$.

Note that $g$ is not continuously differentiable on $\mathbb{ℝ}$ since $g^{\prime }$ is not continuous at $0$. Namely, $\underset{x\to 0}{\lim }{g}^{\prime }(x)$ does not exist.

Power Rule

Theorem. $(x^n{)}^{\prime }=n{x}^{n-1}$ for all $x\in \mathbb{ℝ}$ and $n\in \mathbb{\mathbb{N}}$.

Proof by induction. In the case $n=1$, we have $x^{\prime }=1=1x^0$ for all $x\in \mathbb{ℝ}$.

Assume that $(x^n{)}^{\prime }=n{x}^{n-1}$ for some $n\in \mathbb{\mathbb{N}}$ and all $x\in \mathbb{ℝ}$. Using the product rule, we obtain $\left(x^{n+1}\right)=\left(x^{n}x\right)=(x^n{)}^{\prime }x+x^n{x}^{\prime }=n{x}^{n-1}x+x^n=(n+1)x^n$.

quod erat demonstrandum

Note: The theorem can also be proved using the formula $\frac{x^n-a^n}{x-a}=x^{n-1}+x^{n-2}a+\dots +x{a}^{n-2}+a^{n-1}$.

In the same breath, $(x^{-n}{)}^{\prime }=-n{x}^{-n-1}$ for all $x\ne 0$ and $n\in \mathbb{\mathbb{N}}$.

Using reciprocal rule, we obtain $(x^{-n}{)}^{\prime }=\left(\frac{1}{x^n}\right)=\frac{-(x^n{)}^{\prime }}{(x^n{)}^2}=\frac{-n{x}^{n-1}}{x^{2n}}=-n{x}^{-n-1}$

Derivative of Inverse Function

Theorem. Suppose $f$ is an invertible continuous function. If $f$ is differentiable at a point $a$ and $f^{\prime }(a)\ne 0$, then the inverse function is differentiable at the point $b=f(a)$ and

$(f^{-1}{)}^{\prime }(b)=\frac{1}{f^{\prime }(a)}=\frac{1}{f^{\prime }(f^{-1}(b))}$

Proof. Since $f$ is differentiable at $a$, we know that $f$ is defined on an open interval $I=(c,d)$ containing $a$. Since $f$ is continuous and invertible, it follows from the Intermediate Value Theorem that $f$ is strictly monotone on $I$, the image $f(I)$ is an open interval containing $b$, and the inverse function $f^{-1}$ is continuous and strictly monotone on $f(I)$.

We have $\underset{x\to a}{\lim }\frac{f(x)-f(a)}{x-a}=f^{\prime }(a)$. Since $f^{\prime }(a)\ne 0$, it follows that $\underset{x\to a}{\lim }\frac{x-a}{f(x)-f(a)}=\frac{1}{f^{\prime }(a)}$. Since $f^{-1}$ is continuous and monotone on the interval $f(I)$, we obtain that $f^{-1}(y)\to a$ and $f^{-1}(y)\ne a$ when $y\to b$ and $y\ne b$.

Therefore $\underset{y\to b}{\lim }\frac{f^{-1}(y)-a}{y-b}=\underset{y\to b}{\lim }\frac{f^{-1}(y)-a}{f(f^{-1}(y))-b}=\underset{x\to a}{\lim }\frac{x-a}{f(x)-f(a)}=\frac{1}{f^{\prime }(a)}$.

quod erat demonstrandum

Remark. In the case $f^{\prime }(a)=0$, the inverse function $f^{-1}$ is not differentiable at $f(a)$.

Indeed, if $f^{-1}$ is differentiable at $b=f(a)$, the chain rule implies that $(f^{-1}\circ f)(a)=(f^{-1}{)}^{\prime }(b)\cdot f^{\prime }(a)$. Obviously the LHS is the identity function, so $(f^{-1}\circ f{)}^{\prime }(a)=1\ne 0$, so that $f^{\prime }(a)\ne 0$.

Example

$f(x)=\arccos x$, $x\in [-1,1]$.

The function $g(y)=\cos y$ is strictly decreasing on $[0,\pi ]$ and maps this interval onto $[-1,1]$. By definition, the function $f(x)=\arccos x$ is the inverse of the restriction of $g$ to $[0,\pi ]$. Notice that $g^{\prime }(0)=g^{\prime }(\pi )=0$ and $g^{\prime }(y)\ne 0$ for $y\in (0,\pi )$. It follows that the function $f$ is continuous on $(-1,1)$ and not differentiable at $1$ and $-1$. Moreover, for any $x\in (-1,1)$,

$f^{\prime }(x)=\frac{1}{g^{\prime }(f(x))}=-\frac{1}{\sin \arccos x}$

Let $y=\arccos x$ (hence $x=\cos y$). We have ${\sin }^{2}y={\cos }^{2}y=1$ by the pythagorean identity. Besides, $\sin y>0$ since $y\in (0,\pi )$. Consequently, $\sin y=\sqrt{1-{\cos }^{2}y}=\sqrt{1-x^2}$. Thus $f^{\prime }(x)=-\frac{1}{\sqrt{1-x^2}}$.

quod erat demonstrandum

Homework hint; use similar method to prove for $\arcsin x$ and $\arctan x$, just use same identity (divide by ${\cos }^{2}y$ for $\arctan x$)

Exponential and Logarithmic Functions

Theorem. The sequence $x_n={(1+\frac{1}{n})}^n$ for $n\in \mathbb{\mathbb{N}}$ is increasing and bounded, hence convergent.

The limit is the number $\mathrm{e}=2.718281828\dots$ (number of letters in each word in "I'm forming a mnemonic to remember a constant in analysis")

Corollary. $\underset{x\to 0}{\lim }{(1+x)}^{\frac{1}{x}}=\mathrm{e}$.

Not proved here... (Ain't nobody got time for that!)

for any $a>0$ and $a\ne 1$, the exponential function $f(x)=a^x$ is strictly monotone and continuous on $\mathbb{ℝ}$. It maps $\mathbb{ℝ}$ onto $(0,\mathrm{\infty })$. Therefore the inverse function $g(y)={\log }_{a}y$ is strictly monotone and continuous on $(0,\mathrm{\infty })$. The natural logarithm ${\log }_{e}y$ is also denoted just $\log y$.

Since ${(1+h)}^{\frac{1}{h}}\to \mathrm{e}$ as $h\to 0$, it follows that $h^{-1}\log (1+h)=\log {(1+h)}^{\frac{1}{h}}\to \log \mathrm{e}=1$ as $h\to 0$. In other words, ${(\log y)|}_{y=1}=1$

Examples

$f(x)={\mathrm{e}}^x$ for $x\in \mathbb{ℝ}$.

$\frac{f(x+h)-f(x)}{h}=\frac{e^x(e^h-1)}{h}$ for all $x,h\in \mathbb{ℝ}$. Therefore for any $x\in \mathbb{ℝ}$, $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{e^h-1}{h}=e^x{f}^{\prime }(0)=e^x$

$f(x)=a^x$ for $x\in \mathbb{ℝ}$, where $a>0$.

Equivalently, $f(x)={\mathrm{e}}^{\log a^x}={\mathrm{e}}^{x\log a}$. So $f^{\prime }(x)={\mathrm{e}}^{x\log a}\log a=a^x\log a$.

$f(x)=\log x$ for $x\in (0,\mathrm{\infty })$.

Since $f$ is the inverse function $g(y)={\mathrm{e}}^y$, we obtain $f^{\prime }(x)=\frac{1}{g^{\prime }(\log x)}=\frac{1}{{\mathrm{e}}^{\log x}}=\frac{1}{x}$ for all $x>0$.

Power Rule: General Case

$\left(x^{\alpha }\right)=\alpha x^{\alpha -1}$ for all $x>0$ and $\alpha \in \mathbb{ℝ}$.

Proof. Let us fix a number $\alpha \in \mathbb{ℝ}$ and consider $f(x)=x^{\alpha }$ for $x\in (0,\mathrm{\infty })$. For any $x>0$, we obtain $f(x)={\mathrm{e}}^{\log x^{\alpha }}=a^{\log x}$, where $a={\mathrm{e}}^{\alpha }$. Hence $f=h\circ g$, where $g(x)=\log x$ for $x>0$ and $h(y)=a^y$ for $y\in \mathbb{ℝ}$. By the chain rule, $f^{\prime }(x)=\alpha x^{\alpha -1}$.

quod erat demonstrandum