MATH 409 Lecture 15

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Lecture Slides

Review

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

Notations:

• Lagrange: $f'$
• 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 $\left.(f(x))'\right|_{x=a}$.

Differentiability Theorems

• Sum and Difference Rules
• Product Rule
• Quotient Rule

Derivatives of Elementary Functions

Higher-Order Derivatives

Defined inductively:

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

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 $n\in \mathbb {N}$, 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)

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

Examples

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

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

For $x\neq 0$.

$f'(x)=\left(x\,\sin {\frac {1}{x}}\right)'=\sin {\frac {1}{x}}+x\,\left(\sin {\frac {1}{x}}\right)=\sin {\frac {1}{x}}-{\frac {1}{x}}\,\cos {\frac {1}{x}}$

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 $x\neq 0$.

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

For $x\neq 0$,

${\begin{aligned}g'(x)&=\left(x\cdot x\sin {\frac {1}{x}}\right)'\\&=2x\,\sin {\frac {1}{x}}-\cos {\frac {1}{x}}\end{aligned}}$

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 $g'$ is not continuous at $0$. Namely, $\lim _{x\to 0}g'(x)$ does not exist.

Power Rule

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

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

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

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, $(x^{-n})'=-n\,x^{-n-1}$ for all $x\neq 0$ and $n\in \mathbb {N}$.

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

Derivative of Inverse Function

Theorem. Suppose $f$ is an invertible continuous function. If $f$ is differentiable at a point $a$ and $f'(a)\neq 0$, then the inverse function is differentiable at the point $b=f(a)$ and

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

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

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

quod erat demonstrandum

Remark. In the case $f'(a)=0$, 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 $\left(f^{-1}\circ f\right)'(a)=(f^{-1})'(b)\cdot f'(a)$. Obviously the LHS is the identity function, so $(f^{-1}\circ f)'(a)=1\neq 0$, so that $f'(a)\neq 0$.

Example

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

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

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

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

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 $x_{n}=\left(1+{\frac {1}{n}}\right)^{n}$ for $n\in \mathbb {N}$ is increasing and bounded, hence convergent.

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

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

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

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

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

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 $x\in \mathbb {R}$, $f'(x)=\lim _{h\to 0}{\frac {e^{h}-1}{h}}=e^{x}\,f'(0)=e^{x}$

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

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

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

Since $f$ is the inverse function $g(y)=\mathrm {e} ^{y}$, we obtain $f'(x)={\frac {1}{g'(\log {x})}}={\frac {1}{\mathrm {e} ^{\log {x}}}}={\frac {1}{x}}$ 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 $x\in \left(0,\infty \right)$. 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, $f'(x)=\alpha \,x^{\alpha -1}$.

quod erat demonstrandum