3. Differential calculus

Definition 0

The number

$$f^{\prime}(a)=\lim _{E \ni x \rightarrow a} \frac{f(x)-f(a)}{x-a}$$

is called the derivative of the function $f$ at $a$.

Definition 1

A function $f : E \to R$ defined on a set $E\subset R$ is differentiable at a point x ∈ E that is a limit point of E if $f (x + h) − f (x) = A(x)h + \alpha(x;h)$, where $h \mapsto A(x) h$ is a linear function in $h$ and $\alpha(x;h) = o(h)$as $h \to 0$, $x +h \in E$.

Definition 2

The function $h \mapsto A(x) h$ of Definition 1, which is linear in $h$, is called the differential of the function$f : E \to \mathcal R$ at the point $x\in E$ and is denoted $\mathrm d f (x)$ or $\mathrm D f (x)$. Thus, $\mathrm d f (x)(h) = A(x)h$.

We obtain

$$\frac{\mathrm{d} f(x)(h)}{\mathrm{d} x(h)}=f^{\prime}(x)$$

We denote the set of all such vectors by $T\mathbb R(x_0)$ or $T\mathbb R_{x_0}$. Similarly, we denote by $T\mathbb R(x_0)$ or $T\mathbb R_{y_0}$ the set of all displacement vectors from the point $y_0$ along the y-axis. It can then be seen from the definition of the differential that the mapping

$$\mathrm{d} f\left(x_{0}\right): T \mathbb{R}\left(x_{0}\right) \rightarrow T \mathbb{R}\left(f\left(x_{0}\right)\right)$$

The derivative of an inverse function

If a function $f$ is differentiable at a point x0 and its differential $\mathrm{d} f\left(x_{0}\right): T \mathbb{R}\left(x_{0}\right) \rightarrow T \mathbb{R}\left(y_0\right)a$ is invertible at that point, then the differential of the function $f^{ −1}$ inverse to $f$ exists at the point $y_0 = f (x_0)$ and is the mapping

$$\mathrm{d} f^{-1}\left(y_{0}\right)=\left[\mathrm{d} f\left(x_{0}\right)\right]^{-1}: T \mathbb{R}\left(y_{0}\right) \rightarrow T \mathbb{R}\left(x_{0}\right)$$

inverse to $\mathrm{d} f\left(x_{0}\right): T \mathbb{R}\left(x_{0}\right) \rightarrow T \mathbb{R}\left(y_0\right)a$ .

The derivative of some common function formula

1. $(C)^{\prime}=0$
2. $\left(x^{\mu}\right)^{\prime}=\mu x^{\mu-1}$
3. $(\sin x)^{\prime}=\cos x$
4. $(\cos x)^{\prime}=-\sin x$
5. $(\tan x)^{\prime}=\sec ^{2} x$
6. $(\cot x)^{\prime}=-\csc ^{2} x$
7. $(\sec x)^{\prime}=\sec x \tan x$
8. $(\csc x)^{\prime}=-\csc x \cot x$
9. $\left(a^{x}\right)^{\prime}=a^{x} \ln a \quad(a>0, a \neq 1)$
10. $\left(\mathrm{e}^{x}\right)^{\prime}=\mathrm{e}^{x}$
11. $\left(\log _{a} x\right)^{\prime}=\frac{1}{x \ln a}(a>0, a \neq 1)$
12. $(\ln x)^{\prime}=\frac{1}{x}$
13. $(\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}}$
14. $(\arccos x)^{\prime}=-\frac{1}{\sqrt{1-x^{2}}}$
15. $(\arctan x)^{\prime}=\frac{1}{1+x^{2}}$
16. $(\operatorname{arccot} x)^{\prime}=-\frac{1}{1+x^{2}}$
17. $(\sinh x)'= \cosh x$
18. $(\cosh x)'=\sinh x$
19. $(\tanh x)' =\frac{1}{\cosh ^{2} x}$
20. $(\operatorname{coth} x)' =-\frac{1}{\sinh ^{2} x}$
21. $(\operatorname{arsinh} x)'=\left(\ln \left(x+\sqrt{1+x^{2}}\right)\right)' = \frac{1}{\sqrt{1+x^{2}}}$
22. $(\operatorname{arcosh} x)'=(\ln \left(x \pm \sqrt{x^{2}-1}\right))'= \pm \frac{1}{\sqrt{x^{2}-1}}$
23. $(\operatorname{artanh} x)'=(\frac{1}{2} \ln \frac{1+x}{1-x})' = \frac{1}{1-x^{2}}$
24. $(\operatorname{arcoth} x)'=(\frac{1}{2} \ln \frac{x+1}{x-1})' = \frac{1}{x^{2}-1}$

theorem Leibniz

L'Hôpital's rule

The theorem states that for functions $f$ and $g$ which are differentiable on an open interval $I$ except possibly at a point $c$ contained in $I$, if ${\displaystyle \lim {x\to c}f(x)=\lim {x\to c}g(x)=0{\text{ or }}\pm \infty ,}$${\displaystyle \lim {x\to c}f(x)=\lim {x\to c}g(x)=0}$or $\pm \infty$ ,and $g'(x)\neq 0$$g'(x)\neq 0$ for all $x$ in $I$ with $x \ne c$, and $\lim_{x\to c}{\frac {f'(x)}{g'(x)}}$ exists, then

Suppose $f,g$are the function that defined in the set $X$, and also have the $n_{th}$ derivative, s.t.

$$(f g)^{(n)}(x)=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) f^{(k)}(x) g^{(n-k)}(x)$$

$$\lim_{x\to c}{\frac {f(x)}{g(x)}}=\lim_{x\to c}{\frac {f'(x)}{g'(x)}}$$

mean value theorem

Taylor's theorem

Rolle's theorem

Let $k \geq 1$ be an integer and let the function $f :\mathbb R\to\mathbb R$ be $k$ times differentiable at the point $a \in\mathbb R$. Then there exists a function $R_k : \mathbb R \to\mathbb R$ such that ,

If a real-valued function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f (a) = f (b)$, then there exists at least one $c$ in the open interval $(a, b)$ such that $f'(c)=0$.

$$f(a+x)=f(a)+f'(a)(x)+{\frac {f''(a)}{2!}}x^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}x^{k}+R_k(x;a)$$

Lagrange’s Mean Value Theorem

and,

If a real-valued function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, then there is exists at least one $c$ in the open interval $(a, b)$ such that

$$f(b)−f(a)=f′(c)(b−a)$$

Cauchy’s Mean Value Theorem

If a real-valued function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $g ' ( x ) \ne 0$ for all $x \in ( a , b )$ , then there is exists at least one $c$ in the open interval $(a, b)$ such that

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

$$R_{k}(x;a)=\int_{a}^{a+x} \frac{f^{(k+1)}(t)}{k !}(a+x-t)^{k} \mathrm d t$$

prove:

$$\begin{aligned} R_{k}(x;a)=&\int_{a}^{a+x} \frac{f^{(k+1)}(t)}{k !}(a+x-t)^{k} \mathrm d t\\ =& \frac{-1}{k !}\int_a^{a+x}f^{(k+1)}(t)(a+x-t)^k\mathrm d( a+x-t)\\ =&\frac{-1}{(k+1) !}\int_a^{a+x}f^{(k+1)}(t)\mathrm d( a+x-t)^{k+1}\\ =&\frac{-1}{(k+1) !}f^{(k+1)}(t)( a+x-t)^{k+1}|_{a}^{a+x}+\frac{-1}{(k+1) !}\int_a^{a+x}( a+x-t)^{k+1}\mathrm df^{(k+1)}(t)\\ =&\frac{f^{(k+1)}(a)}{(k+1) !}x^{k+1}+R_{k+1}(x;a) \end{aligned}$$

q.e.d

remainder term

using little $o$ notation, $R_{k}(x;a)=o\left(|x|^{k}\right), \quad x \rightarrow 0$(The Peano remainder term)

The Lagrange form remainder term( Mean-value forms)

$$R_{n}(x;a)=\frac{f^{(n+1)}(\theta)}{(n+1) !}x^{n+1}\quad (\theta\in(a,a+x))$$

$\blacksquare$