Derivatives

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 5 February 2010

First definition

Let $$\frac{d}{dx}:\mathbb{Q}[[x]]\to \mathbb{Q}[[x]]$$ such that if $\beta ,\gamma \in \mathbb{Q}$ and $f,g\in \mathbb{Q}[[x]]$ then

1. ${\displaystyle \frac{d}{dx}\left(\beta f+\gamma g\right)=\beta \frac{df}{dx}+\gamma \frac{dg}{dx}}$,
2. ${\displaystyle \frac{d\left(fg\right)}{dx}=f\frac{dg}{dx}+\frac{df}{dx}g}$, and
3. ${\displaystyle \frac{d}{dx}\left(x\right)=1}$.

Second definition

Let $f:\left[a,b\right]\to ℝ$. The derivative of $f$ at $x=c$ is $$f\prime \left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$$ or equivalently $$f\prime \left(c\right)=\underset{\Delta x\to 0}{\lim }\frac{f\left(c+\Delta x\right)-f\left(c\right)}{\Delta x}$$

Theorem

Let $f:\left[a,b\right]\to ℝ$ and $g:\left[a,b\right]\to ℝ$ and let $\beta ,\gamma \in ℝ$. Assume that $f\prime \left(c\right)$ and $g\prime \left(c\right)$. Then

1. $\left(\beta f+\gamma g\right)\prime \left(c\right)=\beta f\prime \left(c\right)+\gamma g\prime \left(c\right)$,
2. $\left(fg\right)\prime \left(c\right)=f\prime \left(c\right)g\left(c\right)+f\left(c\right)g\prime \left(c\right)$,
3. if $f:\left[a,b\right]\to ℝ$ is given by $f\left(x\right)=x$ then $f\prime \left(c\right)=1$, and
4. if $f\prime \left(c\right)$ exists then $f$ is continuous at $x=c$.

		Proof.
	(d): By assumption $f\prime \left(c\right)$ exists. So ${\displaystyle \underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}}$ exists. There exists $l$ such that ${\displaystyle \underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}}$. To show: $f$ is continuous at $x=c$. To Show: ${\displaystyle \underset{x\to c}{\lim }f\left(c\right)=f\left(c\right)}$. To show: If $ϵ\in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $\left|x-c\right|<\delta$ then $\left|f\left(x\right)-f\left(c\right)\right|<ϵ$. Assume $ϵ\in {ℝ}_{>0}$. We know that there exists ${\delta }_1\in {ℝ}_{>0}$ such that if $\left|x-c\right|<{\delta }_1$, then ${\displaystyle \left|\frac{f\left(x\right)-f\left(c\right)}{x-c}-l\right|<\frac{ϵ}{2}}$. Let $\delta =\min \left(1,{\delta }_1,\frac{ϵ}{2l}\right)$. To show: If $\left|x-c\right|<\delta$ then $\left|f\left(x\right)-f\left(c\right)\right|$. Assume $\left|x-c\right|<\delta$. To show: $\left|f\left(x\right)-f\left(c\right)\right|<ϵ$. ${\displaystyle \begin{array}{rcl}\left|f\left(x\right)-f\left(c\right)\right|& =& \left|\frac{f\left(x\right)-f\left(c\right)}{x-c}\left(x-c\right)\right|\\ & =& \left|\left(\frac{f\left(x\right)-f\left(c\right)}{x-c}-l\right)\left(x-c\right)+l\left(x-c\right)\right|\\ & \le & \left|\frac{f\left(x\right)-f\left(c\right)}{x-c}-l\right|\left|x-c\right|+l\left|x-c\right|\\ & <& \frac{ϵ}{2}·\delta +l\delta \\ & <& \frac{ϵ}{2}+\frac{lϵ}{2l}\\ & =& ϵ\text{.}\end{array}}$ $\square$

Standard derivatives

1. If $n\in \mathbb{Z}$ then ${\displaystyle \frac{d}{dx}{x}^n=n{x}^{n-1}}$.
2. If $c\in ℂ$ then ${\displaystyle \frac{d}{dx}{x}^c=c{x}^{c-1}}$.
3. ${\displaystyle \frac{d}{dx}{e}^x=e^x}$.
4. ${\displaystyle \frac{d}{dx}\log x=\frac{1}{x}}$.
5. ${\displaystyle \frac{d}{dx}\sin x=cosx}$.
6. ${\displaystyle \frac{d}{dx}\arcsin x=\frac{-1}{\sqrt{1-x^2}}}$.

The chain rule

${\displaystyle \frac{d}{dx}\left(f\circ g\right)=\frac{df}{dg}\frac{dg}{dx}}$.

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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