Miscellaneous Proofs

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Cegep1

Mathematics

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2176 words

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14 minutes

$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$

Proof 1

where $x\in (0,\frac{\pi }{2})$ is the angle between the x-axis and the line $\overline{OC}$.
This creates the following inequality:

$$A_{\mathrm{△}DOB}\le A_{\text{sector }DOB}\le A_{\mathrm{△}COB}$$

$A_{\mathrm{△}DOB}$

The base, $\overline{OB}$, and the hypotenuse, $\overline{DO}$, are radii of the unit circle, so they have length 1.

For the height,

$$\overline{DA}=\sin x\cdot \overline{DO}=\sin x\cdot 1=\sin x$$

Then,

$$A_{\mathrm{△}DOB}=\frac{bh}{2}=\frac{\sin x}{2}$$

$A_{\text{sector }DOB}$

$$A_{\text{sector }DOB}=\frac{1}{2}{r}^2\theta =\frac{x}{2}$$

$A_{\mathrm{△}COB}$

$$\overline{CB}=\tan x\cdot \overline{OB}=\tan x\cdot 1=\tan x$$

Then,

$$A_{\mathrm{△}COB}=\frac{bh}{2}=\frac{\tan x}{2}$$

---

Therefore, we have

$$\begin{array}{rlr}{A}_{\mathrm{△}DOB}& \le A_{\text{sector }DOB}& \le A_{\mathrm{△}COB}\\ \frac{\sin x}{2}& \le \frac{x}{2}& \le \frac{\tan x}{2}\end{array}$$

Since $x\in (0,\frac{\pi }{2})$, we consider $x\to 0^{+}$.
Multiplying through by $\frac{2}{\sin x}>0$ yields

$$1\le \frac{x}{\sin x}\le \frac{1}{\cos x}$$

Taking reciprocals gives

$$\begin{array}{rlr}\cos x& \le \frac{\sin x}{x}& \le 1\\ ⟹\underset{x\to 0^{+}}{lim}\cos x& \le \underset{x\to 0^{+}}{lim}\frac{\sin x}{x}& \le \underset{x\to 0^{+}}{lim}1\\ 1& \le \underset{x\to 0^{+}}{lim}\frac{\sin x}{x}& \le 1\end{array}$$

$\therefore$ by the Squeeze Theorem,

$$\underset{x\to 0^{+}}{lim}\frac{\sin x}{x}=1$$

For $x\to 0^{-}$, since $\frac{\sin x}{x}$ is an even function as demonstrated below

$$\frac{\sin -x}{-x}=\frac{-\sin x}{-x}=\frac{\sin x}{x}$$

It follows that $\underset{x\to 0^{-}}{lim}\frac{\sin x}{x}=1$.
Hence,

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$$

$◻$

Derivative of a constant

[!abstract] $\frac{d}{dx}c=0$ for $c\in \mathbb{R}$

Let $f(x)=c$, then

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{c-c}{h}\\ & =\underset{h\to 0}{lim}\frac{0}{h}\\ & =\underset{h\to 0}{lim}0\\ & =0\end{array}$$

$◻$

Derivative of function times a constant

[!abstract]+
Let $c\in \mathbb{R}$ and $f$ be a differentiable function, then

$$\frac{d}{dx}(cf(x))=c\frac{d}{dx}f(x)$$

Let $g(x)=cf(x)$, then

$$\begin{array}{rl}\frac{d}{dx}(cf(x))=g^{\prime }(x)& =\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{cf(x+h)-cf(x)}{h}\\ & =c\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =c{f}^{\prime }(x)\end{array}$$

$◻$

Derivative of sum / difference

[!abstract]+
Let $f$ and $g$ be differentiable functions, then

$$\frac{\mathrm{d}}{\mathrm{d}x}(f(x)\pm g(x))=\frac{\mathrm{d}}{\mathrm{d}x}f(x)\pm \frac{\mathrm{d}}{\mathrm{d}x}g(x)$$

Let $s(x)=f(x)\pm g(x)$, then

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}(f(x)\pm g(x))=s^{\prime }(x)& =\underset{h\to 0}{lim}\frac{s(x+h)-s(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{(f(x+h)\pm g(x+h))-(f(x)\pm g(x))}{h}\\ & =\underset{h\to 0}{lim}\frac{(f(x+h)-f(x))\pm (g(x+h)-g(x))}{h}\\ & =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\pm \underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}\\ & =f^{\prime }(x)\pm g^{\prime }(x)\end{array}$$

$◻$

Derivative of $\sin x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\sin x=\cos x$

Let $f(x)=\sin x$.

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin x}{h}$$

Using the sine rule for sums,

$$\begin{array}{rl}& =\underset{h\to 0}{lim}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ & =\underset{h\to 0}{lim}\frac{\sin x(\cos h-1)+\cos x\sin h}{h}\\ & =\underset{h\to 0}{lim}\frac{\sin x(\cos h-1)}{h}+\underset{h\to 0}{lim}\frac{\cos x\sin h}{h}\\ & =\sin x\underset{h\to 0}{lim}\frac{\cos h-1}{h}+\cos x\underset{h\to 0}{lim}\frac{\sin h}{h}\\ & =\cos x\end{array}$$

$◻$

Derivative of $\cos x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\cos x=-\sin x$

Let $f(x)=\sin x$.

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{lim}\frac{\cos (x+h)-\cos x}{h}$$

Using the cosine rule for sums,

$$\begin{array}{rl}& =\underset{h\to 0}{lim}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\\ & =\underset{h\to 0}{lim}\frac{\cos x(\cos h-1)-\sin x\sin h}{h}\\ & =\underset{h\to 0}{lim}\frac{\cos x(\cos h-1)}{h}-\underset{h\to 0}{lim}\frac{\sin x\sin h}{h}\\ & =\cos x\underset{h\to 0}{lim}\frac{\cos h-1}{h}-\sin x\underset{h\to 0}{lim}\frac{\sin h}{h}\\ & =-\sin x\end{array}$$

$◻$

Derivative of $\tan x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\tan x={\sec }^{2}x$

$$\begin{array}{rl}\frac{d}{dx}\tan x& =\frac{d}{dx}\frac{\sin x}{\cos x}\\ & =\frac{\cos x(\sin x{)}^{\prime }-\sin x(\cos x{)}^{\prime }}{{\cos }^{2}x}\\ & =\frac{\cos x\cos x+\sin x\sin x}{{\cos }^{2}x}\\ & =\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}\\ & =\frac{1}{{\cos }^{2}x}\\ & ={\sec }^{2}x\end{array}$$

$◻$

Derivative of $\sec x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\sec x=\sec x\tan x$

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\sec x& =\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{\cos x}\\ & =\frac{\cos x(1{)}^{\prime }-(\cos x{)}^{\prime }}{{\cos }^{2}x}\\ & =\frac{\cos x\cdot 0-(-\sin x)}{{\cos }^{2}x}\\ & =\frac{\sin x}{{\cos }^{2}x}\\ & =\frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}\\ & =\sec x\tan x\end{array}$$

$◻$

Derivative of $\csc x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\csc x=-\csc x\cot x$

$$\begin{array}{rl}\frac{d}{dx}\csc x& =\frac{d}{dx}\frac{1}{\sin x}\\ & =\frac{\sin x(1{)}^{\prime }-(\sin x{)}^{\prime }}{{\sin }^{2}x}\\ & =\frac{\sin x\cdot 0-\cos x}{{\sin }^{2}x}\\ & =-\frac{\cos x}{{\sin }^{2}x}\\ & =-\frac{1}{\sin x}\cdot \frac{\cos x}{\sin x}\\ & =-\csc x\cot x\end{array}$$

$◻$

Derivative of $\cot x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\cot x=-{\csc }^{2}x$

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\cot x& =\frac{\mathrm{d}}{\mathrm{d}x}\frac{\cos x}{\sin x}\\ & =\frac{\sin x(\cos x{)}^{\prime }-\cos x(\sin x{)}^{\prime }}{{\sin }^{2}x}\\ & =\frac{-{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x}\\ & =-\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}\\ & =-\frac{1}{{\sin }^{2}x}\\ & =-{\csc }^{2}x\end{array}$$

$◻$

Derivative of $\arcsin x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\arcsin x=\frac{1}{\sqrt{1-x^2}}$

Let $y=\arcsin x$ so $x=\sin y$.
Note that the range of $y=\arcsin x$ is $y\in [-\frac{\pi }{2},\frac{\pi }{2}]$.
Using implicit differentiation, we have

$$\begin{array}{rl}x& =\sin y\\ \frac{\mathrm{d}}{\mathrm{d}x}x& =\frac{\mathrm{d}}{\mathrm{d}x}\sin y\\ 1& =\cos y\cdot y^{\prime }\\ y^{\prime }& =\frac{1}{\cos y}\end{array}$$

We use ${\sin }^{2}y+{\cos }^{2}y=1⟹|\cos y|=\sqrt{1-{\sin }^{2}y}$. Since $y\in [-\frac{\pi }{2},\frac{\pi }{2}]$, $\cos y\ge 0$.
Then,

$$\begin{array}{rl}& =\frac{1}{\sqrt{1-{\sin }^{2}y}}\\ & =\frac{1}{\sqrt{1-x^2}}\end{array}$$

$◻$

Derivative of $\arccos x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\arccos x=-\frac{1}{\sqrt{1-x^2}}$

Let $y=\arccos x$ so $x=\cos y$.
Note that the range of $y=\arccos x$ is $y\in [0,\pi ]$.
Using implicit differentiation, we have

$$\begin{array}{rl}x& =\cos y\\ \frac{\mathrm{d}}{\mathrm{d}x}x& =\frac{\mathrm{d}}{\mathrm{d}x}\cos y\\ 1& =-\sin y\cdot y^{\prime }\\ y^{\prime }& =-\frac{1}{\sin y}\end{array}$$

We use ${\sin }^{2}y+{\cos }^{2}y=1⟹|\sin y|=\sqrt{1-{\cos }^{2}y}$. Since $y\in [0,\pi ]$, $\sin y\ge 0$.
Then,

$$\begin{array}{rl}& =-\frac{1}{\sqrt{1-{\cos }^{2}y}}\\ & =-\frac{1}{\sqrt{1-x^2}}\end{array}$$

$◻$

Derivative of $\arctan x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\arctan x=\frac{1}{1+x^2}$

Let $y=\arctan x$ so $x=\tan y$. Using implicit differentiation, we have

$$\begin{array}{rl}x& =\tan y\\ \frac{\mathrm{d}}{\mathrm{d}x}x& =\frac{\mathrm{d}}{\mathrm{d}x}\tan y\\ 1& ={\sec }^{2}y\cdot y^{\prime }\\ y^{\prime }& =\frac{1}{{\sec }^{2}y}\\ y^{\prime }& =\frac{1}{1+{\tan }^{2}y}\\ y^{\prime }& =\frac{1}{1+x^2}\end{array}$$

$◻$

Derivative of $\mathrm{arcsec}x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arcsec}x=\frac{1}{|x|\sqrt{x^2-1}}$

Let $y=\mathrm{arcsec}x$ so $x=\sec y$, equivalently $\cos y=\frac{1}{x}$.
Note that the range and domain of $y=\mathrm{arcsec}x$ are $y\in [0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi ]$ and $x\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$, respectively.
Using implicit differentiation, we have

$$\begin{array}{rl}\cos y& =\frac{1}{x}\\ \frac{\mathrm{d}}{\mathrm{d}x}\cos y& =\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{x}\\ -\sin y\cdot y^{\prime }& =-\frac{1}{x^2}\\ y^{\prime }& =\frac{1}{x^2\sin y}\end{array}$$

We use ${\sin }^{2}y+{\cos }^{2}y=1⟹|\sin y|=\sqrt{1-{\cos }^{2}y}$. Since $y\in [0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi ]$, $\sin y\ge 0$.
Then,

$$\begin{array}{rl}& =\frac{1}{x^2\sqrt{1-{\cos }^{2}y}}\\ & =\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}\\ & =\frac{1}{x^2\sqrt{\frac{x^2-1}{x^2}}}\\ & =\frac{1}{x^2\cdot \frac{1}{|x|}\sqrt{x^2-1}}\end{array}$$

If $x\in (-\mathrm{\infty },-1]$ (i.e. negative values), then $|x|=-x$ and $\frac{x^2}{|x|}=-x$. If $x\in [1,\mathrm{\infty })$ (i.e. positive values), then $|x|=x$ and $\frac{x^2}{|x|}=x$. In both cases, $\frac{x^2}{|x|}=|x|$, hence

$$=\frac{1}{|x|\sqrt{x^2-1}}$$

$◻$

Derivative of $\mathrm{arccsc}x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arccsc}x=-\frac{1}{|x|\sqrt{x^2-1}}$

Let $y=\mathrm{arccsc}x$ so $x=\csc y$, equivalently $\sin y=\frac{1}{x}$.
Note that the range and domain of $y=\mathrm{arccsc}x$ are $y\in [-\frac{\pi }{2},0)\cup (0,\frac{\pi }{2}]$ and $x\in (-\mathrm{\infty },-1]\cup [1,\mathrm{\infty })$, respectively.
Using implicit differentiation, we have

$$\begin{array}{rl}\sin y& =\frac{1}{x}\\ \frac{\mathrm{d}}{\mathrm{d}x}\sin y& =\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{x}\\ \cos y\cdot y^{\prime }& =-\frac{1}{x^2}\\ y^{\prime }& =-\frac{1}{x^2\cos y}\end{array}$$

We use ${\sin }^{2}y+{\cos }^{2}y=1⟹|\cos y|=\sqrt{1-{\sin }^{2}y}$. Since $y\in [-\frac{\pi }{2},0)\cup (0,\frac{\pi }{2}]$, $\cos y\ge 0$.
Then,

$$\begin{array}{rl}& =-\frac{1}{x^2\sqrt{1-{\sin }^{2}y}}\\ & =-\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}\\ & =-\frac{1}{x^2\sqrt{\frac{x^2-1}{x^2}}}\\ & =-\frac{1}{x^2\cdot \frac{1}{|x|}\sqrt{x^2-1}}\end{array}$$

If $x\in (-\mathrm{\infty },-1]$ (i.e. negative values), then $|x|=-x$ and $\frac{x^2}{|x|}=-x$. If $x\in [1,\mathrm{\infty })$ (i.e. positive values), then $|x|=x$ and $\frac{x^2}{|x|}=x$. In both cases, $\frac{x^2}{|x|}=|x|$, hence

$$=-\frac{1}{|x|\sqrt{x^2-1}}$$

$◻$

Derivative of $\mathrm{arccot}x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{arccot}x=-\frac{1}{1+x^2}$

Let $y=\mathrm{arccot}x$ so $x=\cot y$. Using implicit differentiation, we have

$$\begin{array}{rl}x& =\cot y\\ \frac{\mathrm{d}}{\mathrm{d}x}x& =\frac{\mathrm{d}}{\mathrm{d}x}\cot y\\ 1& =-{\csc }^{2}y\cdot y^{\prime }\\ y^{\prime }& =-\frac{1}{{\csc }^{2}y}\\ y^{\prime }& =-\frac{1}{1+{\cot }^{2}y}\\ y^{\prime }& =-\frac{1}{1+x^2}\end{array}$$

$◻$

Derivative of ${\log }_{b}x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}{\log }_{b}x=\frac{1}{x\ln b}$

Note that

$$\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{1}{n})}^n\equiv \underset{m\to 0}{lim}(1+m{)}^{1/m}=e$$

Let $f(x)={\log }_{b}x$, then

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{{\log }_b(x+h)-{\log }_{b}x}{h}\\ & =\underset{h\to 0}{lim}(\frac{1}{h}{\log }_b\frac{x+h}{x})\\ & =\underset{h\to 0}{lim}(\frac{1}{h}{\log }_b(1+\frac{h}{x}))\\ & =\underset{h\to 0}{lim}(\frac{1}{x}\cdot \frac{x}{h}{\log }_b(1+\frac{h}{x}))\\ & =\frac{1}{x}\underset{h\to 0}{lim}{\log }_b{(1+\frac{h}{x})}^{x/h}\end{array}$$

Since ${\log }_b$ is continuous,

$$\begin{array}{rl}& =\frac{1}{x}{\log }_b\underset{h\to 0}{lim}{(1+\frac{h}{x})}^{x/h}\\ & =\frac{1}{x}{\log }_b\underset{h\to 0}{lim}{(1+\frac{h}{x})}^{1/(h/x)}\end{array}$$

Using definition of $e$ with $m=\frac{h}{x}$,

$$=\frac{1}{x}{\log }_{b}e$$

Using the change of base rule,

$$\begin{array}{rl}& =\frac{1}{x}\frac{\ln e}{\ln b}\\ & =\frac{1}{x\ln b}\end{array}$$

$◻$

Derivative of $a^x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}{a}^x=a^x\ln a$

Let $y=a^x$ so $x={\log }_{a}y$. Using implicit differentiation, we have

$$\begin{array}{rl}x& ={\log }_{a}y\\ \frac{\mathrm{d}}{\mathrm{d}x}x& =\frac{\mathrm{d}}{\mathrm{d}x}{\log }_{a}y\\ 1& =\frac{1}{y\ln a}\cdot y^{\prime }\\ y^{\prime }& =y\ln a\\ y^{\prime }& =a^x\ln a\end{array}$$

$◻$

Differentiability implies continuity

We have to show that $f(a)=\underset{x\to a}{lim}f(x)$. We know that $f$ is differentiable at $a$, so

$$f^{\prime }(a)=\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}$$

is certain to exist.
Then,

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}\\ f^{\prime }(x)& =\frac{\underset{x\to a}{lim}(f(x)-f(a))}{\underset{x\to a}{lim}(x-a)}\\ f^{\prime }(x)\cdot \underset{x\to a}{lim}(x-a)& =\underset{x\to a}{lim}(f(x)-f(a))\\ f^{\prime }(x)\cdot 0& =\underset{x\to a}{lim}f(x)-\underset{x\to a}{lim}f(a)\\ \underset{x\to a}{lim}f(a)& =\underset{x\to a}{lim}f(x)\\ f(a)& =\underset{x\to a}{lim}f(x)\end{array}$$

The contrapositive of the theorem says non-continuity $\implies$ non-differentiability.

The converse is not necessarily true, i.e. continuity doesn't imply differentiability.

Constant on interval when derivative is 0

[!abstract] If $f^{\prime }(x)=0\mathrm{\forall }x\in (a,b)$, then $f$ is constant on $(a,b)$.

We want to show that for any $x_1,x_2\in (a,b)$ where $x_1<x_2$, $f(x_1)=f(x_2)$.
Since $f$ is differentiable on $(a,b)$, it must be differentiable on $(x_1,x_2)$ and continuous on $[x_1,x_2]$.
$\therefore$ by M. V. T., $\mathrm{\exists }c\in (x_1,x_2)$ such that

$$\mathrm{\Delta }f(x)=f^{\prime }(c)\mathrm{\Delta }x$$

Since $f^{\prime }(x)=0\mathrm{\forall }x\in (a,b)$, we have

$$\mathrm{\Delta }f(x)=0⟹f(x_2)=f(x_1)$$

Therefore, $f$ has the same value at any two numbers $x_1$ and $x_2$ in $(a,b)$. This means that $f$ is constant on $(a,b)$.

$◻$

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Ajitani Hifumi

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