DOSEN PROFIL LENGKAP

A bit of a scratchpad for rendering math forumlae for articles, etc:

Proof that the differential operator is linear

Let $g(x)=\alpha f(x)\!$

${\frac {d}{dx}}{\Bigg (}\alpha f(x){\Bigg )}$	$={\frac {d}{dx}}{\Bigg (}g(x){\Bigg )}$	By hypothesis
	$=\lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}$	By the definition of differentiaion
	$=\lim _{h\to 0}{\frac {\alpha f(x+h)-\alpha f(x)}{h}}$	By hypothesis
	$=\alpha \lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$	Factoring out the $\alpha$
${\frac {d}{dx}}{\Bigg (}\alpha f(x){\Bigg )}$	$=\alpha {\frac {d}{dx}}{\Bigg (}f(x){\Bigg )}$	By the definition of differentiation

Let $j(x)=f(x)+g(x)\!$

${\frac {d}{dx}}{\Bigg (}f(x)+g(x){\Bigg )}$	$={\frac {d}{dx}}{\Bigg (}j(x){\Bigg )}$	By hypothesis
	$=\lim _{h\to 0}{\frac {j(x+h)-j(x)}{h}}$	By the definition of differentiaion
	$=\lim _{h\to 0}{\frac {f(x+h)+g(x+h)-f(x)-g(x)}{h}}$	By hypothesis
	$=\lim _{h\to 0}{\Bigg [}{\frac {f(x+h)-f(x)}{h}}+{\frac {g(x+h)-g(x)}{h}}{\Bigg ]}$	Reordering
	$=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}+\lim _{h\to 0}{\frac {g(x+h)-g(x)}{h}}$	By the linearity of limits
${\frac {d}{dx}}{\Bigg (}f(x)+g(x){\Bigg )}$	$={\frac {d}{dx}}{\Bigg (}f(x){\Bigg )}+{\frac {d}{dx}}{\Bigg (}g(x){\Bigg )}$	By the definition of differentiation, we obtain the required result.

Consider the case $n>0\!$ with $n\in \mathbb {N}$

${\frac {d}{dx}}x^{n}$	$=\lim _{h\to 0}{\frac {(x+h)^{n}-x^{n}}{h}}$	By the definition of differentiaion.
	$=\lim _{h\to 0}{\frac {\sum _{i=0}^{n}{{n \choose i}x^{n}h^{n-i}}-x^{n}}{h}}$	Expanding the $(x+h)^{n}$ term.
	$=\lim _{h\to 0}{\frac {\sum _{i=0}^{n-1}{{n \choose i}x^{n}h^{n-i}}+x^{n}-x^{n}}{h}}$	Pulling out the highest order term from the sum.
	$=\lim _{h\to 0}{\frac {\sum _{i=0}^{n-1}{{n \choose i}x^{n}h^{n-i}}}{h}}$	We note that $n-i\geq 1\ \forall i$ hence we may factor h from the sum.
	$=\lim _{h\to 0}{\frac {h{\Big (}\sum _{i=0}^{n-1}{{n \choose i}x^{n}h^{n-i-1}}{\Big )}}{h}}$	Factoring out an h from each term in the sum
	$=\lim _{h\to 0}\sum _{i=0}^{n-1}{{n \choose i}x^{n}h^{n-i-1}}$	Canceling the h in the denominator and numerator
	$={n \choose n-1}x^{n-1}$	Applying the limit makes all terms in the sum zero except the $h^{0}$ term
${\frac {d}{dx}}x^{n}$	$=nx^{n-1}\!$	Finally, we apply the identity ${n \choose n-1}=n\!$