#help beanie basic stuff calc
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@woeful glen dms
$\derivative x f(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
wolfqz
formal definition
straight line that intersects the curve at two pts
yeah and what was h again
like
the displacement
or
the
curved
alr
yeah'
thats integration
right
oh
okay
$\derivative x x^2$
wolfqz
this
$\lim_{h\to 0}\frac{(x+h)^2 - x^2}{h}$
$\lim_{h\to 0} \frac{x^2+2xh+h^2-x^2}{h}$
Yeah
alr
that is the derivative of x^2
alr
okay
$\derivative x (ax^n+bx^{n-1}) = \derivative x ax^n + \derivative x bx^{n-1}$
wolfqz
alright
$\derivative x (x^2+2x+1)$ at $x=5$
wolfqz
$\derivative x \sin x$
$\lim_{h \to 0} \frac{\sin(x+h)-\sin(x)}{h}$
now
we do something tricky stuff
$\lim_{h \to 0} \frac{\sin(x)\cos(h) - \sin(x) +\sin(h)\cos(x)}{h}$
wolfqz
$\lim_{h \to 0} \frac{-\sin(x)(1-\cos(h))}{h} + \frac{\sin(h)\cos(x)}{h}$
wolfqz
oh
$1-\cos(h) = 2\sin^2\frac{h}{2}$
wolfqz
$\cosaddition$
wolfqz
$\cos^2x - \sin^2x = 1-\sin^2x-\sin^2x = 1-2\sin^2x = \cos(2x)$
wolfqz
$x = \frac{h}{2}$
$1-2\sin^2 \frac{h}{2} = \cos h$
wolfqz
^
$\lim_{h \to 0} \frac{-2\sin(x) \sin^2\frac{h}{2}}{h} + \frac{\sin(h)\cos(x)}{h}$
$\lim_{h \to 0} \frac{-2\sin(x) \sin^2\frac{h}{2}}{h} + \lim_{h \to 0} \frac{\sin(h)\cos(x)}{h}$
$-2\sin(x) \lim_{h \to 0} \frac{\sin^2 \frac{h}{2}}{h}+ \cos(x)\lim_{h \to 0} \frac{\sin(h)}{h}$
wolfqz
$g(x) \leq f(x) \leq h(x)$\*
$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) =L$
then $\lim_{x \to a} f(x) = L$
wolfqz
$\cos^2x \leq \frac{\sin x}{x} \leq 1$
wolfqz
^ can be proved using sectors and stuff
alr
$\lim_{x \to 0} \cos^2 x = 1$
wolfqz
$\lim_{x \to 0} 1 = 1$
wolfqz
Yeah
okay
back here
$-2\sin(x) \lim_{h \to 0} \frac{\sin^2 \frac h2}{h} + \cos(x)$
$-\sin(x) \lim_{h \to 0} \frac{\sin \frac h2}{\frac h2} \times \sin(0) + \cos(x)$
wolfqz
$0+\cos(x) = \cos(x)$