Hello, Goody!
I'll walk through the first one . . .
Prove that: \(\displaystyle (\cos x)'\,=\,-\sin x,\;(\csc x)'\,=\,-\csc x\cdot\cot x,\;(\cot x)'\,=\,-\csc^2x\) using the limit.
If we must use:
.\(\displaystyle f'(x)\:=\:\lim_{h\to0}\frac{f(x+h) - f(x)}{h}\), then we
must be given these two theorems:
. . . \(\displaystyle [1]\;\lim_{h\to0}\,\frac{\sin(h)}{h}\:=\:1\qquad\qquad\qquad\qquad\qquad[2]\;\lim_{h\to0}\,\frac{1 - \cos(h)}{h}\:=\:0\)
The first one: If \(\displaystyle f(x)\,=\,\cos x\), then \(\displaystyle f'(x)\,=\,-\sin x\)
We have:
.\(\displaystyle f(x+h)\,-\,f(x) \:= \:\cos(x+h)\,-\,cos(x)\)
. . . . . . \(\displaystyle =\;\cos(x)\cos(h)\,-\,\sin(x)\sin(h)\,-\,\cos(x)\)
. . . . . . \(\displaystyle =\;-\sin(x)\sin(h)\,-\,\cos(x)[1\,-\,\cos(h)]\)
Divide by \(\displaystyle h:\;\;\frac{f(x+h)\,-\,f(x)}{h}\;=\;\frac{-\sin(x)\sin(h)\,-\,\cos(x)[1\,-\,\cos(h)]}{h}\)
. . . . . . \(\displaystyle =\;-\sin(x)\cdot\frac{\sin(h)}{h}\,-\,\cos(x)\cdot\frac{1\,-\,\cos(h)}{h}\)
Take the limit:
.\(\displaystyle \lim_{h\to0}\left[-\sin(x)\cdot\frac{\sin(h)}{h}\right]\;-\;\lim_{h\to0}\left[\cos(x)\cdot\frac{1\,-\,\cos(h)}{h}\right]\)
. . . . . . \(\displaystyle =\;-\sin(x)\cdot\left[\lim_{h\to0}\frac{\sin(h)}{h}\right]\:-\:\cos(x)\cdot\left[\lim_{h\to0}\frac{1\,-\,\cos(h)}{h}\right]\)
. . . . . . \(\displaystyle =\;-\sin(x)\cdot1 \;-\;\cos(x)\cdot0\)
. . . . from theorems [1] and [2]
. . . . . . \(\displaystyle =\; -\sin(x)\)
Therefore:
.\(\displaystyle \frac{d}{dx}(\cos x)\;= \;-\sin x\)
