derivitves

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Goody2shooes05

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Nov 3, 2005
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eeeeeekkkkkk!!!!!!! im going insane, i have to proove that the derv. of cosx=-sinx, cscx=-cscxcotx, and cotx=-csc^2x...using the limit and i go that you have to use
f(x+h)-f(x)/
h
thing so i plug it in then i dont know what to do with it...please help![/u]
 
How far have you gotten in applying trigonometric identities?

Eliz.
 
my TA said that we have to use the limit, so i just figured out how to do it for cos(x), and i used the product and quotent rused for the other 2, but i dont think i was supposed to...is there a way to do the other 2 with the limit?
 
Hello, Goody!

I'll walk through the first one . . .

Prove that: (cos⁡x)′ = −sin⁡x,  (csc⁡x)′ = −csc⁡x⋅cot⁡x,  (cot⁡x)′ = −csc⁡2x\displaystyle (\cos x)'\,=\,-\sin x,\;(\csc x)'\,=\,-\csc x\cdot\cot x,\;(\cot x)'\,=\,-\csc^2x using the limit.
Click to expand...
If we must use: .f′(x) = lim⁡h→0f(x+h)−f(x)h\displaystyle f'(x)\:=\:\lim_{h\to0}\frac{f(x+h) - f(x)}{h}, then we must be given these two theorems:

. . . [1]  lim⁡h→0 sin⁡(h)h = 1[2]  lim⁡h→0 1−cos⁡(h)h = 0\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 f(x) = cos⁡x\displaystyle f(x)\,=\,\cos x, then f′(x) = −sin⁡x\displaystyle f'(x)\,=\,-\sin x

We have: .f(x+h) − f(x) = cos⁡(x+h) − cos(x)\displaystyle f(x+h)\,-\,f(x) \:= \:\cos(x+h)\,-\,cos(x)

. . . . . . =  cos⁡(x)cos⁡(h) − sin⁡(x)sin⁡(h) − cos⁡(x)\displaystyle =\;\cos(x)\cos(h)\,-\,\sin(x)\sin(h)\,-\,\cos(x)

. . . . . . =  −sin⁡(x)sin⁡(h) − cos⁡(x)[1 − cos⁡(h)]\displaystyle =\;-\sin(x)\sin(h)\,-\,\cos(x)[1\,-\,\cos(h)]


Divide by h:    f(x+h) − f(x)h  =  −sin⁡(x)sin⁡(h) − cos⁡(x)[1 − cos⁡(h)]h\displaystyle h:\;\;\frac{f(x+h)\,-\,f(x)}{h}\;=\;\frac{-\sin(x)\sin(h)\,-\,\cos(x)[1\,-\,\cos(h)]}{h}

. . . . . . =  −sin⁡(x)⋅sin⁡(h)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: .lim⁡h→0[−sin⁡(x)⋅sin⁡(h)h]  −  lim⁡h→0[cos⁡(x)⋅1 − cos⁡(h)h]\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]

. . . . . . =  −sin⁡(x)⋅[lim⁡h→0sin⁡(h)h] − cos⁡(x)⋅[lim⁡h→01 − cos⁡(h)h]\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]

. . . . . . =  −sin⁡(x)⋅1  −  cos⁡(x)⋅0\displaystyle =\;-\sin(x)\cdot1 \;-\;\cos(x)\cdot0 . . . . from theorems [1] and [2]

. . . . . . =  −sin⁡(x)\displaystyle =\; -\sin(x)


Therefore: .ddx(cos⁡x)  =  −sin⁡x\displaystyle \frac{d}{dx}(\cos x)\;= \;-\sin x
 
Note to : I'm not sure why you think "using trig identities" means "don't use the limit formula". In fact, if you're going to work from the limit formula, you must use identities.

Please review the tutor's solution in detail, and note where trig identities were used.

Eliz.
 
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