Across the Universe in a Yellow Submarine (deleted)....

[kindred_spiritz]

...Laynette, Hi, lol, hey...would you mind taking some time and showing me an entire process for this example, please. It's solved by the length and distance formula

Comments

you must be kidding HAHAHAHA!!

[minor_access]

I think you can handle this on your own WRITE SOME POETRY!!!
*GIVES YOU THREE HUGS*
are you really okay Amy - I am so sorry for what has happened - you know where I'm at but I AM NOT GOING FOR THIS!!!

Re: you must be kidding HAHAHAHA!!

[kindred_spiritz]

lol, I guess not...well I just wanted some online attention from you (if you hadn't noticed this is one of the 12 pre-cal questions I missed in HS)..Yes! I'm that bored....the complex numbers are constructed from ordered pairs of reals..therefore z = x +i*y is a representation for (x,y); the real part and imaginary part of x+i*y are x and y. The arithmetic operations on complex numbers are defined as corresponding operations on ordered pairs..knowing a comlex number came simply be vied as an ordered pair this way..it's possible to plot complex numbers as points on the Cartesian plane and so we talk of the magnitudeof a complex number z=x+i*y which is simply the lenght of the vector (x,y) equal to the Euclidian distance Sqrt(x^2+y^2). The circle x^2(y-1)^2=9 evidently is a cirlce centered about the point (0,1) whilch is in our other view a circle of radius 3 centered about the complexe number i. I suppose I've come along way since then...but how was I to know?

I might write some stuff tonight. I'll be okay......

(:

thought so!!!

[minor_access]

Claim: lim(h->0){f[g(x+h)]-f[g(x)]}/h = f '[g(x)]g(x)(*)

Proof: (*)iff lim(h->0){{f[g(x+h)]-f[g(x)]}/h - f '[g(x)]g '(x)}=0

lim(h->0){{f[g(x+h)]-f[g(x)]}/h-f'[g(x)]g'(x)}

=lim(h->0){{f[g(x+h)]-f[g(x)]}/h-f'[g(x)][g (x+h)-g(x)]}/h

lim(h->0){ f[g(x+h)]-f[g(x)]-f '[g(x)]g(x+h)+ f'[g(x)]g(x)]}/h

lim(h->0)f[g(x+h)]-lim(h->0)f[g(x)]-lim(h->0)f'[g(x)]lim(h->0) g
(x+h)+lim(h->0)f'[g(x)]g(x)}/lim(h->0)h

(f,g diff.=>f,g continuous)

={f[g(x)]-f[g(x)]- f'[g(x)]g(x)+ f '[g(x)]g(x)]}/lim(h->0)h

=0/lim(h->0)h

=0

is this proof of the chain rule right????

Re: thought so!!!

[kindred_spiritz]

im(h->0){ f[g(x+h)]-f[g(x)]-f '[g(x)]g(x+h)+ f'[g(x)]g(x)]}/h

lim(h->0)f[g(x+h)]-lim(h->0)f[g(x)]-lim(h->0)f'[g(x)]lim(h->0) g
(x+h)+lim(h->0)f'[g(x)]g(x)}/lim(h->0)h

That step was "formally" incorrect; first you must be precise why you can distrubute limits....secondly you can't just distrubute this way; the denominator would be 0. What would come next is totally false simply because the expression you've stated is non existant anyway

(:

::thank you for the hugs::