Recent content by ARYT

Thanks for the reply. Fusion is less likely to be controlled at this time. Actually, we really don't know the consequences of using high-energy particle beams in combustion. It could be fission, fusion or totally something different. Now during the process of explosion, we have chemical...

I suppose, I couldn't explain myself that well. By particle accelerator, I don't mean a huge one which creates antimatter and other incontrollable particles. Let’s imagine a huge CRT. As far as I know, the direct effect of particles beam on an explosion (or combustion) has never been...

I'm currently working on a new paper on how to increase the efficiency of space rockets (both in performance speed and power). Getting the idea from a simple CRT (Cathode ray tube) and how the beams make an object move in the tube; I came up with this idea to use the same thing in a typical...

And the second one which compare to the answer given by the Microsoft Math is wrong:

Although it's too long. I won't be able to do it without a software (for multiplication and things like that). Also, We can't use calculator.

OK. Here I've tried to solve the first one.

(fog(x) )'=g' (x) f' (g(x) ) we have this general rule for two functions only. Give me sth for n functions.

I know, I could solve the third one myself, but the first one :(

Homework Statement Homework Equations Answers by Microsoft Math 3.0 The Attempt at a Solution This is confusing. Too many parentheses. We used to solve a composite of two or three functions.

No, It's OK. Thanks for the tips. :D A better dramatic end: lol Although they're killing my innovation here. For one assignment (C++ programming), I've used only one line to compile a rather complicated program without sth useless (i.e. End of File function: eof.). And the tutor told me...

I think I've found the problem. You've cited the equation wrongly. It should be: [f(x+h)-f(x)]/h while h tends to 0. According to equation 1, we have: [f(x)+f(h)+x^2 h + xh^2 - f(x)] / h => lim f(h)/h + lim (x^2+xh) while h maps to zero. So f'(x)=1+x^2 and accordign to this we...

We're under heavy fire C.O., Clarification is needed. :D lol Isn't it a bit stupid to say: We have: f(0) = 0 and Equation 1: we have lim [f(0+h)-f(h)]/h while h maps to 0. Thus (h tends to 0): lim f(h)/h - lim f(h)/h = 1-1 = 0

f'(x)= [f(x+h) - f(h)]/h while h tends to zero. so f'(0)=[f(0+0)-f(0)]/0 f'(0)= 0/0 ?

so f'(0) is also zero? and f'(x)=1? I think I got it. Thanks

well, if we go for 0 and 0, we will have: f(0) = f(0)+f(0) that is f(0)=2f(0) and 1=2 ! Great!