Mathematics Corner :)

the answer is simply (1.8/1.5)*(2)
i think that will help!
the scale is (1.5/2) so by doing the: yourheigth/(scale)=answer
let me ask my question!
how can you calculate fast the derivative of a 100 multipled factors.something like this:

[(2x+1)(4x^2+7)(54x^4)(5x^3)....(9x+3)]'=?
is there any simple way to do it!
hope i have i said my will right!!
:confused:

format_quote Originally Posted by time_spender

the answer is simply (1.8/1.5)*(2)
i think that will help!
the scale is (1.5/2) so by doing the: yourheigth/(scale)=answer
let me ask my question!
how can you calculate fast the derivative of a 100 multipled factors.something like this:

[(2x+1)(4x^2+7)(54x^4)(5x^3)....(9x+3)]'=?
is there any simple way to do it!
hope i have i said my will right!!
:confused:

:confused: Could you copy it [the question] from the textbook (if you have it) please.

First you multiply out the first bracket:
(2x + 1)(4x^2 + 7) which if I'm right gives:
(8x^3 + 14x + 4x^2 + 7) which you then mutiply by the next bracket which in your question is (54x^4) which gives:
(432x^7 + 756x^5 + 216x^6 + 378x^4) which you multiply by the next bracket (5x^3) giving:
(2160x^10 + 3780x^8 + 1080X^9 + 1890x^7)
And now multiply all of this to the final bracket (9x + 3):
(19440x^11 + 34020x^9 + 9720x^10 + 17010x^8 + 6480x^10 + 11340x^8 + 3240x^9 + 5670x^7)
Now lastly collect all like terms:
= 19440x^11 + 16200x^10 + 37260x^9 + 28350x^8 + 5670x^7
......and there you have it. The final answer. Phew that took patience. I sometimes wonder about the people who write these questions. They've probably never been kids!
( I hope its right otherwise I think I'll quit doing math.)

Aww thank you Savdah
Probably you have spent time for it!
But I meant the derivation of the whole braket
Sorry….there is no textbook to paste it from there.
(My english is not good..thats why i fear to understand myself to you!)
Let me write it again, I mean( y’)

-----d/dx of ([(2x+1)(4x^2+7)(54x^4)(5x^3)....(9x+3)]) not the result of the bracket

I found something interesting for it
It goes like this:

Y=([(2x+1)(4x^2+7)(54x^4)(5x^3)....(9x+3)])
Now you use the natural logarithm
Lny=ln([(2x+1)(4x^2+7)(54x^4)(5x^3)....(9x+3)]) so it will be more simple:

Lny=ln(2x+1)+ln(4x^2+7)+ln(54x^4)+ln(5x^3)…+ln(9 x+3)

Now you can get the derivation easily:

d/dx====> y’/y=(2/2x+1)+(8x/4x^2+7)+(216x^3/54x^4)+(15x^2/5x^3)+….(9/9x+3)

now y’ obviously equals to:

y’ = y*[(2/2x+1)+(8x/4x^2+7)+(216x^3/54x^4)+(15x^2/5x^3)+….(9/9x+3)]
which y=([(2x+1)(4x^2+7)(54x^4)(5x^3)....(9x+3)]) (our first whole bracket)

that’s the best way I have found….any other solutions?

I sometimes wonder about the people who write these questions. They've probably never been kids!

Haha…they have been kids for a long time sister…once they played with + and -!!!
Let me tell you something....i,m in the 3rd term of civil engineering and I read this solution one year ago.
I just wanted to know that if there is a better solution I use it…that’s all!
wasalam

heheh maths my flavourite

I really like algebra but some of the parts i really dont get it:confused:

thats a really good thread sister :sister:, if i need any help in algebra then i'll ask you

i think that the first question is rite!

I didn't quite understand the questions you provided. You gave the equations yet what are we supposed to be finding?



P.S Threads merged.