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:
 
:sl:

time_spender said:
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:
Click to expand...

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

:w:
 
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(9x+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!
Click to expand...

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:)
 
:sl:

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:)
:peace:

:w:
 
Maths - need help!!!

:sl:
right here is wat i did 4 the first question n i dunno how to so som of de other questions. i hope som1 will elp me understand...

x2 = x squared

x-2/ x2-4

i did it as...

x-2/(x+2)(x-2) = 1/(x+2).

is it rite?

now the nxt 1 i need help...

remember x2 = x squared.

x2-49/2x+5 (division sign) 4x-28/4 x2-25

plz help...som1 *pleading*
 
Re: Maths - need help!!!

ok dat 1 mi8 be rite, but how bout the 2nd 1? :? :?
dats da 1 i need help on. :rollseyes
 
:sl:

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

:w:

P.S Threads merged.
 
Muhammad said:
:sl:

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

:w:

P.S Threads merged.
Click to expand...

oops sowweee its simplification. it struck me yeaterdy wen i went to bed at nite dat i didnt day wat u r s'pposd to b findin sowwee bout dat :statisfie
 
aZn_pLayGurL said:
i Like linear equations in maths :p they soo easy even A Level lineas ;P
Click to expand...

yeah yeah :rollseyes now som 1s gotta elp me wid dis or i get a detention on monday. its a home work ya no. i gotta get it done other wise

:hiding:

 

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