West Coast Connection Forum
Lifestyle => Train of Thought => Topic started by: coola on April 28, 2006, 07:07:50 AM
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i need help in a differentiation question... if you could show working, that'd be superb... cheers.
1.) (2y - x)^4 + x^2 = y + 3
2.) 2(x^2 + 1)^3 + (y^2 + 1)^2 = 17
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wat does the whole q say
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Damn you're gonna have to use implicit differentiation for this. I would help out but it's been years since I did this, and I would need the weekend to school up.
The only hint I can give you is that both answers will be expressed in terms of x and y, not just x as regular differentiation usually gives you.
Good luck.
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this is the question: find dy/dx by differentiating implicitly. when applicable, express the result in terms of x and y ...
i just goto find dy/dx dont need to solve for x and y on this one...
dammet ! the problem is i never did calculus in high school, i'm learning all this shit myself from the text book... so i have a few gaps in fundamental rules of calculus.. >:(
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1st one
4(2y-x)^3 - 2x / 8(2y-x)^3 - 1 = dy/dx
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i need help in a differentiation question... if you could show working, that'd be superb... cheers.
1.) (2y - x)^4 + x^2 = y + 3
2.) 2(x^2 + 1)^3 + (y^2 + 1)^2 = 17
1. (1 - 8(2y -x)^3)^-1 * (2x - (4(2y - x)^3)) = dy/dx
2. 12x(x^2 + 1)^2 / (-4y(y^2 + 1)) = dy/dx
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Damn, questions like these were the reason why i stuck to business maths at school
What field is this applicable to, overall ?
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steps
1.) (2y - x)^4 + x^2 = y + 3
d((2y - x)^4) / dx = 4(2y - x)^3((d((2y - x))/dx = 4(2y - x)^3(2y * dy/dx - 1)
8y(2y - x)^3dy/dx - 4(2y - x)^3 + 2x = dy/dx
- 4(2y - x)^3 + 2x = dy/dx - 8y(2y - x)^3dy/dx
2x - 4(2y - x)^3 = dy/dx - 8y(2y - x)^3dy/dx
2x - 4(2y - x)^3 = dy/dx(1 - 8y(2y - x)^3)
(2x - 4(2y - x)^3) / (1 - 8y(2y - x)^3) = dy/dx
2.) 2(x^2 + 1)^3 + (y^2 + 1)^2 = 17
12x(x^2 + 1)^2 + 4y(y^2 + 1)dy/dx = 0
12x(x^2 + 1)^2 = -4y(y^2 + 1)dy/dx
12x(x^2 + 1)^2 / -4y(y^2 + 1) = dy/dx
3x(x^2 + 1)^2 / -y(y^2 + 1) = dy/dx
it looks like my answer agrees with za scarab's, so that is good.
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thanks for that nibs and scarab.. i'm just going over the workings now, you got alot closer than i did...
the solution is:
[4(2y-x)^3 - 2x] / [8(2y-x)^3 - 1]
and nibs got the second solution right... props on that.
the one i'm finding real tricky is this one too:
(3x^2) / ((y^2)+1) + y = 3x + 1
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i get the second one !!! ;D
the only area i fucked up is -- 12x(x^2 + 1)^2 = -4y(y^2 + 1)dy/dx -- i dont know why, but i fucked up here.. stupid mistake dammet. instead of isolating the dy/dx, i multiplied one of the brackets out, thats what i mean i have gaps even in algebra and BIMDAS...
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is that for an assingment or a final. shit i got calculus for engineering final on the 4th and am getting this shit wrong
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is that for an assingment or a final. shit i got calculus for engineering final on the 4th and am getting this shit wrong
your answer wasn't "wrong", you simply omitted the parenthesis to indicate the correct terms.
your answer:
(4(2y-x)^3 - 2x) / (8(2y-x)^3 - 1) = dy/dx
i had a -1 / -1 factor in mine, so it looked different but also the same
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cool i didn't check it
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thanks for that nibs and scarab..
Ask my Muslim brothers nibs and scarab (Kassem Fawzy), they'll take care of you. The world has the Muslims to thank for Algebra (invented by Al-Jabeer), Arabic numerals, and the number 0.
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is that for an assingment or a final. shit i got calculus for engineering final on the 4th and am getting this shit wrong
nah man, these are just exercises we have to do each week.. yeah i'm getting alot of it wrong too ;D
what studies have you done nibs ?
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thanks for that nibs and scarab..
Ask my Muslim brothers nibs and scarab (Kassem Fawzy), they'll take care of you. The world has the Muslims to thank for Algebra (invented by Al-Jabeer), Arabic numerals, and the number 0.
yep i appreciate their help... and we have alot of cultures to thank for what the world is today...
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???
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try this one !!
(3x^2) / ((y^2)+1) + y = 3x + 1
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try this one !!
(3x^2) / ((y^2)+1) + y = 3x + 1
3x^2((y^2)+1)^-1 + y = 3x + 1
6x((y^2)+1)^-1 - 3x^2((y^2)+1)^-2(2ydy/dx) + dy/dx = 3
6x((y^2)+1)^-1 - 6x^2(ydy/dx)((y^2)+1)^-2 + dy/dx = 3
6x((y^2)+1)^-1 - (6yx^2((y^2)+1)^-2 + 1)dy/dx = 3
6x((y^2)+1)^-1 - 3 = (6yx^2((y^2)+1)^-2 + 1)dy/dx
(6x / ((y^2) + 1) - 3) / ((6yx^2/((y^2)+1)^2) + 1) = dy/dx
it's been re-corrected. dy/dx didn't have the '3' factor to factor out.
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!
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Calculus II will kill you if you having problems with this... ;D
It's scary how I forgot all this shit :-\
http://www.mathematicshelpcentral.com/lecture_notes/calculus_1.htm
btw fuck geometry
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yeah nibs, i was wondering wtf you changed the 3 to a 1 before... ima go eat some soup then put down on paper your workings... the solution is different in the text :-\
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Calculus II will kill you if you having problems with this... ;D
It's scary how I forgot all this shit :-\
http://www.mathematicshelpcentral.com/lecture_notes/calculus_1.htm
btw fuck geometry
yeah this is only my first year... its basic stuff and i'm struggling >:( might need a tutor next year.. hot blonde will do ;D
props for the web-site !
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heres the solution... i'm hittin the sack, too much for one night..
[ 3((y^2) + 1)((y^2) - 2x + 1) ] / [ ((y^2) + 1)^2 - 6(x^2)y ]
fuck knows hot to get that :'(
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i factored wrong:
3x^2((y^2)+1)^-1 + y = 3x + 1
6x((y^2)+1)^-1 - 3x^2((y^2)+1)^-2(2ydy/dx) + dy/dx = 3
6x((y^2)+1)^-1 - 6x^2(ydy/dx)((y^2)+1)^-2 + dy/dx = 3
6x((y^2)+1)^-1 + (1 - 6yx^2((y^2)+1)^-2)dy/dx = 3
6x((y^2)+1)^-1 - 3 = (6yx^2((y^2)+1)^-2 - 1)dy/dx
(6x / ((y^2) + 1) - 3) / (6yx^2((y^2)+1)^-2 - 1) = dy/dx
then they simplified this further:
(6x / ((y^2) + 1) - 3) / (6yx^2((y^2)+1)^-2 - 1) = dy/dx
(6x((y^2)+1)^2 / ((y^2) + 1) - 3((y^2)+1)^2) / ( 6yx^2 - ((y^2)+1)^2) = dy/dx
(6x((y^2)+1) - 3((y^2)+1)^2) / ( 6yx^2 - ((y^2)+1)^2) = dy/dx
(6x((y^2)+1) - 3((y^2)+1)((y^2)+1)) / ( 6yx^2 - ((y^2)+1)^2) = dy/dx
[ 3((y^2)+1)(2x - (y^2) + 1) ] / [ 6yx^2 - ((y^2)+1)^2 ] = dy/dx
factoring out the minus sign
[ 3((y^2)+1)((y^2) - 2x + 1) ] / [ ((y^2)+1)^2 - 6y(x^2) ] = dy/dx
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thanks for that nibs and scarab..
Ask my Muslim brothers nibs and scarab (Kassem Fawzy), they'll take care of you. The world has the Muslims to thank for Algebra (invented by Al-Jabeer), Arabic numerals, and the number 0.
well fuck him then cause i hate algebra
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thanks for that nibs and scarab..
Ask my Muslim brothers nibs and scarab (Kassem Fawzy), they'll take care of you. The world has the Muslims to thank for Algebra (invented by Al-Jabeer), Arabic numerals, and the number 0.
well fuck him then cause i hate algebra
He was just joking around. "Al-Jabeer", "Al-ge-bra". While the name algebra is devired from the arabic word "al-ğabr" or "al jebr", which I'm told stands for "reunion of broken parts", the idea of agebra as we know it goes as far back as the Babylonian times at least 4000 years ago. 2400 years before anyone was calling themselves a Muslim. Now Calculus as we know it today came about from Newton and his colleagues in Europe about 300 years ago.
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I just failed my calculus test today. ;D
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I can count to 12