macOS Math Problem

[Quboid]

I know programming is math and logic ( and creativity), and i was just wondering if anybody here could help me derive a caculus question that got me pulling my hair out.
x(1-x)^2

thats: x multiplied by 1-x in bracket squared. I am expected to derive this and find the values for x in teh first derivative, i.e, equate it to zero and solve for x.

thanks guys

 

[exabytes18]

Are you looking for a programming solution or a calculus solution?

 

[n-abounds]

Wait, is it:

(x(1-x))^2

or

x((1-x)^2))

?

I think either interpretation is quite easy. Just simplify either expression, then use the power rule. Maybe you don't know the power rule... Anyways, after using the power rule you would set that equal to zero and solve for x.

 

[nordesmic]

I know programming is math and logic ( and creativity), and i was just wondering if anybody here could help me derive a caculus question that got me pulling my hair out.
x(1-x)^2

thats: x multiplied by 1-x in bracket squared. I am expected to derive this and find the values for x in teh first derivative, i.e, equate it to zero and solve for x.

thanks guys

If I understand you correctly, and I can still remember my chain rule and product rule, the answer would be..

x*2(1-x)*-1 + 1*(1-x)^2 =0

that is use the chain rule on the bit with the brackets and then apply the product rule.

Then the answer would be 2x^2+2x+1-2x+x^2=0
therefore 3x^2-4x+1=0
therefore x=.333 or 1

I hope that is right.

 

[n-abounds]

Hmm I get either:

1 - 4x + 3x^2 = 0

or

2x - 6x^2 + 4x^3 = 0

depending on how that exponent is interpreted. The answer should be the same whether simplified first, or whether the product and chain rules are used first...

Nordesmic, you did it right except in the first line where you have "1*2(1-x)*-1 + 1*(1-x)^2 =0" You shouldn't have taken the derivative of x on the left-hand side of the expression...so you should have had "x*2(1-x)*-1 + 1*(1-x)^2 =0" But this mistake throws off the answer you gave...

 

[nordesmic]

Hmm I get either:

1 - 4x + 3x^2 = 0

Nordesmic, you did it right except in the first line where you have "1*2(1-x)*-1 + 1*(1-x)^2 =0" You shouldn't have taken the derivative of x on the left-hand side of the expression...so you should have had "x*2(1-x)*-1 + 1*(1-x)^2 =0" But this mistake throws off the answer you gave...

Yep I fixed it now, should be perfect.

 

[Mord]

you just use the product and chain rule.

 

[TBi]

The exponent always takes precedence over a multiplication.

 

[dmcxii]

more description.....

finding derivative of x(1-x)^2

first get rid of parenthesis
x(1-x)(1-x)
multiply it all out
(x-x^2)(1-x)
multiply it out again
x-x^2-x^2+x^3
combine like terms
x-2x^2+x^3
take derivative.....
1-4x+3x^2
put in correct form
3x^2-4x+1
factor
(3x-1)(x-1) and solve for zeroes..... x=1, 1/3

 

[Eraserhead]

Solve x(1-x)^2

dmcxii is completely correct

A solution as an image from Pages+LaTeX (which looks pretty)

 

[AJ Muni]

Reminds me how much I HATED calculus.

 

[dmcxii]

thanks for cleaning it up....

I knew someone would clean up my typing.

thanks

 

[Quboid]

finding derivative of x(1-x)^2

first get rid of parenthesis
x(1-x)(1-x)
multiply it all out
(x-x^2)(1-x)
multiply it out again
x-x^2-x^2+x^3
combine like terms
x-2x^2+x^3
take derivative.....
1-4x+3x^2
put in correct form
3x^2-4x+1
factor
(3x-1)(x-1) and solve for zeroes..... x=1, 1/3

Wish it was that easy guys. Thanks for all the help though. I made a mistake with the question. I just got from school and realise that the power that i gave the expression was wrong. It was the square root of everything in the bracket. That would be:
x(1-x)^1/2.
taking into consideration that everything raised to the power of one half is the root. Deriving using the chainerule would be perfect, its the simplifing and finding "x" values that got me. Any help??

 

[TequilaBoobs]

youre violating the ethics code at your school - do your own homework.

 

[Eraserhead]

Wish it was that easy guys. Thanks for all the help though. I made a mistake with the question. I just got from school and realise that the power that i gave the expression was wrong. It was the square root of everything in the bracket. That would be:
x(1-x)^1/2.
taking into consideration that everything raised to the power of one half is the root. Deriving using the chainerule would be perfect, its the simplifing and finding "x" values that got me. Any help??

Follow it through in the same way, except replace the 2 with 1/2, should be pretty easy, especially if you follow through my method, think of it as an example .

 

[Quboid]

Follow it through in the same way, except replace the 2 with 1/2, should be pretty easy, especially if you follow through my method, think of it as an example .

I have alraedy deferenciated it, its just the the simplification and solving for x. You see, when we deferenciate, we would be left with -1/2 as one of the powers, forcing us into a fraction and that's where things get messy.

This is not homework, its practice for upcomming finals. Its one of the only sums i cant seem to simplyfy, or better yet, i am not sure of my answer.

 

[Eraserhead]

I have alraedy deferenciated it, its just the the simplification and solving for x. You see, when we deferenciate, we would be left with -1/2 as one of the powers, forcing us into a fraction and that's where things get messy.

This is not homework, its practice for upcomming finals. Its one of the only sums i cant seem to simplyfy, or better yet, i am not sure of my answer.

Well try and get one side of the equation (assuming the other side is 0) to only have a single denominator (number on the bottom), but multiplying the top of any other parts by the bottom, then as it equals 0 you can just cancel the denominator away.

 

[Quboid]

Well try and get one side of the equation (assuming the other side is 0) to only have a single denominator (number on the bottom), but multiplying the top of any other parts by the bottom, then as it equals 0 you can just cancel the denominator away.

thats what i did.And wasn't sure of. thanks alot man.

 

[xov]

2/3 ?

 

[wms121]

...just don't attempt to perform any ol' integral

...some solutions of:

dy^2/dx^2 = x(x-1)^2 give the "elliptic logarithm"..see:

http://mathworld.wolfram.com/EllipticLogarithm.html

...they have a "general purpose D.E. solver" on that site too.

WW

 

[Am3822]

A question to the OP -- what/where are you studying (hs/college)?

If this forum is about to become a math questions' forum, maybe someone should install the TeX module