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**ninjaman****Member**- Registered: 2013-10-15
- Posts: 55

hello,

i have this, y = (x+1)(2x+1)^3/(x-3)^1/2

i got

dy/dx = (x+1)(2x+1)^3/(x-3)^1/2 [1/x+1 + 3/2x+1 - 1/2(x-3)]

im not sure about this?

thanks

simon:)

*Last edited by ninjaman (2014-01-27 00:35:46)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,290

Hi;

Is this what you had to start?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**ninjaman****Member**- Registered: 2013-10-15
- Posts: 55

no the bottom bit is (x-3)^1/2

i dont understand how that square root comes about

cheers

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,651

hi Simon

Those two expressions are the same.

because

Now back to the question:

This expression has 'everything' ; a product, a quotient, and a function of a function so I'll split up the problem a little.

Let's call that y = P/Q where P and Q are the above expressions that make up the fraction.

By the quotient rule:

So lets work out dP/dx

By the product rule

Substituting this in the above:

factorising

That was very tricky and may contain one or more errors. I will take a break and then come back and check it (if Stefy doesn't beat me to it of course )

Checked at 21.09 gmt and one error found and corrected.

Bob

*Last edited by bob bundy (2014-01-27 09:45:19)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,651

hi Nehushtan

You may have a point.

Exercise for anyone with the will to try it:

Show that method leads also to my answer.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Well, I think the OP made a slight mistake.

ninjaman wrote:

dy/dx = (x+1)(2x+1)^3/(x-3)^1/2 [1/x+1 + 3/2x+1 - 1/2(x-3)]

I think it should be

dy/dx = (x+1)(2x+1)^3/(x-3)^1/2 [1/x+1 +

6/2x+1 - 1/2(x-3)]

*Last edited by Nehushtan (2014-01-28 00:36:07)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,651

hi Nehushtan

Thanks. You have answered his original question and cleared up for me what that first post said. If it's not been 'Latexed' I find it very hard to follow, especially when the equation is a long one. I just ignored that and worked from bobbym's post 2.

OK, I've learned my lesson: "Read the question"; but I have also paid the price: I slogged through it the other way. Made my brain hurt.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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