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#1 2014-01-27 00:35:01

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

logarithmic differentiation

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)]

thanks
simon:)

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

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#2 2014-01-27 02:35:58

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: logarithmic differentiation

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.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2014-01-27 06:34:18

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

Re: logarithmic differentiation

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

i dont understand how that square root comes about

cheers

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#4 2014-01-27 07:44:11

bob bundy
Registered: 2010-06-20
Posts: 8,337

Re: logarithmic differentiation

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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#5 2014-01-27 10:02:16

Nehushtan
Member
Registered: 2013-03-09
Posts: 914
Website

Re: logarithmic differentiation

@bob bundy

logarithmic differentiation

Now carry on from here

Last edited by Nehushtan (2014-01-27 10:56:26)

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#6 2014-01-27 23:48:30

bob bundy
Registered: 2010-06-20
Posts: 8,337

Re: logarithmic differentiation

hi Nehushtan

You may have a point.

Exercise for anyone with the will to try it:

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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#7 2014-01-28 00:35:39

Nehushtan
Member
Registered: 2013-03-09
Posts: 914
Website

Re: logarithmic differentiation

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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#8 2014-01-28 01:14:48

bob bundy
Registered: 2010-06-20
Posts: 8,337

Re: logarithmic differentiation

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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