logarithmic differentiation

ninjaman · 2014-01-27 00:35:01

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)

bobbym · 2014-01-27 02:35:58

Hi;

Is this what you had to start?

ninjaman · 2014-01-27 06:34:18

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

i dont understand how that square root comes about

cheers

Bob · 2014-01-27 07:44:11

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 (2014-01-27 09:45:19)

Nehushtan · 2014-01-27 10:02:16

@bob bundy

Thread title wrote:

logarithmic differentiation

Now carry on from here

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

Bob · 2014-01-27 23:48:30

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

Nehushtan · 2014-01-28 00:35:39

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)

Bob · 2014-01-28 01:14:48

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

Math Is Fun Forum

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

logarithmic differentiation

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

Re: logarithmic differentiation

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

Re: logarithmic differentiation

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

Re: logarithmic differentiation

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

Re: logarithmic differentiation

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

Re: logarithmic differentiation

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

Re: logarithmic differentiation

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

Re: logarithmic differentiation

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