[maths] Re: calculus; more derivatives
- From: "Ned Granic" <ngranic@xxxxxxx>
- To: <maths@xxxxxxxxxxxxx>
- Date: Tue, 7 Nov 2006 17:00:30 -0700
Thanks Nelson, yes it is -3 for question 2; I did it again earlier today.
As for the plague of the first question, I get stuck when I get derivative of
sqrt(x), which is
1/2x^-1/2, read as:
one half x to the power of negative one half power.
Also, the derivatives of sqrt(x) and sqrt(x)-1 are the same. Right?
What do I get after multiplying:
sqrt(x) * 1/2x^-1/2
Back to question 3, the derivative of:
-3/(x^2-2x+1)
I get is:
(-2(x+1))/3. Correcto?
Cheers!
Ned
----- Original Message -----
From: Nelson Blachman
To: maths@xxxxxxxxxxxxx
Sent: Tuesday, November 07, 2006 4:12 PM
Subject: [maths] Re: calculus; more derivatives
Ned,
The answer to your first question is: No. Please show your work to
indicate how you've misused the hints I gave you.
The answer you give for the second question llooks almost right; I get -3
where you got 3.
The answer to your 3rd question is that the derivative with respect to x of
3/z is
-3(dz/dx)/z^2.
HTH
Nelson
----- Original Message -----
From: Ned Granic
To: maths@xxxxxxxxxxxxx
Sent: Tuesday, November 07, 2006 12:17 PM
Subject: [maths] calculus; more derivatives
SO, here I go again:
Is 2/x the derivative of [(sqrt(x)-1)/sqrt(x)]+1?
And:
Is 3/((x-1)^2) the derivative of:
(2x+1)/(x-1)?
And:
How to find the derivative of:
3/(x^2 - 2x + 1)?
Many thanks in advance!
Ned
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