Finding the second derivative.

Feb 2013
11
0
Nebraksa
(x-2)3/x2 I know you have to use the quotient rule, however I'm getting lost within my own work :[
 

chiro

MHF Helper
Sep 2012
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Australia
Hey Skaskia.

Show us what you have tried.
 
Feb 2013
11
0
Nebraksa
For the first derivative I got:
x2(3x-2)2-2x(x-2)3/x4

Then after staring at it for a little bit:
x2(9x-4)-2x(x-8)/x4
9x3-2x2-16x/x4

Then for the second derivative:
x4(27x2-4x)-4x(9x3-2x2-16x)/x6

Simplified into:
27x6-4x5-36x4-8x3-72x2/x6
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
The first thing I will do is expand out the relationship to have:

(x^3 - 3*2*x^2 + 3*2^2*x - 2^3)/x^2 using the Binomial Theorem which gives

x - 6 + 12/x - 8/x^2.

Differentiating once gives us

1 - 0 - 12/x^2 +16/x^3.

Differentiating again gives us

24/x^3 - 48/x^4
 
Feb 2013
11
0
Nebraksa
The answer I'm told I am supposed to get is 24(x-2)/x^4
I don't understand how you got the two fractions.

However, going from what you did, I should expand the (x-2)^3 then differentiate using the quotient rule?