S Skaskia Feb 2013 11 0 Nebraksa Mar 5, 2013 #1 (x-2)^{3}/x^{2 I know you have to use the quotient rule, however I'm getting lost within my own work :[ }

(x-2)^{3}/x^{2 I know you have to use the quotient rule, however I'm getting lost within my own work :[ }

C chiro MHF Helper Sep 2012 6,608 1,263 Australia Mar 5, 2013 #2 Hey Skaskia. Show us what you have tried.

S Skaskia Feb 2013 11 0 Nebraksa Mar 5, 2013 #3 For the first derivative I got: x^{2}(3x-2)^{2}-2x(x-2)^{3}/x^{4} Then after staring at it for a little bit: x^{2}(9x-4)-2x(x-8)/x^{4} 9x^{3}-2x^{2}-16x/x^{4} Then for the second derivative: x^{4}(27x^{2}-4x)-4x(9x^{3}-2x^{2}-16x)/x^{6} Simplified into: 27x^{6}-4x^{5}-36x^{4}-8x^{3}-72x^{2}/x^{6}

For the first derivative I got: x^{2}(3x-2)^{2}-2x(x-2)^{3}/x^{4} Then after staring at it for a little bit: x^{2}(9x-4)-2x(x-8)/x^{4} 9x^{3}-2x^{2}-16x/x^{4} Then for the second derivative: x^{4}(27x^{2}-4x)-4x(9x^{3}-2x^{2}-16x)/x^{6} Simplified into: 27x^{6}-4x^{5}-36x^{4}-8x^{3}-72x^{2}/x^{6}

C chiro MHF Helper Sep 2012 6,608 1,263 Australia Mar 5, 2013 #4 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

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

S Skaskia Feb 2013 11 0 Nebraksa Mar 5, 2013 #5 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?

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?

C chiro MHF Helper Sep 2012 6,608 1,263 Australia Mar 5, 2013 #6 That answer is the same as mine if you expand it out. You can use the quotient rule but I expanded it out using the Binomial Theorem because its easier to follow. Here is the theorem: Binomial theorem - Wikipedia, the free encyclopedia Reactions: 1 person

That answer is the same as mine if you expand it out. You can use the quotient rule but I expanded it out using the Binomial Theorem because its easier to follow. Here is the theorem: Binomial theorem - Wikipedia, the free encyclopedia