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avatar+117 

F(x) = \(f'(x)= {x^5-5x^4+x \over x^3}\)

I know I have to use u'v*uv' / v^2 

In this case my u =  x^5-5x^4+x              v= x^3

                         u' =      5x^4-20x^3 +1           v'= 3x^2 

 

but I don't seem to get the correct answer. Could anyone please help me understand how to get to the correct answer which is 2x-5 -2/x^3

 Dec 2, 2015

Best Answer 

 #1
avatar+128475 
+10

I don't particularly like using the quotient rule, but here goes

 

[ (5x^4-20x^3 +1)(x^3) - ( x^5-5x^4+x )(3x^2)] /  x^6

 

[ 5x^7 - 20x^6 + x^3  - 3x^7 + 15x^6 - 3x^3]  / x^6  

 

[2x^7 - 5x^6 -2x^3] / x^6  =

 

[2x^7]/ x^6   -  [5x^6]/x^6   -  [2x^3]/ x^6  

 

2x   -  5     -  2 / x^3

 

 

And there you are.....!!!!

 

 

cool cool cool

 Dec 2, 2015
 #1
avatar+128475 
+10
Best Answer

I don't particularly like using the quotient rule, but here goes

 

[ (5x^4-20x^3 +1)(x^3) - ( x^5-5x^4+x )(3x^2)] /  x^6

 

[ 5x^7 - 20x^6 + x^3  - 3x^7 + 15x^6 - 3x^3]  / x^6  

 

[2x^7 - 5x^6 -2x^3] / x^6  =

 

[2x^7]/ x^6   -  [5x^6]/x^6   -  [2x^3]/ x^6  

 

2x   -  5     -  2 / x^3

 

 

And there you are.....!!!!

 

 

cool cool cool

CPhill Dec 2, 2015
 #2
avatar+117 
0

Hi CPhill! 

Thank you so much for answering my question. However, I just have one question. Why do you divide by x^6? 

 Dec 2, 2015
 #3
avatar+128475 
+10

Remember....in the quotient rule, the denominator is the "v" term squared.......so

 

v = x^3

 

And

 

v^2   = [x^3]^2   = x^6

 

 

 

cool cool cool

 Dec 2, 2015
edited by CPhill  Dec 2, 2015

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