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(a) work the problem as written applying the quotient rule to each fraction,    
(b) rewrite f(x) using negative exponents and then find the derivative,  
(c) rewrite f(x) as a single fraction and then find the derivative, and  
(d) verify that the answers for (a) - (c) are equivalent. 

f(x)=10/x^4 + 3/x^2


 

 Feb 9, 2016

Best Answer 

 #1
avatar+129852 
+5

f(x)=10/x^4 + 3/x^2

 

1. quotient rule   ........     y = f(x)/g(x)    → y ' =  [ f'(x)g(x)  - g'(x)f(x)] / [ g(x)]^2

 

   [0*x^4 - 4x^3 *10] / [x^8]    +  [0*x^2 - 2x *3]/ [ x^4]   =

 

[ -40x^3] / [x^8]   + [ -6x]/ [x^4]   =

 

-40 / x^5   - 6 / x^3

 

2.   f(x)  = 10x^-4  + 3x^-2   ....use the power rule.....

 

f ' (x)  =  -40x^-5   - 6x^-3    =   -40/x^5 - 6/x^3

 

3. as a single fraction  =   f(x)  =  [10 + 3x^2] / x^4

 

f ' (x)  =   ( 6x * x^4  -  4x^3 [10 + 3x^2] ) / x^8  =

 

(6x^5 - 40x^3 - 12x^5) /x^8  =

 

[-6x^5 - 40x^3] / x^8  =

 

[-6x^5]/ x^8   -   [40x^3] / x^8  =

 

-6/x^3  - 40/x^5  =       -40/x^5 - 6/x^3

 

4. All are the same

 

Note, Yura_chan......I rarely [if ever] use the quotient rule......much each easier to write the expression with negative exponents and use the power rule.....IMHO......!!!!

 

 

 

cool cool cool

 Feb 9, 2016
edited by CPhill  Feb 10, 2016
 #1
avatar+129852 
+5
Best Answer

f(x)=10/x^4 + 3/x^2

 

1. quotient rule   ........     y = f(x)/g(x)    → y ' =  [ f'(x)g(x)  - g'(x)f(x)] / [ g(x)]^2

 

   [0*x^4 - 4x^3 *10] / [x^8]    +  [0*x^2 - 2x *3]/ [ x^4]   =

 

[ -40x^3] / [x^8]   + [ -6x]/ [x^4]   =

 

-40 / x^5   - 6 / x^3

 

2.   f(x)  = 10x^-4  + 3x^-2   ....use the power rule.....

 

f ' (x)  =  -40x^-5   - 6x^-3    =   -40/x^5 - 6/x^3

 

3. as a single fraction  =   f(x)  =  [10 + 3x^2] / x^4

 

f ' (x)  =   ( 6x * x^4  -  4x^3 [10 + 3x^2] ) / x^8  =

 

(6x^5 - 40x^3 - 12x^5) /x^8  =

 

[-6x^5 - 40x^3] / x^8  =

 

[-6x^5]/ x^8   -   [40x^3] / x^8  =

 

-6/x^3  - 40/x^5  =       -40/x^5 - 6/x^3

 

4. All are the same

 

Note, Yura_chan......I rarely [if ever] use the quotient rule......much each easier to write the expression with negative exponents and use the power rule.....IMHO......!!!!

 

 

 

cool cool cool

CPhill Feb 9, 2016
edited by CPhill  Feb 10, 2016
 #2
avatar+209 
+5

Thank you so much! I believe on our test it'll be specified that we need to use the quotient method, just to check if we know it..but if it isn't I will use that trick! Thanks! I like the power rule better too. 

 Feb 10, 2016

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