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find  `dy/dx|_(x=1)`  when `y= (1+x+...x^15+x^16)/x^8`  

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user7230927 | Salutatorian

Posted June 25, 2013 at 11:10 PM via web

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find  `dy/dx|_(x=1)`  when `y= (1+x+...x^15+x^16)/x^8`

 

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted June 25, 2013 at 11:54 PM (Answer #1)

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`y = (1+x+x^2+......x^15+x^16)/x^8`

 

`x^8y = 1+x+x^2+......x^15+x^16`

`x^8xxy'+yxx8x^7 = 1+2x+3x^2+.......15x^14+16x^15`

 

`y' = [(1+2x+3x^2+.......15x^14+16x^15)-yxx8x^7]/x^8`

 

`y_(x=1) = (1+(1+1^2+.......+1^15+1^16))/1^8 = (1+16)/1 = 17`

 

`y'_(x = 1) = [(1+2+3+.......15+16)-17xx8]/1`

 

1+2+3+.....15+16 is a arithmaic series with initial  term a = 1,  common difference d = 1 and last term l = 16.

So the sum of this series is `S_n = n/2(a+l)`


`1+2+3+.....15+16 = 16/2(1+16) = 8xx17`

 

`y'_(x = 1) = [8xx17-17xx8]/1 = 0`

 

So the answer is;

 `|(dy)/dx|_(x=1) = 0`

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