power rule
f ' (x) = 10 x -11
f ' (5) = 10 * 5-11=39
2007-09-26 09:16:53
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answer #1
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answered by Dokta T 2
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Remember that:
f'(x^n) = n(x)^(n-1)
So, f'(x^2) = 2x and f'(x) = 1.
Also, the derivative of a constant is 0.
So, f'(x) = 5f'(x^2) - 11f'(x) - f'(1) = 10x - 11
And, f'(5) = 10*5 - 11 = 39
2007-09-26 16:21:18
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answer #2
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answered by pyaarmusafir 2
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once you've done the first part the second is simple.
use the power rules on the first problem to find f'(x): if f(x)=x^y then f'(x) = y*(x^(y-1)), and recall that the derivative of a constant is just zero.
f'(x) = 5*2*[x^(2-1)]-11*1*[x^(1-1)]+0
f'(x)=10x-11
now, to find f'(5), just plug in 5 for x
f'(x)=10x-11
f'(5)=10(5)-11
f'(5)=39
2007-09-26 16:19:44
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answer #3
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answered by grompfet 5
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1.) take the power, and multiply the coeficcient by it. then reduce the power on the x factor by 1.
If there is no x, delete that term.
2*5x -11 -0
10x-11=f'(x)
2.) put 5 instead of x. solve.
2007-09-26 16:18:05
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answer #4
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answered by Darkwolf 5
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39
2007-09-26 16:21:35
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answer #5
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answered by roadrunner426440 6
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Take the derivative of each term and get
f'(x) = 10x -11
Then plug in 5 for x to get
f'(5) = 10*5 - 11 = 39
Doug
2007-09-26 16:18:31
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answer #6
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answered by doug_donaghue 7
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f(x) = 5x^2-11x-1
f'(x) = 10x-11 (basic derivative)
f'(5) = 10(5)-11
f'(5) = 39
2007-09-26 16:17:52
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answer #7
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answered by jwbyrdman 4
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