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4 answers

f(x)=(82.8)/(1+21.8(7.14)^-x)

Use: dy/dx = Vδu/δx - Uδv/δx] / V²
Let u = 82∙8 and v = 1 + 21∙8(7∙14)^-x
dy/dx = (1 + 21∙8(7∙14)^-x)(0) - (82∙8)(0 - x(21∙8(7∙14)^-x^0 /[1 + 21∙8(7∙14)^-x]²
dy/dx = 0 - 0 + 1805∙04(7∙14)^-1 /[1 + 21∙8(7∙14)^-x]²
dy/dx = 436 /[1 + 21∙8(7∙14)^-x]²
dy/dx = 436 / 1² + (21∙8)²(7∙14)^-2x
dy/dx = 436 /[1 + 21∙8(7∙14)^-x]²
dy/dx = 436 [1² + (21∙8)²(7∙14)^-2x]^-1
dy/dx = 436 [1^-² + (21∙8)^-²(7∙14)^2x]
dy/dx = 436 [1 + (0∙002 104 199..)(7∙14)^2x]
dy/dx = 436 + 0∙917 431 192(7∙14)^2x

Now stick in your value for x (not given).

2006-12-12 22:49:37 · answer #1 · answered by Brenmore 5 · 0 0

Let a = 82.8.
Let b = 21.8.
Let c = 7.14.

Then f'(x) = ab(ln c)c^(-x)/(1 + bc^(-x))^2.

2006-12-13 03:12:41 · answer #2 · answered by I know some math 4 · 0 0

0 as the derivative of a constant is zero

2006-12-13 03:12:21 · answer #3 · answered by raj 7 · 1 0

f(x)=a/(b+c(d^(-x)))
where a=82.8
b=1
c=21.8
d=7.14
is this your question?
if so, then ask me at http://cbsemaths.co.in or e-mail

please always ask properly

2006-12-13 03:36:08 · answer #4 · answered by anubhav2k 2 · 0 0

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