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(b) f'(x): f(x) = 5/ cube root x^7
(c) dx/dy where y=x^2- 1/x^2

2007-06-08 15:43:03 · 2 answers · asked by funkmistress_flex 1 in Science & Mathematics Mathematics

2 answers

b)
ƒ(x) = 5 / x^(7/3) = 5 * x^(-7/3)
ƒ(x) = -7/3 * 5x^(-7/3 - 1)
ƒ'(x) = -35 / [3x^(10/3)]

c)
y = x² - x^-2
dy/dx = 2x + 2x^-3
dy/dx = 2x + 2/x³

2007-06-08 15:47:31 · answer #1 · answered by MathGuy 6 · 0 0

Question b)
f(x) = 5.x^(-7/3)
f ` (x) = (-35/3).x^(-10/3)
f `(x) = (-35) / 3x^(10/3)

Question c)
y = x² - x^(-2)
dy / dx = 2x + 2x ^(-3)
dy/dx = (2x^4 + 2) / x³
dx/dy = x³ / (2x^4 + 2)
dx/dy = x³ / (2.(x^4 + 1))

2007-06-09 07:54:43 · answer #2 · answered by Como 7 · 0 0

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