Asked by Haymi
first derivative of ((5x-2)^-3)/(3x-2)
Answers
Answered by
oobleck
the quotient rule says that if y = u/v then y' = (u'v - uv')/v^2
So, if
y = ((5x-2)^-3)/(3x-2)
y' = ((-3(5x-2)^-4)5)(3x-2) - (5x-2)^-3(3))/(3x-2)^2 = 4(9-15x)(5x-2)^-4/(3x-2)^2 = (36-60x) / ((5x-2)^4 * (3x-2)^2)
or, you could use the product rule
y = (5x-2)^-3 * (3x-2) ^-1
y' = -3(5x-2)^-4(5)(3x-2) + (-1)(3x-2)^-3(3) = -3(5x-2)^-3 * (3x-2) (5(5x-2)^-1 + 3)
So, if
y = ((5x-2)^-3)/(3x-2)
y' = ((-3(5x-2)^-4)5)(3x-2) - (5x-2)^-3(3))/(3x-2)^2 = 4(9-15x)(5x-2)^-4/(3x-2)^2 = (36-60x) / ((5x-2)^4 * (3x-2)^2)
or, you could use the product rule
y = (5x-2)^-3 * (3x-2) ^-1
y' = -3(5x-2)^-4(5)(3x-2) + (-1)(3x-2)^-3(3) = -3(5x-2)^-3 * (3x-2) (5(5x-2)^-1 + 3)
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