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How do you differentiate #(4x^-2) - 8x + 1#?

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1 Answer

Oct 18, 2015

#y'=-8x^-3-8#

Explanation:

#y'=4(x^-2)'-8(x)'+(1)' = 4*(-2)*x^(-2-1) - 8*1 + 0#

#y'=-8x^-3-8#

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