How do you find the derivative of #3/(y^3)#?

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Answer 1 · 2017-02-08T01:57:40

#(3/y^3)'=-9/x^4#

Since #(cf(x))'=cf'(x)#

We can say

#(3/y^3)'=3(1/y^3)'#

Since #1/y^3=y^-3# then,

#3(1/y^3)'=3(y^-3)'#

The Power Rule states #(x^n)'=nx^(n-1)#, then

#3(y^-3)'=3(-3)x^(-3-1)=-9x^-4=-9/x^4#