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

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

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