How do you differentiate $y=(v^3-2vsqrtv)/(v)$ ?

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Answer 1 · 2017-03-04T20:12:05

$2v-1/sqrtv$

Both terms on top have a $v$ in them, which is the denominator too, so you can actually rewrite quite simply:

$(v^3-2vsqrt(v))/v=v^2-2sqrtv=v^2-2v^(1/2)$

Now we have a simple situation of using the power rule, where

$d/dx ax^n = n*ax^(n-1)$

so, in the case above,

$d/dx[v^2-2v^(1/2)]=2v-1/2*2v^(1/2-1)$

$= 2v-v^(-1/2)$

$= 2v-1/sqrtv$

How do you differentiate y=(v^3-2vsqrtv)/(v)y=(v^3-2vsqrtv)/(v)?