How do you differentiate $y = \frac{v^{3} - 2 v \sqrt{v}}{v}$ $y=(v^3-2vsqrtv)/(v)$ ?

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Answer 1 · 2017-03-04T20:12:05

$2 v - \frac{1}{\sqrt{v}}$ $2v-1/sqrtv$

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

$\frac{v^{3} - 2 v \sqrt{v}}{v} = v^{2} - 2 \sqrt{v} = v^{2} - 2 v^{\frac{1}{2}}$ $(v^3-2vsqrt(v))/v=v^2-2sqrtv=v^2-2v^(1/2)$

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

$\frac{d}{d x} a x^{n} = n \cdot a x^{n - 1}$ $d/dx ax^n = n*ax^(n-1)$

so, in the case above,

$\frac{d}{d x} [v^{2} - 2 v^{\frac{1}{2}}] = 2 v - \frac{1}{2} \cdot 2 v^{\frac{1}{2} - 1}$ $d/dx[v^2-2v^(1/2)]=2v-1/2*2v^(1/2-1)$

$= 2 v - v^{- \frac{1}{2}}$ $= 2v-v^(-1/2)$

$= 2 v - \frac{1}{\sqrt{v}}$ $= 2v-1/sqrtv$

How do you differentiate y=v3−2v√vvy=(v^3-2vsqrtv)/(v)?