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

Question

Emma Aldridge · Accepted Answer

#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#

Axel Alvarez · Answer

undefined To differentiate the function $ y = \frac{v^3 - 2v\sqrt{v}}{v} $, you can use the quotient rule, which states that if you have a function of the form $ \frac{u}{v} $, then its derivative is given by $ \frac{u'v - uv'}{v^2} $. Here's how you apply the quotient rule to the given function:

1. Let $ u = v^3 - 2v\sqrt{v} $ and $ v = v $.
2. Find the derivatives of $ u $ and $ v $ with respect to $ v $.
3. Apply the quotient rule to get the derivative of the function.

Derivative of $ u $ with respect to $ v $:
$$ u' = 3v^2 - 2\sqrt{v} - \frac{v}{\sqrt{v}} $$

Derivative of $ v $ with respect to $ v $:
$$ v' = 1 $$

Now, apply the quotient rule:
$$ \frac{d}{dv}\left(\frac{u}{v}\right) = \frac{(u'v - uv')}{v^2} $$

$$ \frac{d}{dv}\left(\frac{v^3 - 2v\sqrt{v}}{v}\right) = \frac{(3v^2 - 2\sqrt{v} - \frac{v}{\sqrt{v}})(v) - (v^3 - 2v\sqrt{v})(1)}{v^2} $$

Simplify the expression:
$$ \frac{d}{dv}\left(\frac{v^3 - 2v\sqrt{v}}{v}\right) = \frac{(3v^3 - 2v^{\frac{3}{2}} - v) - (v^3 - 2v\sqrt{v})}{v^2} $$

$$ \frac{d}{dv}\left(\frac{v^3 - 2v\sqrt{v}}{v}\right) = \frac{3v^3 - 2v^{\frac{3}{2}} - v - v^3 + 2v\sqrt{v}}{v^2} $$

$$ \frac{d}{dv}\left(\frac{v^3 - 2v\sqrt{v}}{v}\right) = \frac{2v\sqrt{v} - v - v^{\frac{3}{2}}}{v^2} $$

$$ \frac{d}{dv}\left(\frac{v^3 - 2v\sqrt{v}}{v}\right) = \frac{-v^{\frac{3}{2}} + 2v\sqrt{v} - v}{v^2} $$

So, the derivative of $ y = \frac{v^3 - 2v\sqrt{v}}{v} $ with respect to $ v $ is $ \frac{-v^{\frac{3}{2}} + 2v\sqrt{v} - v}{v^2} $.