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#