How do you differentiate #y = ((2x+3)^4)/ x#?

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Answer 1 · 2018-07-16T10:48:42

#(dy)/(dx)=(3(2x-1)(2x+3)^3)/x^2#

Here,

#y=(2x+3)^4/x#

Using Quotient Rule ,we get

#(dy)/(dx)=(xd/(dx)((2x+3)^4)-(2x+3)^4d/(dx)(x))/(x)^2#

#=>(dy)/(dx)=(x*4(2x+3)^3 xx2-(2x+3)^4*1)/x^2#

#=>(dy)/(dx)=((2x+3)^3{8x-2x-3})/x^2#

#=>(dy)/(dx)=((2x+3)^3(6x-3))/x^2#

#=>(dy)/(dx)=(3(2x+3)^3(2x-1))/x^2#