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

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

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

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

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

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

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