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(x+1)^(2)(x+2)^(3)(x+3)^(4)...

`(x+1)^(2)(x+2)^(3)(x+3)^(4)`

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The correct Answer is:
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Let ` y = (x+1)^(2)(x+2)^(3)(x+3)^(4)`
`:. logy = log{(x+1)^(2).(x+2)^(3)(x+3)^(4)}`
`= log (x+1)^(2)+log(x+2)^(3)+log(x+3)^(4)`
and `(d)/(dy)logy.(dy)/(dx)= (dy)/(dx)=(d)/(dx)[2log(x+1)]+(d)/(dx)[3log(x+2)]+(d)/(dx)[4log(logx)=1/x]`
`1/y.(dy)/(dx) = (2)/((x+1)).(d)/(dx)(x+1)+3.(1)/((x+2)).(d)/(dx)(x+2)+4.(1)/((x+3)).(d)/(dx)(x+3)[:'(d)/(dx)(logx)=1/x]`
`=[(2)/(x+1)+(3)/(x+2)+(4)/(x+3)]`
`:. (dy)/(dx) = y [(2)/((x+1))+(3)/((x+2))+(4)/((x+3))]`
`= (x+1)^(2).(x+2)^(3).(x+3)^(4)[(2)/((x+1))+(3)/((x+2))+(4)/((x+3))]`
` = (x+1)^(2).(x+2)^(3).(x+3)^(4)`
`[(2(x+2)(x+3)+3(x+1)(x+3)+4(x+1)(x+2))/((x+1)(x+2)(x+3))]`
`= ((x+1)^(2)(x+2)^(3)(x+3)^(4))/((x+1)(x+2)(x+3))`
`[2(x^(2)+5x+6)+3(x^(2)+4x+3)+4(x^(2)+3x+2)]`
`= (x+1)(x+2)^(2)(x+3)^(3)`
`[2x^(2)+10x+12+13x^(2)+12+9+4x^(2)+12x+8]`
`=(x+1)(x+2)^(2)(x+3)^(3)[9x^(2)+34x+29]`
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