y=x^(3)+2x^(2)+7x+8 then (dy)/(dx) will ...

`y=x^(3)+2x^(2)+7x+8` then `(dy)/(dx)` will be-

`3x^(2)+2x+15`

`3x^(2)+4x+7`

`x^(3)+2x^(2)+15`

`x^(3)+4x+7`

Verified by Experts

The correct Answer is:

`y=x^(3)+2x^(2)+7x+8`
`(dy)/(dx)=3x^(2)+4x+7`

If y= x^(x^(2x) ) ,then (dy)/(dx) =

` x^(x^(2x)) x^(2x) log x( 2+ 2log x +( 1)/(xlog x ))`

` x^(x^(2x)) x^(2x) log x( 2+ log x +( 1)/(xlog x ))`

` x^(x^(2x)) x^(2x) ( 2+ 2log x +( 1)/(xlog x ))`

` x^(x^(2x)) x^(2x) ( 2+ log x +( 1)/(xlog x ))`

`log_(e)(x^(2)+4).{(14x-5)/(7x^(2)-5x+1)-(2xy)/(x^(2)+4)}`

`(1)/(log_(e)(x^(2)+4)){(14x-5)/(7x^(2)-5x+1)-(2xy)/(x^(2)+4)}`

`log_(e)(7x^(2)-5x+1){(2x)/(x^(2)+4)-((14x-5)y)/(7x^(2)-5x+1)}`

`(1)/(log_(e)(7x^(2)-5x+1)){(2x)/(x^(2)+4)-((14x-5)y)/(7x^(2)-5x+1)}`

`(x^(2) +2xy +y^(2))/( y^(2)- 2xy -x^(2))`

`(x^(2) +2xy +y^(2))/( y^(2)+ 2xy -x^(2))`

`(x^(2) -2xy -y^(2))/( y^(2)+ 2xy +x^(2))`

`(x^(2) + 2xy -y^(2))/( y^(2)+ 2xy +x^(2))`