Question

If y= 1-x+x²/1+x-x² than how about y'=?
This question is about derived function.

Answer 1

as

`(d)/(dx)\left(1-x+(x^2)/(1)+x-x^2\right)=0`

so

`y'=0`

Explanation:

Given the function

`y=\:1-x+(x^2)/(1)+x-x^2`

Taking derivative

`(d)/(dx)\left(1-x+(x^2)/(1)+x-x^2\right)`

`\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'`

`=(d)/(dx)\left(1\right)-(d)/(dx)\left(x\right)+(d)/(dx)\left((x^2)/(1)\right)+(d)/(dx)\left(x\right)-(d)/(dx)\left(x^2\right)`

as

`(d)/(dx)\left(1\right)=0` ∵
`\mathrm{Derivative\:of\:a\:constant}:\quad (d)/(dx)\left(a\right)=0`

`(d)/(dx)\left(x\right)=1` ∵
`\mathrm{Apply\:the\:common\:derivative}:\quad (d)/(dx)\left(x\right)=1`

`(d)/(dx)\left((x^2)/(1)\right)=2x` ∵
`\mathrm{Apply\:the\:Power\:Rule}:\quad (d)/(dx)\left(x^a\right)=a\cdot x^(a-1)`

`(d)/(dx)\left(x\right)=1` ∵
`\mathrm{Apply\:the\:common\:derivative}:\quad (d)/(dx)\left(x\right)=1`

`(d)/(dx)\left((x^2)/(1)\right)=2x` ∵
`\mathrm{Apply\:the\:Power\:Rule}:\quad (d)/(dx)\left(x^a\right)=a\cdot x^(a-1)`

substituting all the values in the expression

`=(d)/(dx)\left(1\right)-(d)/(dx)\left(x\right)+(d)/(dx)\left((x^2)/(1)\right)+(d)/(dx)\left(x\right)-(d)/(dx)\left(x^2\right)`

`=0-1+2x+1-2x`

`=0`

Hence,

`(d)/(dx)\left(1-x+(x^2)/(1)+x-x^2\right)=0`

Therefore,

`y'=0`