Question

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

Answer 1

Answer:

as

$\frac{d}{dx}\left(1-x+\frac{x^2}{1}+x-x^2\right)=0$

so

$y'=0$

Step-by-step explanation:

Given the function

$y=\:1-x+\frac{x^2}{1}+x-x^2$

Taking derivative

$\frac{d}{dx}\left(1-x+\frac{x^2}{1}+x-x^2\right)$

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

$=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{1}\right)+\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(x^2\right)$

as

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

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

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

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

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

substituting all the values in the expression      

$=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{1}\right)+\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(x^2\right)$

$=0-1+2x+1-2x$

$=0$

Hence,

$\frac{d}{dx}\left(1-x+\frac{x^2}{1}+x-x^2\right)=0$

Therefore,

$y'=0$