Answer:
as
[tex]\frac{d}{dx}\left(1-x+\frac{x^2}{1}+x-x^2\right)=0[/tex]
so
[tex]y'=0[/tex]
Step-by-step explanation:
Given the function
[tex]y=\:1-x+\frac{x^2}{1}+x-x^2[/tex]
Taking derivative
[tex]\frac{d}{dx}\left(1-x+\frac{x^2}{1}+x-x^2\right)[/tex]
[tex]\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'[/tex]
[tex]=\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)[/tex]
as
[tex]\frac{d}{dx}\left(1\right)=0[/tex] ∵ [tex]\mathrm{Derivative\:of\:a\:constant}:\quad \frac{d}{dx}\left(a\right)=0[/tex]
[tex]\frac{d}{dx}\left(x\right)=1[/tex] ∵ [tex]\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1[/tex]
[tex]\frac{d}{dx}\left(\frac{x^2}{1}\right)=2x[/tex] ∵ [tex]\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}[/tex]
[tex]\frac{d}{dx}\left(x\right)=1[/tex] ∵ [tex]\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1[/tex]
[tex]\frac{d}{dx}\left(\frac{x^2}{1}\right)=2x[/tex] ∵ [tex]\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}[/tex]
substituting all the values in the expression
[tex]=\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)[/tex]
[tex]=0-1+2x+1-2x[/tex]
[tex]=0[/tex]
Hence,
[tex]\frac{d}{dx}\left(1-x+\frac{x^2}{1}+x-x^2\right)=0[/tex]
Therefore,
[tex]y'=0[/tex]
