y = [itex]\sqrt{x^2- 2x + 1}[/itex](adsbygoogle = window.adsbygoogle || []).push({});

The y' is always the slope right?

Then if we simplify the equation

y =[itex]\sqrt{(x-1)^2 }[/itex]----or---y = [itex]\sqrt{(1-x)^2}[/itex]

y = x - 1 y = 1-x

Checking:---------------Checking:

[itex]\sqrt{(x-1)^2}[/itex]-------------[itex]\sqrt{(1-x)^2}[/itex]

[itex]\sqrt{x^2-2x+1}[/itex]-----------[itex]\sqrt{1-2x + x^2}[/itex]

y'= -----------------------------------y' = -1

Using the law of derivatives then

[itex]\frac{2x-2}{2\sqrt{x^2-2x+1}}[/itex]

and simplifying this will also gave 2 answers....

So the problem is:

Is there a posibility that there will be 2 y' ?

And also the graph of y=[itex]\sqrt{x}[/itex]

Is always in the first quadrant but don't you think it could be in the 4th quadrant because if x = 1 then y = 1 and y=-1

(-1)(-1) = 1 and (1)(1) = 1

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# The y' of a square root

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