question if these derivative are equal

Question

Starting fuction: $y= \sqrt{x^2-2x+1}$

derivative A: $y^{\prime} = (1/2)(2x-2)(x^2-2x+1)^{-1/2}$

derivative B: rewrote the starting function as: $\sqrt{\left(x-1\right)^2} = \vert x -1 \vert$ thus

$y^{\prime} = $ either 1 if $x > 0$ or -1 if $x < 0$

First, are these the same answer? If they are different please tell me where I went wrong, if they are the same, please explain how and why I can have two different answers?

The starting point of derivative $B$ should be $y =\sqrt{(x-1)^2}$. — Umberto P., Sep 17 '13 at 12:21
@UmbertoP. thanks I missed the typo in the question and have hopefully improved it. — yiyi, Sep 17 '13 at 12:23

score 2 · Accepted Answer · answered Sep 17 '13 at 12:24

2

Derivative $A$ is $$ y' = \frac 12 (2x-2) (x^2 - 2x + 1)^{-1/2} = \frac{x-1}{\sqrt{x^2 - 2x + 1}} = \frac{x-1}{|x-1|}.$$

Derivative $B$ should be $1$ if $x > 1$ and $-1$ if $x < 1$.

With this correction observe they agree.

answered Sep 17 '13 at 12:24

Umberto P.

52,165

could you explain how $\frac{x-1}{|x-1|}$ equals the rewrite derivative B? – yiyi Sep 17 '13 at 12:29
do you know the definition of $|x - 1|$? – Umberto P. Sep 17 '13 at 12:30
its the absolute value. it is the numerator that I am having trouble with. Oh. I now get it as I typed this. They divide out. – yiyi Sep 17 '13 at 12:31

question if these derivative are equal

1 Answers1