Understanding Simplification of Arc Length Calculations

[Screen]

Homework Statement

This is probably very simple, but I'm teaching myself arc length via Paul's Online Calculus Notes and there's a simplification on the page:

I was wondering why the first Δx^2 was simplified to 1? I understand the other Δx^2 came out of the square root.

 

[vela]

What you're thinking is akin to \sqrt{a^2+b^2} = \sqrt{a^2+1}\ b, which isn't correct. You can only pull something out of a radical if it's a factor. What Paul is doing is this:
\sqrt{\Delta x^2 + [f'(x_i^*)]^2 \Delta x^2} = \sqrt{\Delta x^2(1 + [f'(x_i^*)]^2)}Now because Δx² is a factor under the radical, you can say
\sqrt{\Delta x^2(1 + [f'(x_i^*)]^2)} = \sqrt{\Delta x^2}\sqrt{1 + [f'(x_i^*)]^2}= \Delta x \sqrt{1 + [f'(x_i^*)]^2}

 

[Screen]

What you're thinking is akin to \sqrt{a^2+b^2} = \sqrt{a^2+1}\ b, which isn't correct. You can only pull something out of a radical if it's a factor. What Paul is doing is this:
\sqrt{\Delta x^2 + [f'(x_i^*)]^2 \Delta x^2} = \sqrt{\Delta x^2(1 + [f'(x_i^*)]^2)}Now because Δx² is a factor under the radical, you can say
\sqrt{\Delta x^2(1 + [f'(x_i^*)]^2)} = \sqrt{\Delta x^2}\sqrt{1 + [f'(x_i^*)]^2}= \Delta x \sqrt{1 + [f'(x_i^*)]^2}

Oh you are completely right, sorry for such a silly question and thank you for explaining.