## anonymous 4 years ago ALGEBRA in Calculus proof... Can you help me understand the basis for bringing delta x upstairs into only the numerator in this step of a problem?

1. anonymous

$1/\Delta x * (\Delta u)v - u(\Delta v)/(v + \Delta v)v$ $((\Delta u / \Delta x)v - u(\Delta v/\Delta x))/(v + \Delta v)v$

2. anonymous

This is from Lecture Session 10 of MIT OCW Math 18.01 Scholar - Single Variable Calculus.

3. anonymous

The lecture and lecture notes show what I typed above. I don't understand why the Delta x would not have to be multiplied against the (v + Delta v)v term in the denominator.

4. anonymous

How does the Delta x term get into the numerator of the numerator?

5. TuringTest

${{1\overΔx}(Δu)v−u(Δv)\over(v+Δv)v} \to {{Δu \overΔx}v−u{Δv \overΔx}\over(v+Δv)v}$is the step?

6. anonymous

Almost. The 1/Delta x is a separate fraction. It started out as simply Delta x under everything else.

7. anonymous

And then ended up as you have on the right hand side.

8. anonymous

see attached for more clarity. It's near the bottom third of the page...

9. TuringTest

oh this is simple: multiply out the bottom by delta x then divide the top and bottom by delta x much easier when I can see it!

10. TuringTest

or divide top and bottom of ${1\over \Delta x}$ by $\Delta x$you get${{1\over \Delta x}\over1}{(\Delta u)v-u (\Delta v)\over(v+\Delta v)v}$

11. anonymous

I need to go to Algebra Tricks Boot Camp. Thanks.

12. anonymous

I still think that is really funky. I will need to do some test problems to convince my rock-like brain this works just like you show it does.