Question

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?

Answer 1

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

\[((\Delta u / \Delta x)v - u(\Delta v/\Delta x))/(v + \Delta v)v\]

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

Almost.

The 1/Delta x is a separate fraction.

It started out as simply Delta x under everything else.

Answer 7

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

Answer 8

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

Answer 9

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!

Answer 10

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

Answer 11

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

Answer 12

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.