Tilted Forum Project Discussion Community - View Single Post - I Can't Understand dy/dx when you throw in "u"

Stompy · 03-10-2005, 12:30 PM

The only time I've ever used "u" with derivatives was like so:

Find the derivative of 1/(4-x^2)

u = x^2
du/dx = 2x
du = 2x dx

the problem then becomes du/(4-u)^2

Since u is x^2, and du is 2x, it becomes 2x/(4-x^2)^2.

In other words... u substitution. (I skipped a few steps, but you get the idea)

03-10-2005, 12:30 PM	#5 (permalink)
Stompy Banned from being Banned Location: Donkey	The only time I've ever used "u" with derivatives was like so: Find the derivative of 1/(4-x^2) u = x^2 du/dx = 2x du = 2x dx the problem then becomes du/(4-u)^2 Since u is x^2, and du is 2x, it becomes 2x/(4-x^2)^2. In other words... u substitution. (I skipped a few steps, but you get the idea) __________________ I love lamp. Last edited by Stompy; 03-10-2005 at 12:36 PM..