Derivative of product & quotient of functions - method of increments

[physicsnnewbie]

In my calculus book, the method of increments is used to find the derived function of the product and quotient of two functions. For example for the derivative of the product of functions u and v where u0 and v0 are the values of u and v at x = x0:

y = uv
y' = u0(dv/dx) + v0(du/dx) + du(dv/dx)

where du in the last term is said to approach 0 as x approaches 0. However if the function u contains a constant, wouldn't it approach the constant? This is probably a dumb question, but i must be missing something.

 

[tiny-tim]

hi physicsnnewbie!

(try using the X₂ icon just above the Reply box )

… where du in the last term is said to approach 0 as x approaches 0. However if the function u contains a constant, wouldn't it approach the constant? This is probably a dumb question, but i must be missing something.

no, du = u - u₀, so if u is constant then du is zero

 

[Goongyae]

>y = uv
>y' = u0(dv/dx) + v0(du/dx) + du(dv/dx)

That formula is ugly. I'd prefer this:

Suppose y changes by dy, u by du, and v by dv.
Then y+dy = (u+du)(v+dv)=uv+udv+vdu+du dv

Subtract y = uv to obtain
dy=udv+vdu+dudv

Now let the change (dy, du, and dv) be infinitely small. Since dudv contains TWO infintesimal quantities, it is infinitely smaller than the rest of the terms and can be dropped:

dy=udv+vdu

Since y, u, and v are presumed to change with x, divide by the final infinitesimal dx to obtain a formula without any infinitesimal terms:

dy/dx=udv/dx + vdu/dx or y'=uv'+vu'

 

[physicsnnewbie]

Thanks tim, I knew it was something simple. I was just forgetting that u0 has been subtracted. Thanks for the explanation goongyae.