by determining :
lim f(x+h)-f(x)
h->0 ---------------
h
show that if f(x)=2x^2+2 then dy/dx=4x.
i figured out that f'(x) is 4x...but i am confused about what to do next...someone help?
"i figured out that f'(x) is 4x..."
That's the answer! dy/dx is just a notation for the derivative. If the function f(x) is dentoted by y, then
f(x+h)-f(x) can be called Delta y (i.e. a change in y) and h is, of course, a change in x, which we can denote as Delta x. So, the ratio
[f(x+h)-f(x)]/h can also be written as:
Delta y/Delta x
The derivative is obtained by taking the limit Delta x--->0. The notation for this limiting case is dy/dx
This notation dates back from the time before limits were invented. dy and dx were taken to be infinitesimal quantities, i.e. smaller than any positive number, yet larger than zero. When mathematics was further developed in the 19-th century it was found that infinitesimal numbers were an ill defined concept and limits were invented to do calculus.
The notation dy/dx for the derivative was kept but you could no longer interpret it as a ratio of two well defined quantities dy and dx.
However, in the 1960s Abraham Robinson showed how one can rigorously define infinitesimal numbers, see here:
http://en.wikipedia.org/wiki/Hyperreal_number
To show that dy/dx = 4x, we need to evaluate the limit as h approaches 0 of [f(x+h) - f(x)]/h, where f(x) = 2x^2 + 2.
Let's start by finding f(x+h) - f(x):
f(x+h) = 2(x+h)^2 + 2
= 2(x^2 + 2xh + h^2) + 2
= 2x^2 + 4xh + 2h^2 + 2
Now, subtract f(x) from f(x+h):
f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 2) - (2x^2 + 2)
= 4xh + 2h^2
Now we can find [f(x+h) - f(x)]/h:
[f(x+h) - f(x)]/h = (4xh + 2h^2)/h
= 4x + 2h
As h approaches 0, the term 2h approaches 0, so we can write the limit as:
lim(h->0) [f(x+h) - f(x)]/h = 4x + 0 = 4x
Therefore, dy/dx = 4x for the function f(x) = 2x^2 + 2.