delta x is a small change in x.
What you are doing is drawing a short line from (x,y) to (x+delta x, y at x + delta x)
Then you are finding the slope of that line
the derivative of f(x) is the limit of the slope of that line as delta x goes to zero.
f(x+delta x) = 8 (x+delta x)^2 +1
f(x) = 8 x^2 +1
so
f(x+delta x) = 8 x^2 +16 x delta x + 8 delta x^2 +1
subtract f(x) = 8 x^2 + 1 from that
then
f(x+delta x)-f(x) = 16 x delta x + 8 delta x^2
now divide by delta x
[f(x+delta x)-f(x)]/delta x =
16 x + 8 delta x
that is the answer to your question
going one step beyond, look what happens when delta x goes to zero
the derivative then f'(x) = 16 x
Find f(x+ (delta)x)- f(x)
---------------------
(delta)x
if f(x)=8x(squared)+ 1
This is so confusing! What is all the delta x stuff?
4 answers
Please tell me if you did not follow this and I will try again some other way. It is really, really important.
delta x is a small change in x. I'm just going to call it dx. dx/x <<1
For this particular f(x) funtion,
[f(x + dx)f f(x)]/dx
= [8*(x + dx)^2 +1 - 8*x^2-1]/dx
= [16x*dx + 8(dx)^2]/dx
= 16x + 8*dx
As dx becomes infinitiely small compared to x, the above expression becomes 16x and is known as the derivative of the function.
For this particular f(x) funtion,
[f(x + dx)f f(x)]/dx
= [8*(x + dx)^2 +1 - 8*x^2-1]/dx
= [16x*dx + 8(dx)^2]/dx
= 16x + 8*dx
As dx becomes infinitiely small compared to x, the above expression becomes 16x and is known as the derivative of the function.
http://www.sosmath.com/calculus/diff/der00/der00.html