Hi
Please have a look on the attachment. You can find find my question there. Thank you for the help.
Regards
PG
Hi,
dx and dy indicate infinitesimally small changes in their respective variable, x or y.
For example:
f(x)=y=x^2
dy/dx=2*x
A small change in x (dx) produces a small change in y (dy) and the division produces the derivative at that point.
To show how this works, say we make x=1 to start with. We then calculate y from the definition:
y=x^2=1
So for x1=1 we got y1=1.
Now we calculate using a small increment in x:
y=x^2
y+dy=(x+dx)^2
Say we make the increment in x equal to 0.1, we then get:
y+dy=1.1^2=1.21
Now that y is 1.21 and the last y was 1.00, subtracting we see the increment in y was 0.21, and with the increment in x being 0.1, we get:
dy/dx=0.21/0.1=2.1
which is approximate, but as we allow dx to approach zero we get an exact expression.
Interestingly, when we make the increment (which is dx and was 0.1 above) tend to zero, we get:
y+dy=f(x+dx)
y/dx+dy/dx=f(x+dx)/dx
or
dy/dx=f(x+dx)/dx-y/dx
or
dy/dx=(f(x+dx)-f(x))/dx
When we make dx go to zero we get dy/dx equal to the first derivative.
So in the above we handled dy and dx as if they were variables, but they also hold other significance in that they show the change in a variable. They dont stand alone however like variables do though, as they have a respective variable (dx has x, dy has y, etc.).
Thus when we say "Take the derivative of y=x^2 with respect to x" we mean dy/dx which is read "The derivative of y with respect to x".
So most of the idea is to find out what happens to dy when dx goes to zero in the limit.
Often the variable 'h' is used in place of dx:
f'(x)=lim[h-->0](f(x+h)-f(x))/h