Hello everyone,
The derivative of functionis , what i understood is if function represents the graph of time and displacement, its derivative at a point on graph tells me the instantaneous velocity.
Question 1: since we find derivative by making theas small as we can lim->0 but we never make it a point, so why do we call it derivative at a point ? or instantaneous velocity at a point. it better be called velocity as x approaches at that particular point.
Ifrepresents the area of square its derivative represents change in area when side is changed by small amount. but when i put in numbers in it just tells me the area at that particular point.
Question 2: if it tells me the area of square at particular point where is the notion of change coming in ?
and it is not accurate as it ignores thepart, i could have found out the area of square much more accurately by simply putting the value in the function.
TrueHello everyone,
The derivative of functionis , what i understood is if function represents the graph of time and displacement, its derivative at a point on graph tells me the instantaneous velocity.
The above sentence does not make sense. What does lim->0 mean. What variable is approaching zero? With respect to what variable are you calculating the function?Question 1: since we find derivative by making theas small as we can lim->0 but we never make it a point,
They are both names for the same thing.so why do we call it derivative at a point ? or instantaneous velocity at a point. it better be called velocity as x approaches at that particular point.
No, you just gave the definition of a differential, not a derivative.Ifrepresents the area of square its derivative represents change in area when side is changed by small amount.
No it doesn't. 2x is the rate of area change as the side of the square changes.but when i put in numbers init just tells me the area at that particular point.
x^2 is the area at a particular point. The rate of change at a particular point is 2xQuestion 2: if it tells me the area of square at particular point where is the notion of change coming in ?
What dx^2 part? You are not doing a second order derivative, are you?and it is not accurate as it ignores thepart,
You need to get straight about rates and amounts.i could have found out the area of square much more accurately by simply putting the value in the function.
but why to calculate derivative in first place when i can tell the the exact change brought to area when i change x by 0.1Going back to x=3, we get A=9 and the area changes at the rate of 2*x at that point, so if we change x by 0.1 unit we get 2*3*dx=6*0.1=0.6 square units, so lets see how close this is:
At x=3.1 we get area A=9.61, so we got close with the estimate, 9.6 vs 9.61 which is exact.
The reason it is not exact here is because the increment is still finite. As we make the increment dx smaller, we get a better an better approximation, until when we get to dx=0 we get the exact value but we must do it analytically we can not do it numerically without some tricks.
Correction as lim of x ->0, it means we never approach 0, as lim 0/0 is indeterminate, we are just approximating curved graphs by a straight line at every point on the curve. perhaps we should never call that straight line tangent to a point because we never reach that point.The above sentence does not make sense. What does lim->0 mean. What variable is approaching zero? With respect to what variable are you calculating the function?
what is the difference between differential and a derivative ?No, you just gave the definition of a differential, not a derivative.
My mistake, it was supposed to beWhat dx^2 part? You are not doing a second order derivative, are you?
I was also thinking that i am confusing rates and amounts, if A=You need to get straight about rates and amounts.
but why to calculate derivative in first place when i can tell the the exact change brought to area when i change x by 0.1
for example at x=3; A=3*3=9 but A=3.1*3.1=9.61 ; just subtract and you get change as 0.61; clearly we can deduce from it that when we x is changed by 0.1 at x=3 the A changes by 0.61 square unit
according to my calculation 3.0000000000000001^2-3^2=0.00000000000000063.0000000000000001^2-3^2=0.00000000000000000
Why do you call analytical as exact when it ignoresanalytical derivative is exact while calculating the numerical derivative is not exact
according to my calculation 3.0000000000000001^2-3^2=0.0000000000000006
but you gave me a new topic to see (analytical and numerical derivative), which i never knew earlier, seems to me what i was analyzing was correct.
Why do you call analytical as exact when it ignorespart in and gives dA =0.6 as opposed to 0.61 given by numeric method.
question is still unanswered, that why do we need to create a formula to find the slope of tangent.
is it because derivative of a function gives us equation of the slope of function all along it more readily.
Instead of slow process of finding slope at every point using numeric method and then finding the equation of from it.
The analytical result tells us the actual derivative right at the target x value, which was 3 in our examples
it is also approximation as we can never put h=0, as notion of change would go haywire. better we call it infinitesimal averaging of the slope of the curve.The one sided derivative is:
(f(x+h)-f(x))/h
it is also approximation as we can never put h=0, as notion of change would go haywire. better we call it infinitesimal averaging of the slope of the curve.
I guess if someone doesn't like the notion of the derivative at a point of a function, he likely cannot like the notion of the derivative of the whole function as well. But perhaps I am wrong.
For instance, what about the dx (also close to zero but not zero) in finding the integral of a function?
The formal definition of a derivative, as you've pointed out, is
However, the subtle point here is what the limit sign actually means.
Formally, (This is 1st year uni level Real Analysis, and is not easy to get your head around) the limit as x tents to a point a is defined as follows:
Let f(x) be an arbitrary function. We want to prove that this function has limit L.
Let an arbitrary number ε > 0 be given.
If we can find a δ such that for any x in the interval
a - δ < x < a + δ
|f(x) - L| < ε,
then we say that
Any good Real Analysis textbook will explain this a lot better than I have, but in terms of a general description:
We're choosing a 'target precision' ε, and we figure out how close x has to be to a in order to get f(x) within ε of L.
For a 'simple' example, consider proving that the limit of f(x) = -x as x -> 1.
let ε >0 be given. we want to prove that f(x) = -x tends to -1 as x tends to +1.
Then |f(x) - (-1)| = |f(x)+1| = |1-x|
If |x-1| < δ, then |1-x| = |x-1| < ε
if we choose any δ ≤ ε .
Don't worry, it took me a whole semester to wrap my head around it.
Anyway the point of all this arcane pure maths is to say that the limit has to be the same regardless of which direction you approach the function from - it has to be true for any a - δ < x < a + δ . If we had a less nice function, like a Heaviside step function H(x) = 0 if x<0, 1 if x>0, 1/2 at x=0 and tried to take the limit as x-> 0, we run into a problem. Approaching x=0 from the left (negative) side, H(x) tends to 0, but approaching from the right says that H(x) tends to 1. No matter what δ>0 we choose, f(0-δ) = 0, but f(0+δ) = 1. We require, when we decrease delta, that f(x) gets closer to the limiting value, but no convergence is seen.
In plainer terms, the limit as x-> 0 of H(x) does not exist.
What does this have to do with derivatives?
Your question seems to stem from being worried about the value of the derivative at point a f'(a) being the gradient 'just to the right of' x=a, but because the derivative is a limit, not a value, it is in fact the value at the point. When you differentiate, you shrink h down to be close to zero from both sides: i.e. we require that
is satisfied. If the above equation is not satisfied, the derivative is not defined.
Isn't analysis fun?
All of the above definitions can be shown to be exactly equivalent with some careful arguments. The most 'correct' definition of the derivative is the gradient of the tangent plane at a point a - i.e. in any neighborhood of the point a,
(remember, f '(a) is the derivative evaluated at a, i.e. a constant with respect to x) where, the error term, tends to zero as x tends to a.
I should stress that this isn't an application of differentiation, but a definition of it.
Substituting x = x + a, we find
Rearranging,
Taking the limit of both sides (remembering f'(a) is just a constant)
So we see that, as long as
(*)
we have the familiar formula for the derivative, but with x here replacing h and a here replacing x
i.e. let x= h to get
f' =
Similarly, we are perfectly entitled to let x = -h
f' =
the slightly weirder (symmetric) formula is derived as follows:
This is sometimes tricky to prove! For example, for f(x) = |x|:
Assume
then,
i.e. xi(x) is a piecewise defined function - it's (-f'(0) -1)x for negative x and (-f'(0) + 1)x for positive x. Like the Heaviside up above, it's discontinuous at x=0 regardless of the value of f'(0) - the limit (*) does not exist.
For this reason, |x| is not differentiable at x=0.
This also gives some insight into where Taylor polynomials come from, but this post is already too long.
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