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Does the derivative (the slope value) only give us the instantaneous rate change at a point and nothing else? Is this the only significance of that value?

For example I took the function $f(x)=x^2$,lets consider at $x=3$. $f(x)=x^2=9$ and $f'(x)=2x=6$. Does the value $6$ only tells us rate of change at point $x=3$?

Lets consider average rate change, e.g. suppose the average rate of change in y with respect to x over some interval is $7$; that is, for every single unit by which x changes, "y" on average changes by $7$ units. Here the average value "$7$" is related to the y value of the function means y on average changes by $7$ units.

Some of my friends said that the value $f'(x)=6$ in the above example only gives the rate of change at that point, i.e it is just the slope of the tangent at that point and nothing else, it has no effect on the y value ($x^2=9$) of the function. But if it doesn't have to do anything with 'y' value, then why is the slope at point called "the instantaneous rate of change of y with respect to x"?can anyone explain to me this, I really need help.

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    $\begingroup$ You are going to need to format your question in a more accessible way if you want a response and to use LaTex for your equations. $\endgroup$ Commented Dec 16, 2020 at 9:40
  • $\begingroup$ ok thanks for the suggestion,i edited my query.btw do u have any idea about it? $\endgroup$ Commented Dec 16, 2020 at 9:49

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Let's consider a slight variation on your example. What if $f(x) = x^2 + 1$? Then at $x=3$ we have $f(x) = 10$ and $f'(x) = 2x = 6$. We could also try $f(x) = x^2 - 2$, and then at $x=3$ we have $f(x)= 7$ but $f'(x) = 6$. These functions are just vertical shifts of your original function, so changing the $y$-value didn't change the derivative. By definition:

$$f'(x) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$

and notice for a given function this equation only depends on $x=a$, so the $y$-value of a function is irrelevant.

The reason we say the "instantaneous rate of change of $y$ with respect to $x$" is because in these scenarios $y$ is a function of $x$. At this point in your mathematical career I would guess you have only taken the derivatives of functions so the wording for the derivative of $y$ with respect to $x$ could also be stated as the derivative of the function $y=f(x)$, with respect to the variable $x$ (because the function depends on $x$).

You will eventually learn about implicit differentiation which will allow you to find the derivative of relations that are not functions. For example consider the equation of the circle $x^2 +y^2 = 5$. At the point $x=1$ there are two associated $y$-values: $y=-2$ and $y=2$ (hence not a function) and each has a different instantaneous rate of change. So for this example when you do not have a function the $y$-value will matter.

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  • $\begingroup$ thanks a lot,just want to clear myself regarding that when we calculate the derivative at a point for example the function y=x^2 at x=3 x^2=9 and f '(x)=6 the value 6 here has no effect on the y value 9. am i right??consider average rate change, but when we consider average rate change,e.g. suppose the average rate of change in y with respect to x over some interval is 7; that is, for every single unit by which x changes, "y" on average changes by 7 units. Here the average value "7" is related to the y value of the function means y on average changes by 7 units. $\endgroup$ Commented Dec 16, 2020 at 15:08
  • $\begingroup$ The value of 6 and the $y$-value 9 have no connection in this case. It is only the $x$-value that has a connection to the 6. When doing the average rate of change you are correct you need to use the $y$-values but remember these $y$-values are determined by the given $x$-values. So in order to find the change in $y$ we only need to be given the $x$-values to do that. Everything is based off of the $x$-values. $\endgroup$ Commented Dec 16, 2020 at 15:48
  • $\begingroup$ ok thanks a lot,i know i am annoying but just one last help.wat is the connection then between 3 and 6,i have understood but still have doubts, i know i am annoying,thanks for all your help still i didnt get i mean wat exactly the value 6 does??what is its significance? instantaneous rate of change of y w.rt x means wat does it do to the y value of the function,instantaneous rate change means wat exactly??wat does the value 6 do,rate change in wat sense, i know m sounding silly but here u are the only one who can help me. $\endgroup$ Commented Dec 16, 2020 at 18:34
  • $\begingroup$ The $6$ explains how $y$ changes respect to $x$ locally, for example take a function $f$ such that $f(0)=0$ and $f'(0)=3$, then for an $x$ close to $0$, $f(x)\approx 3x$, so intead of instantaneous rate of change you could see it as an approximation of local rate of change $\endgroup$ Commented Dec 17, 2020 at 8:14
  • $\begingroup$ Hey i didn't get wat u want to say can you explain how y changes and wat the value 6 does to change y wrt x. Use this example f(x) =x^2 at x=3 where f'(x) =2x=6.plz help $\endgroup$ Commented Dec 17, 2020 at 9:40
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Does the derivative (the slope value) only give us the instantaneous rate change at a point and nothing else? Is this the only significance of that value?

The expressions "only" and "nothing else" in mathematical statements generally require a considerate effort in finding the proof, which is often nontrivial.

I'll attempt to show the inverse: derivatives can be used in an unexpected way to solve problems. For example, to prove Young's inequality $$\forall\,a,b > 0, p \ge 1, q:= \frac{p}{p-1}, ab \le \frac{a^p}{p} + \frac{b^q}{q},$$ one might consider the function $f(x) = \dfrac1p x^p − x +\dfrac1q$ and its derivatives.

Apart from giving "instantaneous rate change" at the points, it actually provides a way for identifying extrema, which help solving lots of optimization problems.

Derivatives help in exploring repeated roots of a polynomial defined on (structures defined over) a field with characteristic zero. If you loosen the meaning of "derivatives" in your question context, you might find that "algebraic derivatives" (defined explicitly for $x^n$ with respect to $x$) have that functionality as well on (structures defined over) a field with nonzero characteristic, where one might not talk about the order '$<$' that the definition of limits replies on.

If you further allow me to move even further away from the topic, the application of $q$-derivatives in quantum calculus in the proof of Ramanujan's summation of $_1\Psi_1$ might amaze you.

References:

  1. Kac V., Cheung P. (2002) Ramanujan Product Formula. In: Quantum Calculus. Universitext. Springer, New York, NY. https://doi.org/10.1007/978-1-4613-0071-7_15
  2. Andrews, G. E., & Askey, R. (1976). A simple proof of Ramanujan’s summation of the $_1\Psi_1$. University of Wisconsin–Madison Mathematical Research Center Technical Summary Report #1669
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  • $\begingroup$ thank you. i get what you are saying.but this is way too advanced for me,i am not that much into advanced mathematics.i have posted my answer below.but really i appreciate u answering and ur efforts to help me.thank you $\endgroup$ Commented Dec 26, 2020 at 14:33
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lets take a function x^2 at x=3, f(x)=9 and here the derivative 2x=6. the value 6 is called as 𝐢𝐧𝐬𝐭𝐚𝐧𝐭𝐚𝐧𝐞𝐨𝐮𝐬 𝐫𝐚𝐭𝐞 𝐨𝐟 𝐜𝐡𝐚𝐧𝐠𝐞 𝐨𝐟 "𝐲" w.r.t x (why rate change of y? thats the reason i started thinking if 6 and 9 are connected,and that change in y meant slope causes change in y.) how does 6 cause rate change of y?how does it change the rate of the function?

Explanation:

There really is no relationship between the 6 in f'(x) = 6 and the 9 in f(x) = 9.

If the rate of change of y wrt x is 6 at that point, then roughly and informally speaking, if x changes by a small amount, then y will change by approximately 6 times that small amount. For example, if x changes by 0.001, then y will change by approximately 0.006.

More specifically, if x changes from 3 to 3.001, then y will change from 9 to approximately 9.006. i ll explain this below.

Now consider a different x value, which I'll call x' (pronounced "x prime"). This doesn't mean the derivative of x, it just means a different x value. Since it's close to the original x value, we can use it to approximate the value of the function at that new x' value.

Δx = x' - x = 0.001

Δf ≈ f'(x) Δx = (6) (0.001) = 0.006

f(x') ≈ f(x) + Δf = 9 + 0.006 = 9.006 (Approximate value).

So in a way, the 6 is related to the y values near the point we're looking at. If we look at the single point where x=3, then y=9 at that point. If we look at nearby values of x (very close to 3), then the change in y will be approximately 6 times the change in x.

now, when x changes from 3 to 5 i.e x changes by 2 units then y should change 2*6=12, therefore 9+12=21 but at x=5, y=25(x^2) so there is a lot of error. as gap between points increases the error increases. now if i say x changes from 4.9 to 5 the error will be less i get 24.99 and x^2=25.

so f '(x) does have a relationship with the y values but in a small region around x=3.

my main problem was i used to think if a slope is 6 then the y value(at x=3 which here is 9) should increase by 6,but now i get the meaning of "𝐰𝐢𝐭𝐡 𝐫𝐞𝐬𝐩𝐞𝐜𝐭 𝐭𝐨 𝐱". "with respect to x" the meaning lies in this sentence. i hadnt concentrated on the topic of linearization and differentials that much so i guess i got stuck with this query. thank you.

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