Derivatives Explained from First Principles by understanding motion studied by Galileo!

[@ciaranmcevoy9875]

If you have any questions or feedback, I'd love to hear from you below!

[@ajayaagri6432]

Your experimental point of view is good but mathematically you should derive the first principal atleast one expression

[@ciaranmcevoy9875]

Hey Ajay, thanks for your feedback. If I understand you correctly, do you mean that you would have liked to have seen the derivation of the equations of motion? The purpose of this video was not to get into the mechanics of the problem but to observe the pattern enough to allow us to predict the displacement as a function of time. This then enabled us to differentiate a simple function from first principles, i.e. applying the limit of time tending to zero on the average rate of change of displacement. Thanks again for your thoughts, let me know if I've misunderstood your comment.

[@rituvardani333]

Sir can I ask you a chemistry question

[@rituvardani333]

Why you have evaluated the expression s(t+∆t)=c(t+∆t)^2? is the displacement s(t)=ct^2 does it always works in all cases.the ratios wont be different for displacement x and x+∆x . Sir please answer☺️☺️

[@ciaranmcevoy9875]

Hi Ritu, as you can see from the video we noticed a relationship between the displacement of the cylinder and the square of the time elapsed, then we went on to realise the velocity of the cylinder has a linear relationship with time and therefore, finally, the acceleration is constant. After noticing this pattern I wanted to demonstrate how differentiation of the displacement function (s(t) = ct^2 which represents what we noticed when we first rolled the cylinder) would get us to that same realisation using first principles of differentiation. In an effort to answer your question: I evaluated s(t+∆t)=c(t+∆t)^2 because we were differentiating with respect to time (because the displacement was varying with the passage of time), not with respect to a variable x. We needed to evaluate the change in displacement over a small interval ∆t and let ∆t tend towards zero. If by "does it always works in all cases", you are referring to the method of differentiating from first principles shown in this video, I can safely tell you that this is the base of differentiation that the majority of patterns and rules, that we know and learn today, come from (e.g. the power rule, trigonometric differentiation, product rule etc.). I'm not too sure I fully understood your query/confusion, but I hope this clarifies things for you pal.

[@rituvardani333]

I meant that is that expression s(t) =ct^2 works every time no matter what I do with time even for a small instant

[@ciaranmcevoy9875]

@@rituvardani333 Okay, I think I'm beginning to understand your confusion. The function s(t) will work for all of its' domain: time, t. From 0 to whenever suits, including the time of t + ∆t. For example, the displacement at 3 seconds would be s(3) = c*3^2, the displacement at 1000 seconds would be s(1000) = c*1000^2. Hence, the displacement at (t + ∆t) seconds is s(t + ∆t) = c(t + ∆t)^2. This is how we evaluate functions. We can then multiply out the brackets in c(t + ∆t)^2 to reveal what terms involve an increment of time, ∆t, in them. That way when we let ∆t tend to zero for the whole ratio of (s(t + ∆t) - s(t))/∆t, the terms with ∆t in them tend to zero also.

[@rituvardani333]

Thanks sir for your help

[@ciaranmcevoy9875]

@@rituvardani333 You're most welcome pal. Glad I could help.