Can you please explain in layman terms. I don't know if this is a duplicate, but if I find one I will delete this question.
The thing I don't get the most is differentials.
Can you please explain in layman terms. I don't know if this is a duplicate, but if I find one I will delete this question.
The thing I don't get the most is differentials.
On
I'm not sure how it was actually discovered historically -- I think Archimedes came close to doing some integrals when he was calculating the areas of various geometric figures -- but integral calculus could have been discovered like this:
Suppose we want to know how much a quantity $f$ changes during a certain (large) interval of time, and we know the instantaneous rate of change of the quantity at each time $t$. (For example, we've been watching the speedometer and we want to know how far we've gone.) It's natural to chop the large interval of time into extremely short subintervals of time, and compute the change in our quantity during each subinterval. The total change is the sum of all the little changes, which is an intuitive way of stating the fundamental theorem of calculus.
We can approximate the change during an extremely short interval of time just by mutiplying instantaneous rate of change (at some instant chosen arbitrarily in our short time interval) by the duration of the short interval of time. For example, if we're moving at $10$ m/s at one instant, then during the next $.1$ seconds we will move approximately $1$ meter.
To get a better approximation to the total change, repeat this process but using even shorter intervals of time.
This strategy leads us immediately to integration and to the fundamental theorem of calculus.
Check out the Wikipedia article on the history of calculus, as well as the one about its ancient fore-runner, the method of exhaustion. Unless you mean how the formulas for derivatives and integrals were deduced ? In which case, I will explain to you why $\Big(x^n\Big)'=n\cdot x^{n-1}$ and $\Big(\sqrt[n]x\Big)'=\displaystyle\frac{\sqrt[n]x}{n\cdot x}$
$$\Big(x^n\Big)'=\lim_{x\to x_0}\frac{x^n-x_0^n}{x-x_0}=\lim_{x\to x_0}\frac{\Big(x-x_0\Big)\Big(x^{n-1}+x^{n-2}x_0+x^{n-3}x_0^2+\ldots+x\cdot x_0^{n-2}+x_0^{n-1}\Big)}{x-x_0}$$
$$=\lim_{x\to x_0}\Big(x^{n-1}+x^{n-2}x_0+x^{n-3}x_0^2+\ldots+x\cdot x_0^{n-2}+x_0^{n-1}\Big)=n\cdot x_0^{n-1}$$
$$\Big(\sqrt[n]x\Big)'=\lim_{x\to x_0}\frac{\sqrt[n]x-\sqrt[n]x_0}{x-x_0}=\lim_{x\to x_0}\frac{\sqrt[n]x-\sqrt[n]x_0}{\Big(\sqrt[n]x\Big)^n-\Big(\sqrt[n]x_0\Big)^n}=$$
$$=\lim_{x\to x_0}\frac{\sqrt[n]x-\sqrt[n]x_0}{\Big(\sqrt[n]x-\sqrt[n]x_0\Big)\cdot\bigg[\Big(\sqrt[n]x\Big)^{n-1}+\Big(\sqrt[n]x\Big)^{n-2}\cdot\sqrt[n]x_0+\Big(\sqrt[n]x\Big)^{n-3}\Big(\sqrt[n]x_0\Big)^2+\ldots+\Big(\sqrt[n]x_0\Big)^{n-1}\bigg]}$$
$$=\lim_{x\to x_0}\frac1{\Big(\sqrt[n]x\Big)^{n-1}+\Big(\sqrt[n]x\Big)^{n-2}\cdot\sqrt[n]x_0+\Big(\sqrt[n]x\Big)^{n-3}\Big(\sqrt[n]x_0\Big)^2+\ldots+\sqrt[n]x\cdot\Big(\sqrt[n]x_0\Big)^{n-2}+\Big(\sqrt[n]x_0\Big)^{n-1}}$$
$$=\frac1{n\cdot\Big(\sqrt[n]x_0\Big)^{n-1}}=\frac{\sqrt[n]{x_0}}{n\cdot x_0}$$