I mean practically integration means sum of all components, and the integral can be visualized as the area below a curve.
Is there a similar intuition or geometric meaning of the derivative?
I mean practically integration means sum of all components, and the integral can be visualized as the area below a curve.
Is there a similar intuition or geometric meaning of the derivative?
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Interestingly the history of Calculus is a bit backwards from the way it is taught in schools. It began with the investigation into finding areas of certain geometric shapes, a field of Mathematics called Quadrature. This led the way to integration theory. One landmark result which is by Fermat, when he demonstrated: $$\int_0^a x^n dx = \frac{a^{n+1}}{n+1}.$$
This was done using a geometric series and methods for factoring polynomials.
It wasn't until half a century later before the derivative was invented.
What does the derivative represent? Well, the quick answer is that it measures the "instantaneous rate of change" of a function at a point $x$.
A better way to think about it, is that the derivative is the slope of the line that best approximates a function at a point $x$. This is evident when we consider the definition of the derivative given as follows.
Let $f:\mathbb{R}\to\mathbb{R}$ be a function. We say that $f$ is differentiable at a point $x$ if there exists an $M$ for which $$\lim_{h\to0} \frac{|f(x+h)-f(x) - Mh|}{h} \to 0$$ as $h \to 0$. We then say $f'(x)=M$. This says that when $h$ is small, $Mh$ becomes a good approximation of $f(x+h)-f(x)$. In other words, $$f(x+h)-f(x) \approx Mh = f'(x)[(x+h) - x].$$ Thus we can approximate the deviation of $f(x+h)$ from $f(x)$ by using the derivative, $f(x+h) \approx f'(x) h + f(x)$.
It was later found that integration and differentiation were intimately connected. This is known as the Fundamental Theorem of Calculus, and it is what connects the two main ideas of Calculus, Integration and Differentiation. It tells us that if a function is written as $$F(x) = \int_0^x f(t) dt$$ then $F'(x) = f(x)$. In other words: $$\frac{d}{da} \frac{a^{n+1}}{n+1} = \frac{d}{da} \int_0^a x^n dx = a^n.$$