If \(\int_{0}^{x} e^{-s^{2}} d s=u,\) find \(\frac{d x}{d u}.\)

The Fundamental Theorem of Calculus is a critical concept in integral calculus. It connects the concept of differentiation and integration in two main parts.

The first part states that if you have a continuous function \(f(x)\), and you create a new function from it by integrating \(f(x)\), say \(F(x) = \int_{a}^{x} f(t) dt\), then \(F(x)\) is differentiable, and its derivative is the original function \(f(x)\). Mathematically, this is written as:
\( \frac{d}{dx} \left( \int_{a}^{x} f(t) dt \right) = f(x)\).

This explains why differentiating \(\int_{0}^{x} e^{-s^2} ds\) gives us \(e^{-x^2}\): because differentiation undoes the process of integration.

The second part of the Fundamental Theorem of Calculus states that if \(F(x)\) is an antiderivative of \(f(x)\), then the definite integral of \(f(x)\) from \(a\) to \(b\) is given by:
\(\int_{a}^{b} f(x) dx = F(b) - F(a)\).

Understanding these connections makes it easier to handle problems involving both differentiation and integration, linking them together seamlessly.

The Chain Rule is a fundamental technique in calculus for finding the derivative of composite functions. When we have a function composed of two smaller functions, like \(y = g(f(x))\), the chain rule helps us differentiate it. The rule states:
\( \frac{d}{dx} g(f(x)) = g'(f(x)) \cdot f'(x)\).

In our exercise, after using the Fundamental Theorem of Calculus to find \(\frac{du}{dx} = e^{-x^2}\), we need to invert this derivative to find \(\frac{dx}{du}\). By the chain rule, we can express the relationship between \(x\) and \(u\) in reverse.

By taking the reciprocal of \(\frac{du}{dx}\), we apply the chain rule for the inverse function. When we say \(\frac{dx}{du}\), we are essentially 'reversing' the differentiation step:
\( \frac{dx}{du} = \frac{1}{\frac{du}{dx}} = \frac{1}{e^{-x^2}} = e^{x^2}\).

This process highlights how integration, differentiation, and their inverses relate through the chain rule.

The concept of the reciprocal is crucial in calculus, especially when dealing with derivatives. In general, the reciprocal of a function \(f(x)\) is given by \(\frac{1}{f(x)}\). In our exercise, once we recognize \(\frac{du}{dx}\), finding \(\frac{dx}{du}\) involves taking its reciprocal.

Since we have \(\frac{du}{dx} = e^{-x^2}\), to find \(\frac{dx}{du}\), we take:
\(\frac{dx}{du} = \frac{1}{e^{-x^2}}\).

This simplifies to \(e^{x^2}\) due to the properties of exponentials. By understanding the interchange between a function and its reciprocal, we can solve many differential and integral problems efficiently.

This idea is important because it tells us that understanding one form of a problem (like \(\frac{du}{dx}\)) can directly give us insights to solve its inverse \(\frac{dx}{du}\).

Reciprocals simplify calculations, making them an essential tool in your calculus toolkit.