The Proof of the Derivative of exe^xex

In the vast landscape of calculus, few concepts are as pivotal as the derivative. The function $e^x$ex, where $e$e is Euler's number (approximately 2.71828), holds a unique position. What makes it so special? The derivative of $e^x$ex is not just any derivative—it’s equal to $e^x$ex itself. But why is this the case? To understand this, we delve into a series of concepts that highlight the inherent properties of the exponential function.

Starting from the definition of the derivative, we recall that the derivative of a function $f\left(x\right)$f(x) at a point $x=a$x=a is given by:

$f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}$f′(a)=h→0lim​hf(a+h)−f(a)​

Applying this to our function $f\left(x\right)=e^x$f(x)=ex, we aim to find $f^{\prime }\left(x\right)$f′(x).

Let’s calculate:

$f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{e^{x+h}-e^x}{h}=\underset{h\to 0}{\lim }\frac{e^x{e}^h-e^x}{h}$f′(x)=h→0lim​hex+h−ex​=h→0lim​hexeh−ex​

Factoring out $e^x$ex:

$=e^x\underset{h\to 0}{\lim }\frac{e^h-1}{h}$=exh→0lim​heh−1​

Now, we need to evaluate the limit:

$\underset{h\to 0}{\lim }\frac{e^h-1}{h}$h→0lim​heh−1​

This limit is fundamental and often requires additional techniques, such as L'Hôpital's rule or series expansion, to solve. Let's explore both.

Using L'Hôpital’s Rule: This rule states that if a limit produces an indeterminate form, we can take the derivative of the numerator and the denominator. Thus, we rewrite the limit as:

$\underset{h\to 0}{\lim }\frac{e^h-1}{h}$h→0lim​heh−1​

Taking the derivative of the numerator $e^h$eh gives us $e^h$eh, and the derivative of $h$h is 1. Therefore, we have:

$=\underset{h\to 0}{\lim }{e}^h=e^0=1$=h→0lim​eh=e0=1

Substituting back, we get:

$f^{\prime }\left(x\right)=e^x\cdot 1=e^x$f′(x)=ex⋅1=ex

But there’s more to the story! The beauty of $e^x$ex is deeply rooted in its definition through a limit as well:

$e=\underset{n\to \infty }{\lim }{(1+\frac{1}{n})}^n$e=n→∞lim​(1+n1​)n

This limit allows us to explore the behavior of $e^x$ex from a binomial expansion perspective.

Let’s take a step back and understand this through series: The Taylor series expansion of $e^x$ex around $x=0$x=0 is:

$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}$ex=n=0∑∞​n!xn​

Differentiating term-by-term, we have:

$\frac{d}{dx}\left(\sum _{n=0}^{\infty }\frac{x^n}{n!}\right)=\sum _{n=1}^{\infty }\frac{n{x}^{n-1}}{n!}=\sum _{n=0}^{\infty }\frac{x^n}{n!}=e^x$dxd​(n=0∑∞​n!xn​)=n=1∑∞​n!nxn−1​=n=0∑∞​n!xn​=ex

This reinforces our earlier findings and highlights the intrinsic connection between the function and its derivative.

In essence, the derivative of $e^x$ex is a reflection of the function itself, which makes it unique in the realm of calculus. This property not only simplifies calculations but also enriches our understanding of exponential growth in various applications, from finance to population dynamics.

To sum up, the derivative of $e^x$ex being $e^x$ex illustrates a critical insight into the nature of exponential functions. It emphasizes that $e^x$ex does not merely grow; it grows at a rate proportional to its own value. Isn’t that fascinating?