How do I prove that the exponential function $e^x$ has gradient $e^x$ from first principles?

Question

Furthermore why is it that $e^x$ is used in exponential modelling? Why aren't other exponential functions used, i.e. $2^x$, etc.?

Thank you!

Answer 1

If you define $$\exp(x) = \sum_{n = 0}^{\infty} {x^n \over n!} = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots$$ then if you differentiate term by term you get back what you started with.

Answer 2

In some contexts it can be prudent to define the exponential function $y=\exp(x)$ as the solution of the differential equation

$\displaystyle \frac{dy}{dx}=y(x)$ such that $y(0)=1$

so that the exponential function is it's own derivative because you define it as such.

By Picard's Existence Theorem the solution of this differential equation is unique. This seems like cheating but depending on what properties of $e^x$ you want to use it is perfectly valid. For example, this doesn't tell you how to even calculate $\exp(2)=e^2$ without a little more work...

It is used in modelling because if you have a quantity $y=f(x)$ that changes in proportion to it's size:

$\displaystyle \frac{dy}{dx}=ky(x)$,

then the solution is $f(x)=y(0)e^{kx}$. Lots of quantities change in proportion to their size; e.g. radioactive decay.

Answer 3

The main reason for using $\exp$ rather than other exponentials is that its derivative (and consequently its Taylor series, antiderivative, etc) is simpler.

Answer 4

Apply the usual definition of the derivative:

$$\frac{d}{dx}\operatorname{e}^x = \lim_{h \to 0} \frac{\operatorname{e}^{x+h}-\operatorname{e}^x}{h} \, . $$

Next, apply the usual laws of indices:

$$\lim_{h \to 0} \frac{\operatorname{e}^{x+h}-\operatorname{e}^x}{h} = \lim_{h \to 0} \frac{\operatorname{e}^x(\operatorname{e}^h-1)}{h}=\operatorname{e}^x \, \lim_{h \to 0}\frac{\operatorname{e}^h-1}{h} \, . $$

If we're able to show that $(\operatorname{e}^h-1)/h \to 1$ as $h \to 0$ then we're done. This is easier said than done. We need to employ a little trickery. Let us first start by defining

$$t = \frac{1}{\operatorname{e}^h-1}$$

so that $1/t = \operatorname{e}^h-1$, and hence $1 + 1/t = \operatorname{e}^h$. Finally, we have: $h = \ln(1+1/t)$. Hence:

$$\begin{array}{ccc} \lim_{h \to 0} \frac{\operatorname{e}^h-1}{h} &=& \lim_{t \to \infty}\frac{1/t}{\ln(1+1/t)} \\ \\ &=&\lim_{t \to \infty}\frac{1}{t\ln(1+1/t)} \\ \\ &=&\lim_{t \to \infty}\frac{1}{\ln[(1+1/t)^t]} \\ \\ &=&\frac{1}{\ln(\operatorname{e})} \\ \\ &=& 1 \end{array}$$

This proof uses the definition that $\operatorname{e} = \lim_{t \to \infty}(1+1/t)^t$. To appreciate this definition, consider compound interest added yearly, monthly, weekly, daily, hourly, etc. Adding less and less interest, more and more frequently corresponds to letting $t$ tends towards infinity in $(1+1/t)^t$. Allowing a system to grow and decay on an infinitesimally small time scale gives rise to $\operatorname{e}^{\tau}$. This definition also benefits from a lack of circular reasoning: to be able to define $\operatorname{e}^x$ as a power series required us knowing how to differentiate it and get its Taylor series. Either that, or it was an amazingly, infinitely-lucky guess.