Editing E (mathematical constant) (section)

=== Calculus ===
{{See also|Characterizations of the exponential function}}
[[File:Exp derivative at 0.svg|thumb|right|The graphs of the functions {{math|''x'' ↦ ''a''{{sup|''x''}}}} are shown for {{math|1=''a'' = 2}} (dotted), {{math|1=''a'' = ''e''}} (blue), and {{math|1=''a'' = 4}} (dashed). They all pass through the point {{math|(0,1)}}, but the red line (which has slope {{math|1}}) is tangent to only {{math|''e''{{sup|''x''}}}} there.]]

[[File:Ln+e.svg|thumb|right|The value of the natural log function for argument {{mvar|e}}, i.e. {{math|ln ''e''}}, equals {{math|1.}}]]

The principal motivation for introducing the number {{mvar|e}}, particularly in [[calculus]], is to perform [[derivative (mathematics)|differential]] and [[integral calculus]] with [[exponential function]]s and [[logarithm]]s.<ref name="kline">{{cite book | last = Kline | first = M. | year = 1998 | title = Calculus: An intuitive and physical approach | page = [https://books.google.com/books?id=YdjK_rD7BEkC&pg=PA337 337 ff] | publisher = Dover Publications | isbn = 0-486-40453-6}}</ref> A general exponential {{nowrap|function {{math|''y'' {{=}} ''a''<sup>''x''</sup>}}}} has a derivative, given by a [[limit of a function|limit]]:

:<math>\begin{align}
  \frac{d}{dx}a^x
    &= \lim_{h\to 0}\frac{a^{x+h} - a^x}{h} = \lim_{h\to 0}\frac{a^x a^h - a^x}{h} \\
    &= a^x \cdot \left(\lim_{h\to 0}\frac{a^h - 1}{h}\right).
\end{align}</math>

The parenthesized limit on the right is independent of the {{nowrap|variable {{mvar|x}}.}} Its value turns out to be the logarithm of {{mvar|a}} to base {{mvar|e}}. Thus, when the value of {{mvar|a}} is set {{nowrap|to {{mvar|e}},}} this limit is equal {{nowrap|to {{math|1}},}} and so one arrives at the following simple identity:
:<math>\frac{d}{dx}e^x = e^x.</math>

Consequently, the exponential function with base {{mvar|e}} is particularly suited to doing calculus. {{nowrap|Choosing {{mvar|e}}}} (as opposed to some other number) as the base of the exponential function makes calculations involving the derivatives much simpler.

Another motivation comes from considering the derivative of the base-{{mvar|a}} logarithm (i.e., {{math|log<sub>''a''</sub> ''x''}}),{{r|kline}} for&nbsp;{{math|''x'' > 0}}:

:<math>\begin{align}
  \frac{d}{dx}\log_a x
    &= \lim_{h\to 0}\frac{\log_a(x + h) - \log_a(x)}{h} \\
    &= \lim_{h\to 0}\frac{\log_a(1 + h/x)}{x\cdot h/x} \\
    &= \frac{1}{x}\log_a\left(\lim_{u\to 0}(1 + u)^\frac{1}{u}\right) \\
    &= \frac{1}{x}\log_a e,
\end{align}</math>

where the substitution {{math|''u'' {{=}} ''h''/''x''}} was made. The base-{{mvar|a}} logarithm of {{mvar|e}} is 1, if {{mvar|a}} equals {{mvar|e}}. So symbolically,
:<math>\frac{d}{dx}\log_e x = \frac{1}{x}.</math>

The logarithm with this special base is called the [[natural logarithm]], and is usually denoted as {{math|ln}}; it behaves well under differentiation since there is no undetermined limit to carry through the calculations.

Thus, there are two ways of selecting such special numbers {{mvar|a}}. One way is to set the derivative of the exponential function {{math|''a''<sup>''x''</sup>}} equal to {{math|''a''<sup>''x''</sup>}}, and solve for {{mvar|a}}. The other way is to set the derivative of the base {{mvar|a}} logarithm to {{math|1/''x''}} and solve for {{mvar|a}}. In each case, one arrives at a convenient choice of base for doing calculus. It turns out that these two solutions for {{mvar|a}} are actually ''the same'': the number {{mvar|e}}.

[[File:Area under rectangular hyperbola.svg|thumb|right|The five colored regions are of equal area, and define units of [[hyperbolic angle]] along the {{nowrap|[[hyperbola]] <math>xy=1.</math>}}]]

The [[Taylor series]] for the exponential function can be deduced from the facts that the exponential function is its own derivative and that it equals 1 when evaluated at 0:<ref name="strangherman">{{cite book|first1=Gilbert |last1=Strang |first2=Edwin |last2=Herman |title=Calculus, volume 2 |publisher=OpenStax |display-authors=etal |chapter=6.3 Taylor and Maclaurin Series |chapter-url=https://openstax.org/books/calculus-volume-2/pages/6-3-taylor-and-maclaurin-series |year=2023 |isbn=978-1-947172-14-2}}</ref>
<math display="block">e^x = \sum_{n=0}^\infty \frac{x^n}{n!}.</math>
Setting <math>x = 1</math> recovers the definition of {{mvar|e}} as the sum of an infinite series.

The natural logarithm function can be defined as the integral from 1 to <math>x</math> of <math>1/t</math>, and the exponential function can then be defined as the inverse function of the natural logarithm. The number {{mvar|e}} is the value of the exponential function evaluated at <math>x = 1</math>, or equivalently, the number whose natural logarithm is 1. It follows that {{mvar|e}} is the unique positive real number such that
<math display="block">\int_1^e \frac{1}{t} \, dt = 1.</math>

Because {{math|''e''<sup>''x''</sup>}} is the unique function ([[up to]] multiplication by a constant {{mvar|K}}) that is equal to its own [[derivative]],

<math display="block">\frac{d}{dx}Ke^x = Ke^x,</math>

it is therefore its own [[antiderivative]] as well:<ref>{{cite book|first1=Gilbert |last1=Strang |first2=Edwin |last2=Herman |title=Calculus, volume 2 |publisher=OpenStax |display-authors=etal |chapter=4.10 Antiderivatives |chapter-url=https://openstax.org/books/calculus-volume-1/pages/4-10-antiderivatives |year=2023 |isbn=978-1-947172-14-2}}</ref>

<math display="block">\int Ke^x\,dx = Ke^x + C .</math>

Equivalently, the family of functions

<math display="block">y(x) = Ke^x</math>

where {{mvar|K}} is any real or complex number, is the full solution to the [[differential equation]]

<math display="block">y' = y .</math>