In this section, we define what is arguably the single most important function in all of mathematics. We have already noted that the function $\ln x $ is injective, and therefore it has an inverse.

Definition 9.3.1 The inverse function of $\ln(x) $ is $y=\exp(x)$, called the natural exponential function.

The domain of $\exp(x) $ is all real numbers and the range is $(0, \infty)$. Note that because $\exp(x)$ is the inverse of $\ln(x)$, $\exp (\ln x) =x$ for $x>0$, and $\ln (\exp x) = x$ for all $x$. Also, our knowledge of $\ln(x)$ tells us immediately that $\exp(1) = e$, $\exp(0) = 1$, $\ds\lim _{x\to\infty} \exp x =\infty$, and $\ds\lim_{x\to -\infty } \exp x = 0$.

Theorem 9.3.2 $\ds {d\over dx}\exp(x) = \exp(x)$.

By the Inverse Function Theorem (9.1.17), $\exp(x)$ has a derivative everywhere. The theorem also tells us what the derivative is. Alternately, we may compute the derivative using implicit differentiation: Let $y=\exp x $, so $\ln y =x $. Differentiating with respect to $x$ we get $\ds {1\over y} {dy\over dx} =1$. Hence, ${dy\over dx} = y =\exp x$.


Corollary 9.3.3 Since $\exp x >0 $, $\exp x $ is an increasing function whose graph is concave up.

The graph of the natural exponential function is indicated in figure 9.3.1. Compare this to the graph of $\ln x$, figure 9.2.2.

Figure 9.3.1. The graph of $\exp(x)$.

Corollary 9.3.4 The general antiderivative of $\exp x $ is $\exp x + C $.

Of course, the word "exponential'' already has a mathematical meaning, and this meaning extends in a natural way to the exponential function $\exp(x)$.

Lemma 9.3.5 For any rational number $q$, $\exp(q) = e^q$.

Let $y=e^q $. Then $\ln y = \ln (e^q ) = q \ln e = q$, and so $y= \exp(q)$.


In view of this lemma, we usually write $\exp(x)$ as $\ds e^x$ for any real number $x$. Conveniently, it turns out that the usual laws of exponents apply to $\ds e^x$.

Theorem 9.3.6 For every $x,y \in \R$ and $q\in\Q$:

    (a) $\ds e^{x+y} = e^x e^y $

    (b) $\ds e^{x-y} = e^x/e^y$

    (c) $\ds (e^x )^q = e^{xq} $

Parts (b) and (c) are left as exercises. For part (a), $\ln (e^x e^y) =\ln e^x + \ln e^y = x +y$, so $e^x e^y = e^{x+y }$.


Example 9.3.7 Solve $\ds e^{4x+5} - 3 =0$ for $x$.

If $\ds e^{4x+5} - 3 =0$ then $\ds 4x+5 =\ln 3$ and so $\ds x={\ln 3 -5\over 4}$.

Example 9.3.8 Find the derivative of $f(x) =e^{x^3 } \sin (4x)$.

By the product and chain rules, $f'(x) =3x^2 e^{x^3 } \sin (4x) + 4 e^{x^3 } \cos(4x)$.

Example 9.3.9 Evaluate $\int x e^{x^2 } dx $.

Let $u=x^2$, so $du = 2x\,dx$. Then $$\int x e^{x^2}\,dx={1\over2}\int e^u\,du= {1\over2}e^u= {1\over2}e^{x^2}+C.$$

Exercises 9.3

Ex 9.3.1 Prove parts (b) and (c) of theorem 9.3.6.

Ex 9.3.2 Solve $\ln (1+ \sqrt{x} ) = 6 $ for $x$.

Ex 9.3.3 Solve $\ds e^{x^2} = 8$ for $x$.

Ex 9.3.4 Solve $\ln (\ln (x) ) = 1 $ for $x$.

Ex 9.3.5 Sketch the graph of $\ds f(x) = e^{4x-5 }+ 6 $.

Ex 9.3.6 Sketch the graph of $f(x) =3e^{x+6} -4 $.

Ex 9.3.7 Find the equation of the tangent line to $f(x) =e^x $ at $x= a $.

Ex 9.3.8 Compute the derivative of $f(x) = 3x^2 e^{5x-6} $.

Ex 9.3.9 Compute the derivative of $\ds f(x)= e^x -\left( 1+ x + {x^2\over 2} + {x^3\over3!} + \cdots + {x^n\over n!}\right)$.

Ex 9.3.10 Prove that $e^x > 1 $ for $x\geq 0$. Then prove that $e^x > 1+ x $ for $x\geq 0 $.

Ex 9.3.11 Using the previous two exercises, prove (using mathematical induction) that $\ds e^x > 1+ x + {x^2\over 2} + {x^3\over3!} + \cdots + {x^n\over n!} =\sum_{k=0 }^n {x^k\over k!}$ for $x\geq 0 $.

Ex 9.3.12 Use the preceding exercise to show that $e> 2.7$.

Ex 9.3.13 Differentiate $\ds {e^{kx}+ e^{-kx}\over 2} $ with respect to $x$.

Ex 9.3.14 Compute $\ds\lim_{x\to\infty} {e^x + e^{-x}\over e^x -e^{-x}}$.

Ex 9.3.15 Integrate $5x^4 e^{x^5}$ with respect to $x$.

Ex 9.3.16 Compute $\ds \int_0^{\pi/3} \cos (2x) e^{\sin 2x}\,dx$.

Ex 9.3.17 Compute $\ds \int {e^{1/x^2}\over x^3}\,dx$.

Ex 9.3.18 Let $\ds F(x) = \int_0^{e^x} e^{t^4}\,dt$. Compute $F'(0)$.

Ex 9.3.19 If $f(x) =e^{kx}$ what is $f^{(940)}(x)$?