derivative of exponential function


In this entry, we shall compute the derivative of the exponential functionDlmfDlmfMathworldPlanetmathPlanetmath from its definition as a limit of powers.

Theorem 1.

If 0≤x<1, then

1+x≤exp⁡x≤11-x
Proof.

By the inequalities for differences of powers, we have

x≤(1+xn)n-1≤x1-(n-1n)⁢x.

Since n-1<n, and x>0, we have 0<(n-1/n)⁢x<x. Because x<1, this implies 1-(n-1/n)⁢x>1-x, so

x1-(n-1n)⁢x<x1-x.

Hence

1+x≤(1+xn)n≤11-x.

Taking the limit as n→∞, we obtain our result. ∎

Theorem 2.
limx→0⁡exp⁡(x)-1x=1
Proof.

Assume 0<x<1. By our bound, we have

1≤exp⁡(x)-1x≤11-x.

Suppose that -1<x<0. Then, since exp⁡(x)=1/exp⁡(-x), we have

exp⁡(x)-1x=1exp⁡(-x)⋅1-exp⁡(-x)x.

From the inequalityMathworldPlanetmath above, we have

1≤1-exp⁡(-x)x≤11+x.

Hence

1exp⁡(-x)≤exp⁡(x)-1x≤1(1+x)⁢exp⁡(-x).

By theorem 1, we have 1-x≤exp⁡(-x)≤1/(1+x), so

1+x≤1-exp⁡(-x)x≤1(1+x)⁢(1-x)=11-x2.

By the squeeze rule, we conclude that

limx→0⁡1-exp⁡(-x)x=1

whether we approach the limit from the left or the right. ∎

Theorem 3.
dd⁢x⁢exp⁡(x)=exp⁡(x)
Proof.

By definition,

dd⁢x⁢exp⁡(x)=limy→x⁡exp⁡(y)-exp⁡(x)y-x.

By the addition theorem for the exponentialMathworldPlanetmathPlanetmath, we have

exp⁡(y)-exp⁡(x)y-x=exp⁡(x)⋅exp⁡(y-x)-1y-x,

so

limy→x⁡exp⁡(y)-exp⁡(x)y-x=exp⁡(x)⁢limy→x⁡exp⁡(y-x)-1y-x=exp⁡(x)⁢limy→0⁡exp⁡y-1y.

By theorem 2, the limit on the right-hand side equals 1, so we have

limy→x⁡exp⁡(y)-exp⁡(x)y-x=exp⁡(x).

∎

Title derivative of exponential function
Canonical name DerivativeOfExponentialFunction
Date of creation 2013-03-22 17:01:39
Last modified on 2013-03-22 17:01:39
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 15
Author rspuzio (6075)
Entry type Theorem
Classification msc 32A05
Related topic ExponentialFunction
Related topic ComplexExponentialFunction
Related topic DerivativeOfTheNaturalLogarithmFunction