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 0x<1, then

1+xexpx11-x
Proof.

By the inequalities for differences of powers, we have

x(1+xn)n-1x1-(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)n11-x.

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

Theorem 2.
limx0exp(x)-1x=1
Proof.

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

1exp(x)-1x11-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

11-exp(-x)x11+x.

Hence

1exp(-x)exp(x)-1x1(1+x)exp(-x).

By theorem 1, we have 1-xexp(-x)1/(1+x), so

1+x1-exp(-x)x1(1+x)(1-x)=11-x2.

By the squeeze rule, we conclude that

limx01-exp(-x)x=1

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

Theorem 3.
ddxexp(x)=exp(x)
Proof.

By definition,

ddxexp(x)=limyxexp(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

limyxexp(y)-exp(x)y-x=exp(x)limyxexp(y-x)-1y-x=exp(x)limy0expy-1y.

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

limyxexp(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