# derivative of exponential function

###### Theorem 1.

If $0\leq x<1$, then

 $1+x\leq\exp x\leq{1\over 1-x}$
###### Proof.

By the inequalities for differences of powers, we have

 $x\leq\left(1+{x\over n}\right)^{n}-1\leq{x\over 1-\left({n-1\over n}\right)x}.$

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

 ${x\over 1-\left({n-1\over n}\right)x}<{x\over 1-x}.$

Hence

 $1+x\leq\left(1+{x\over n}\right)^{n}\leq{1\over 1-x}.$

Taking the limit as $n\to\infty$, we obtain our result. ∎

###### Theorem 2.
 $\lim_{x\to 0}{\exp(x)-1\over x}=1$
###### Proof.

Assume $0. By our bound, we have

 $1\leq{\exp(x)-1\over x}\leq{1\over 1-x}.$

Suppose that $-1. Then, since $\exp(x)=1/\exp(-x)$, we have

 ${\exp(x)-1\over x}={1\over\exp(-x)}\cdot{1-\exp(-x)\over x}.$
 $1\leq{1-\exp(-x)\over x}\leq{1\over 1+x}.$

Hence

 ${1\over\exp(-x)}\leq{\exp(x)-1\over x}\leq{1\over(1+x)\exp(-x)}.$

By theorem 1, we have $1-x\leq\exp(-x)\leq 1/(1+x)$, so

 $1+x\leq{1-\exp(-x)\over x}\leq{1\over(1+x)(1-x)}={1\over 1-x^{2}}.$

By the squeeze rule, we conclude that

 $\lim_{x\to 0}{1-\exp(-x)\over x}=1$

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

###### Theorem 3.
 ${d\over dx}\exp(x)=\exp(x)$
###### Proof.

By definition,

 ${d\over dx}\exp(x)=\lim_{y\to x}{\exp(y)-\exp(x)\over y-x}.$
 ${\exp(y)-\exp(x)\over y-x}=\exp(x)\cdot{\exp(y-x)-1\over y-x},$

so

 $\lim_{y\to x}{\exp(y)-\exp(x)\over y-x}=\exp(x)\lim_{y\to x}{\exp(y-x)-1\over y% -x}=\exp(x)\lim_{y\to 0}{\exp y-1\over y}.$

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

 $\lim_{y\to x}{\exp(y)-\exp(x)\over y-x}=\exp(x).$

Title derivative of exponential function DerivativeOfExponentialFunction 2013-03-22 17:01:39 2013-03-22 17:01:39 rspuzio (6075) rspuzio (6075) 15 rspuzio (6075) Theorem msc 32A05 ExponentialFunction ComplexExponentialFunction DerivativeOfTheNaturalLogarithmFunction