derivative of exponential function

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

Theorem 1.

If , then

$$1+x\le \exp x\le \frac{1}{1-x}$$

Proof.

By the inequalities for differences of powers, we have

$$x\le {\left(1+\frac{x}{n}\right)}^n-1\le \frac{x}{1-\left(\frac{n-1}{n}\right)x}.$$

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

Hence

$$1+x\le {\left(1+\frac{x}{n}\right)}^n\le \frac{1}{1-x}.$$

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

Theorem 2.

$$\underset{x\to 0}{lim}\frac{\exp (x)-1}{x}=1$$

Proof.

Assume . By our bound, we have

$$1\le \frac{\exp (x)-1}{x}\le \frac{1}{1-x}.$$

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

$$\frac{\exp (x)-1}{x}=\frac{1}{\exp (-x)}\cdot \frac{1-\exp (-x)}{x}.$$

From the inequality above, we have

$$1\le \frac{1-\exp (-x)}{x}\le \frac{1}{1+x}.$$

Hence

$$\frac{1}{\exp (-x)}\le \frac{\exp (x)-1}{x}\le \frac{1}{(1+x)\exp (-x)}.$$

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

$$1+x\le \frac{1-\exp (-x)}{x}\le \frac{1}{(1+x)(1-x)}=\frac{1}{1-x^2}.$$

By the squeeze rule, we conclude that

$$\underset{x\to 0}{lim}\frac{1-\exp (-x)}{x}=1$$

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

Theorem 3.

$$\frac{d}{dx}\exp (x)=\exp (x)$$

Proof.

By definition,

$$\frac{d}{dx}\exp (x)=\underset{y\to x}{lim}\frac{\exp (y)-\exp (x)}{y-x}.$$

By the addition theorem for the exponential, we have

$$\frac{\exp (y)-\exp (x)}{y-x}=\exp (x)\cdot \frac{\exp (y-x)-1}{y-x},$$

so

$$\underset{y\to x}{lim}\frac{\exp (y)-\exp (x)}{y-x}=\exp (x)\underset{y\to x}{lim}\frac{\exp (y-x)-1}{y-x}=\exp (x)\underset{y\to 0}{lim}\frac{\exp y-1}{y}.$$

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

$$\underset{y\to x}{lim}\frac{\exp (y)-\exp (x)}{y-x}=\exp (x).$$

∎