# Why is d/dxe^x=e^x?

Dec 10, 2015

This follows from the definition of natural logarythm and its inverse.

#### Explanation:

$\ln {e}^{x} = x$

$\frac{d}{\mathrm{dx}} \ln {e}^{x} = \frac{d}{\mathrm{dx}} x$

$\frac{1}{e} ^ x \frac{d}{\mathrm{dx}} {e}^{x} = 1$

$\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

Dec 10, 2015

The "why" depends on how you've defined ${e}^{x}$.

#### Explanation:

Define $\ln x$ first

One approach is to define $\ln x = {\int}_{1}^{x} \frac{1}{t} \mathrm{dt}$ and
then to define $\exp \left(x\right)$ to be the inverse of $\ln x$ and,
finally, define ${e}^{x} = \exp \left(x\right)$.

In this case $y = {e}^{x}$ if and only if $\ln y = x$.

Differentiating implicitly gets us $\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$.

So, $\frac{\mathrm{dy}}{\mathrm{dx}} = y = {e}^{x}$

Define ${e}^{x}$ independently of $\ln x$

Definition 1

For positive $a$, define

${a}^{x} = {\lim}_{r \rightarrow x} {a}^{x}$ with $r$ in the rational numbers
(We owe you a proof that this is well-defined.)

Then, using the definition of derivative:

$\frac{d}{\mathrm{dx}} \left({a}^{x}\right) = {\lim}_{h \rightarrow 0} \frac{{a}^{x + h} - {a}^{x}}{h} = \lim \left(h \rightarrow 0\right) \left({a}^{x} \frac{{a}^{h} - 1}{h}\right)$

We then define $e$ to be the number that satisfies ${\lim}_{h \rightarrow 0} \frac{{e}^{h} - 1}{h} = 1$

With this definition we get
$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {\lim}_{h \rightarrow 0} \frac{{e}^{x + h} - {e}^{x}}{h} = {\lim}_{h \rightarrow 0} \left({e}^{x} \frac{{e}^{h} - 1}{h}\right)$

$= {e}^{x} {\lim}_{h \rightarrow 0} \left(\frac{{e}^{h} - 1}{h}\right) = {e}^{x}$

Definition 2

e^x = 1+x/1+x^2/(2*1)+x^3/(3*2*1) + x^4/(4!)+x^5/(5!)+ * * *

(n! = n(n-1)(n-2)* * * (1))
(We owe you a proof that this is well defined.)

Differentiating term by term (we owe you a proof that this is possible), we get

d/dx(e^x) = 0+1+(cancel(2)x)/(cancel(2)*1)+(cancel(3)x^2)/(cancel(3)*2*1) + x^3/(3!)+x^4/(4!)+* * *

Which simplifies to 1+x/1+x^2/(2*1)+x^3/(3*2*1) + x^4/(4!)+x^5/(5!)+ * * *  which is again ${e}^{x}$