1-page summary sheet:

The Exponential Function

• Essential Property: The rate of change [slope] of the function is proportional to the function itself:
  
  $${dy \over dx} \; = \; k \; y$$
  
  

• Solution: The function satisfying this property is
  
  $$y(x) \; = \; y_0 \; e^{kx} \qquad \hbox{\rm [$y_0$ is the value of $y$ at $x = 0$] }$$
  
  
  
  
  $$\hbox{\rm Thus} \qquad {d(e^{kx}) \over dx} \; = \; k \; e^{kx}$$
  
  

  Note: the derivation on the handout defines the function  $\exp(x)$  in terms of a power series, but it does not justify the claim that this function is equivalent to the number   $e \equiv 2.718\cdots$  raised the the   $x^{\rm th}$  power, of which we make much use. Consult your Math text for a proof of this equivalence.

• Power of a Product: ${\displaystyle e^{kx} \; = \; [e^x]^k }$ so it suffices to consider just the simplest case, ${\displaystyle e^x }$

• Negative Powers: ${\displaystyle e^{-x} \; = \; {1 \over e^x} }$

• Limits:
  
  $$e^0 \; = \; 1$$
  
  
  
  
  $$e^{x \to \infty} \longrightarrow \infty$$
  
  
  
  
  $$e^{x \to -\infty} \longrightarrow 0$$
  
  

• Natural Logarithms:
  
  $$e^{\ln(x)} = x \qquad \qquad \ln(e^x) = x \qquad \qquad {d [\ln(x)] \over dx} \; = \; {1 \over x}$$
  
  

• Ratios:
  
  $$e^{-x} = {1\over2} \qquad \hbox{\rm when} \qquad x = \ln(2) = 0.693$$
  
  
  
  
  $$e^{-x} = {1\over4} \qquad \hbox{\rm when} \qquad x = \ln(4) = 2 \ln(2) = 2 \times 0.693 = 1.386$$
  
  
  • ${\displaystyle e^{-x} }$ decreases by a (multiplicative) factor of ${\displaystyle {1 \over 2} }$ each time the "argument" x increases by an (additive) amount 0.693
  • ${\displaystyle e^{-x} }$ decreases by a multiplicative factor of ${\displaystyle {1 \over e} = 0.368 }$ each time the "argument" x increases by 1.