The formula for the derivative of the natural exponential function
f(x)=ex was worked out in
previous parts of this exploration, and
the formula says that f '(x)=ex.
This is a curious relationship, though -- f(x) and
f '(x) are both the same function,
f(x)=f '(x)=ex.
This relationship
can be expressed as a differential equation by noticing that
if y=f(x)=ex,
then y '=f '(x)=ex
=f(x)=y, or in a more concise form,
y '=y. For this differential equation,
y '=y, the function
y=f(x)=ex is a
particular solution, but there are other solutions. For example,
if y=g(x)=3ex, then
y '=g '(x)=3ex=y
also. In fact, the general solution to the differential
equation y '=y has the form
y=Cex, where C is a constant.
The value of C can be determined by knowing the position of
one point on the graph of the solution function, as can be
seen with the differential equation y '=y in
the following applet:
How to use this applet
However, in the first part of this exploration,
the graphs of exponential functions required two points to
determine the graph -- this was because the equation there,
y=Cekx also included the undetermined
constant k, and a second point was needed to determine both
C and k. For a function
h(x)=ekx, the derivative
h '(x) can be computed using the above formula
for the derivative of f(x)=ex,
along with the Chain Rule for derivatives, to get
h '(x)=kekx, which
satisfies the differential equation y '=ky.
The general solution of the differential equation y '=ky
is y=Cekx, the same equation as in the
first part of this exploration.
As examples, look at
y '=2y (k=2) and
y '=y/2 (k=1/2)
in the above applet.
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