General exponential functions, of the form f(x)=Ce^kx, can be graphed using just two points (x₀,y₀) and (x₁,y₁) on the graph of the function.

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If the two points are on opposite sides of the x-axis, then there is no graph drawn. This is because of the relationship between the constants, C and k, and the points, (x₀,y₀) and (x₁,y₁), on the graph.

In the equation y=Ce^kx, the two constants C and k determine the shape of the graph. These two values are fixed, given two points (x₀,y₀) and (x₁,y₁) on the graph. To see how these two points determine the values of C and k, plug the two points into the equation separately:

Now solve each of these equations for C:

Since both equations represent C in terms of the other variables, set the right hand sides equal --- this eliminates the constant C from the equations:

Now solve this equation for k -- first, rewrite the equation a bit:

Now take natural logarithms on both sides:

This is where it is important that the two points (x₀,y₀) and (x₁,y₁) be on the same side of the x-axis -- in other words, that y₀ and y₁ have the same sign, so that y₁/y₀ is positive and the logarithm on the left hand side is defined. If y₀ and y₁ have the same sign (which can be checked with the inequality y₀y₁>0) then ln(y₁/y₀) =ln(|y₁|)-ln(|y₀|) by properties of logarithms, with the absolute values guaranteeing that the logarithms will be defined even if y₀ and y₁ are negative. Some care must be taken at this point to check the inequality y₀y₁>0, though, since this expression with the absolute values no longer requires y₀ and y₁ to have the same sign.

Continuing to solve for k, the equation now looks like:

Solving this algebraically for k:

It is interesting to notice that the above equation resembles the slope of a line.

Now that the value of k is known (from the points (x₀,y₀) and (x₁,y₁)), finding C is easier: plug either point into the equation y=Ce^kx -- plugging in (x₀,y₀) gives y₀=Ce^kx₀. Now divide both sides by e^kx₀ to solve for C:

These formulas for C and k depend only on the two points (x₀,y₀) and (x₁,y₁).

© 2001 by Tom Leathrum -- additional copyright information

Published January 2001

To compute the derivative of the function f(x)=e^x, use the definition of the derivative as a limit of a difference quotient:

This limit can be simplified using properties of exponents:

Now, in the numerator, factor out the common e^x:

Since the limit is with respect to the variable h, and the variable x is independent of h, the e^x may be treated as a constant factor and removed completely from the limit:

The remaining limit (not including the e^x factor) is the critical limit for finding the derivative of f(x)=e^x -- if this limit exists and is equal to K, then f '(x)=K e^x. It remains only to find K (if it exists). For the purposes of plugging into the applet below to find K, rewrite the limit in terms of the variable x instead of h:

Now use the tables in the applet below to approximate the value of K.

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From these tables, it seems that K=1. This is an approximation to K using the values in the tables from the applet, but it seems to be a good approximation, given the trends in the tables. If this is the case, then (from above)

This formula is important enough to bear repeating:

With the formula for the derivative of f(x)=e^x giving f '(x)=e^x, the derivative can be used to find slopes of tangent lines to the graph of the function f(x)=e^x. At a point (x₀,y₀) on the graph of f(x) (so that y₀=f(x₀)), the line tangent to the graph will have slope m=f '(x₀). Plugging into the Point-Slope form equation for a line, then, the equation for the tangent line at (x₀,y₀) will be:

This is the equation used to find the tangent lines to the graph in the applet below. Using the functions f(x)=e^x and f '(x)=e^x, the applet shows the tangent line for any given value of x₀ in the graph, starting below with x₀=1.

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Other values for x₀ (shown in the "x=" field in the applet) can be chosen either by entering a new value into the "x=" field or by clicking and dragging the mouse on the graph.

One very common mistake in computing the derivative of the exponential f(x)=e^x is to use the Power Rule for derivatives, rather than the formula f '(x)=e^x. To see why the Power Rule does not work, consider what the resulting derivative would be, using the Power Rule: f '(x)=x e^x-1 (please keep in mind that this is incorrect). Using the functions f(x)=e^x and f '(x)=x e^x-1 in the same applet as on the previous page, it is easy to see that the values of this "derivative" do not give slopes of tangent lines to the graph of f '(x):

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So be careful, when computing the derivative of an exponential function, to use the correct formula:

The formula for the derivative of the natural exponential function f(x)=e^x was worked out on Page 2  of this exploration, and the formula says that f '(x)=e^x. This is a curious relationship, though -- f(x) and f '(x) are both the same function, f(x)=f '(x)=e^x. This relationship can be expressed as a differential equation by noticing that if y=f(x)=e^x, then y '=f '(x)=e^x =f(x)=y, or in a more concise form, y '=y. For this differential equation, y '=y, the function y=f(x)=e^x is a particular solution, but there are other solutions. For example, if y=g(x)=3e^x, then y '=g '(x)=3e^x=y also. In fact, the general solution to the differential equation y '=y has the form y=Ce^x, 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:

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However, on Page 1 of this exploration, the graphs of exponential functions required two points to determine the graph -- this was because the equation there, y=Ce^kx also included the undetermined constant k, and a second point was needed to determine both C and k. For a function h(x)=e^kx, the derivative h '(x) can be computed using the above formula for the derivative of f(x)=e^x, along with the Chain Rule for derivatives, to get h '(x)=ke^kx, which satisfies the differential equation y '=ky. The general solution of the differential equation y '=ky is y=Ce^kx, the same equation as on Page 1 of this exploration. As examples, look at y '=2y (k=2) and y '=y/2 (k=1/2) in the above applet.