1. Derivatives of Exponentials Rates of Change for Quantities that Grow or Decrease in an Exponential Way Dr. E. Fuller Dept of Mathematics WVU

2. Exponential Growth/Decrease  Recall that a quantity is “exponential” if its value increases or decreases as a product of a fixed base multiplied by itself some number of times  Value = (initial amount)b £ b £ b £  £ b

3. Change in e x  So how does change occur in exponential cases?  In class we saw that if f(x) = e x then for every x  So we can approximate change for values of x by approximating this expression

4. Approximating Change  Listed below are approximations of f 0 (x) for x between –3 and 3 (h=.0001)  What do you notice about this graph? (e h -1)/h as before

5. The Derivative of e x  If f(x) = e x, then f 0 (x) = e x  So the value of the slope at any point is equal to the value of the function  e x is the only function for which this is true

6. Proportional Change  What this means is that the change in f is a constant multiple of the value of f  We can change the exponent slightly to see this effect  Ex: f(x) = e 2x f 0 (x) ¼ 2f(x)

7. A General Formula  In general, for f(x) = e kx where k is a constant, f 0 (x) = ke kx.  In other words, for f(x) = e kx, f 0 = kf (the change in f is proportional to the value of f)  Ex: f(x) = e.25x then f 0 (x)=.25e.25x  Ex: f(x) = e -2x, then f 0 (x) = -2e -2x

8. A General Fact  More surprising, it works the other way as well: If f 0 (x) = kf(x) for some constant k, then f(x) = Ae kx for some constant A. Note that A = f(0) (the initial value of f)  So if you know a quantity is changing at a rate proportional to its value, that quantity must be exponential

9. The Chain Rule for e u(x)  The most general rule for exponentials says that if f(x) = e u(x), then f 0 (x) = e u(x) u 0 (x)  This is the Chain Rule for exponential derivatives. All the previous examples are special cases of this.  To test your understanding, work the Exercises Exercises

10. An Example  Ex: If f(x) = e x 2, find f 0 (2).  Solution: f 0 (x) = e x 2 (2x) = 2xe x 2 and so f 0 (2) = 2(2)e 2 2 =4e 4