Friday, February 5, 2010

Exponential Functions

I'm getting feedback that exponential functions are giving you extra trouble. I'd appreciate getting feedback to help know how I can clarify the concepts. Here is a summary of some of the key concepts that I'm wanting you to understand:
  • bx is not just a formula "b to the power x" but is a new function, which I'm asking you to call expb(x).
  • The properties of exponents like bx+y=bx by and (bx)y=bxy become properties of the exponential functions.
    expb(xy)= [expb(x)]y

  • Logarithms are the inverse functions of exponential functions.

    In formula representation, these are written as follows:
  • Whenever you see a formula with an exponential, say b3x-2, you should be able to think in both formula and function modes interchangeably.
    b3x-2 = b3x b-2 = (b3)x b-2
    b3x-2 = expb(3x-2)
    The first mode allows us to recognize that b3x-2t is actually of the form Aqx where A=b-2 and q=b3. The second mode is useful to remind us that we really have a composition when we need to compute a derivative or when dealing with inverse functions.
  • There is a special base e that is most important because the corresponding exponential function is its own derivative.

    The natural exponential is written without a base. The corresponding inverse function is called the natural logarithm, ln(x).

    In formula representation, these are written as follows:
    The x in these formulas, as always, is a placeholder. So any number or formula could be used in place of x.
  • Any exponential can be written in terms of the natural exponential. The key is to use the properties of exponents.
    b = eln(b)
    bx = [eln(b)]x = eln(b) x
    Another way to think of this is using composition of inverse functions: exp(ln( ))
    bx = exp(ln(bx))
    ln(bx) = x ln(b)
    bx = exp(x ln(b)) = ex ln(b)

    That is, by replacing bx by eln(b) x, we can write any exponential A bx in the form A ekx where k is the number ln(b).
  • Derivatives of exponentials use the basic property that exp'(x) = exp(x), or d/dx[ex] = ex. Usually, this also requires the chain rule: d/dx[eu] = eu u'.
    d/dx[e2x] = e2x (2) = 2e2x (u = 2x)
    d/dx[e-3x] = e-3x (-3) = -3e-3x (u = -3x)
    d/dt[e(t2-5t)] = e(t2-5t) (2t-5) = (2t-5)e(t2-5t) (u = t2-5t)

So this was a moderately long list. Perhaps just re-reading it helped you understand something better. Or perhaps you realize something is still confusing. Please post comments explaining why you are finding problems challenging or perhaps explaining what helped you suddenly understand what you had missed earlier.