University:
University of California, Berkeley
Course:
MATH 56 Linear Algebra
Academic year:

2023
Views:

187

Pages:

2
Author:

Clint W.

Derivatives of Exponential and Logarithmic Functions In this section we’d like to consider the derivatives of exponential and logarithmic functions. Consider a dynamical system for bacteria population, with a closed form solution given by b(t) = 2t . In order to find b0 (t), we’ll need to return to the definition of the derivative. 2t+∆t − 2t 2t 2∆t − 2t 2∆t − 1 b(t + ∆t) − b(t) = lim = lim = 2t lim ∆t→0 ∆t→0 ∆t→0 ∆t→0 ∆t ∆t ∆t ∆t b0 (t) = lim Above we were able to factor out the 2t from the limit because the limit only depends on ∆t, so t ∆t is essentially a constant. Now our task is to evaluate the limit lim∆t→0 2 ∆t−1 , which we cannot do yet, without the help of the chain rule. However, we could numerically approximate the limit and see that it approaches ln(2) ≈ 0.693. Thus, we find that b0 (t) = ln(2) · 2t ≈ 0.693 · 2t The derivative of this exponential function is just a constant times the function itself. Using this observation, that the derivative of an exponential function is just a constant times the exponential function, we can make the following, clever definition. Definition of e The number e is defined as the number such that e∆x − 1 =1 ∆x→0 ∆x lim We can follow a procedure completely analagous to that which we did for the exponential function 2t , to find that d x e∆x − 1 e = ex lim = ex ∆x→0 dx ∆x The reason for the above definition of e should now be clear. The derivative of the exponential function is just the function itself. In fact, the only functions with this property are functions that are constant multiples of e. It follows immediately that d2 x e = ex dx2 and in fact dn x e = ex dxn where n ∈ Z+ (is a positive integer). If this is not clear, find the derivative of ex , and then find the derivative of that derived function, and so on. Example 1 Find g 0 (x) if g(x) = Kex , where K is a constant. Solution Since g(x) is just a constant multiple of ex , it follows immediately that g 0 (x) = g(x) = Kex . Example 2 Find p0 (x) if p(x) = xex . Solution In this case we need to apply the product rule with f (x) = x and g(x) = ex . It follows that p0 (x) = f 0 (x)g(x) + f (x)g 0 (x) = 1 · ex + xex = (1 + x)ex Example 3 Find all solutions to the differential equation dg = g(x) dx Solution In order to solve the above equation, we need to find all functions for which the derivative of the function is equal to the function itself. This family of function consists of all constant multiples of ex . Thus, Kex is a solution to the above equation, for all K ∈ R (that are real numbers). Now that we have considered the exponential function, let us consider its inverse, the natural logarithm. We do not yet possess the tools to calculate this derivative, but we soon will. For now it will have to suffice to say that 1 d ln(x) = dx x Since this is a power function, namely x−1 , we can apply the power rule to find that d2 1 ln(x) = − 2 2 dx x ln(x) The graphs of these functions are as follows 3 2 1 0 -1 -2 -3 0 2 4 6 8 10 6 8 10 6 8 10 1/x x 10 8 6 4 2 0 0 2 4 -1/x^2 x 0 -2 -4 -6 -8 -10 0 2 4 x Example 4 Find the derivative of x ln(x). Solution We need to apply the product rule in this situation, to find that d d d 1 (x ln(x)) = x · ln(x) + x · ln(x) = x · + 1 · ln(x) = 1 + ln(x) dx dx dx x