Techniques of Differentiation 11 Techniques of Differentiation with Applications

11.5 Derivatives of Logarithmic and Exponential Functions

Derivatives of Logarithmic and Exponential Functions Derivative of the Natural Logarithm Quick Example Recall that ln x = logex. Derivative of a constant times a function

Derivatives of Logarithmic and Exponential Functions Derivative of the Logarithm with Base b Quick Example Notice that, if b = e, we get the same formula as previously.

Derivation of the formulas [ln x] = and [logbx] =

Derivation of the formulas [ln x] = and [logbx] = To compute [ln x], we need to use the definition of the derivative. We also use properties of the logarithm to help evaluate the limit. Definition of the derivative Algebra Properties of the logarithm

Derivation of the formulas [ln x] = and [logbx] = which we rewrite as As h → 0+, the quantity x/h is getting large and positive, and so the quantity in brackets is approaching e, which leaves us with Algebra Properties of the logarithm

Derivation of the formulas [ln x] = and [logbx] = The rule for the derivative of logbx follows from the fact that logbx = ln x/ln b. If we were to take the derivative of the natural logarithm of a quantity (a function of x), rather than just x, we would need to use the chain rule.

Derivation of the formulas [ln x] = and [logbx] = Derivatives of Logarithms of Functions

Derivation of the formulas [ln x] = and [logbx] = Quick Example u = x2 + 1

Example 1 – Derivative of Logarithmic Function Find Solution: The calculation thought experiment tells us that we have the natural logarithm of a quantity, so

Derivation of the formulas [ln x] = and [logbx] = The chain rule, gives us the following formulas.

Derivation of the formulas [ln x] = and [logbx] = Quick Example u = x2 + 1

Derivation of the formulas [ln x] = and [logbx] = The following shows the correct way of differentiating bx, beginning with a special case. Derivative of ex Quick Example

Derivation of the formulas [ln x] = and [logbx] = For bases other than e, we have the following generalization: Derivative of bx If b is any positive number, then Note that if b = e, we obtain the previous formula. Quick Example

Derivation of the formula [ex] = ex

Derivation of the formula [ex] = ex To find the derivative of ex we use a shortcut. Write g(x) = ex. Then ln g(x) = x. Take the derivative of both sides of this equation to get or g (x) = g(x) = ex.

Derivation of the formula [ex] = ex Derivatives of Exponentials of Functions

Derivation of the formula [ex] = ex Quick Example u = x2 + 1

Applications

Example 4 – Epidemics In the early stages of the AIDS epidemic during the 1980s, the number of cases in the United States was increasing by about 50% every 6 months. By the start of 1983, there were approximately 1,600 AIDS cases in the United States. Had this trend continued, how many new cases per year would have been occurring by the start of 1993? Solution: To find the answer, we find that t years after the start of 1983 the number of cases is A = 1,600(2.25t ).

Example 4 – Solution cont’d We are asking for the number of new cases each year. In other words, we want the rate of change, dA/dt: = 1,600(2.25)t ln 2.25 cases per year. At the start of 1993, t = 10, so the number of new cases per year is = 1,600(2.25)10 ln 2.25 ≈ 4,300,000 cases per year.