Mt020.02 Slide 1 on 4/14/00 Differentiating Exponential Functions Lots of real world processes have behavior which can be modeled only by exponential functions. Growth or depreciation of an investment, population growth, epidemic spread, and radioactive decay all behave according to functions involving exponentials. As we study these processes, we are interested in how rapidly they change and so studying their derivatives seems a natural next step. What is remarkable about the (natural based) exponential function is that this function is its own derivative. So we say (without proof) If f(x) = ex, then f’(x) = ex . You might want to check out the textbook for a more detailed treatment of this result. raj Mt020.02 Slide 2 on 4/14/00 In graphic terms this means that the curve y = ex passes through the point ( x , ex ) which is obvious, but it does so with a slope of ex, since the derivative provides the tangent’s slope. Example: Find the point along the curve y = ex at which the tangent line has slope = 20. Solution: Since the slope of the tangent is the derivative which is just ex, we solve 20 = ex for the variable x. Applying the natural logarithm (ln) to both sides we obtain ln(20) = ln(ex) = x showing that x = ln(20) = 2.995 and the point in question is ( 2.995 , e2.995 ) = ( 2.995 , 20 ). But the biggest impact of this differentiation formula if felt when we mix this fact with the chain rule. Recall that if h(x) = f(g(x)) then we have h’(x) = f’(g(x)) ´ g’(x). raj Mt020.02 Slide 3 on 4/14/00 Earlier applications of the chain rule were limited to the cases where the second function to be applied (the “f” in the expression f(g(x)) ) was a power function. It became our power rule. 3 Example: Suppose that h(x) = e(x - 3) Find h’(x). Solution : Here h(x) = f(x3 – x) where f(x) = ex. So we get h’(x) = f’(x)( x3 – x) ´ ( x3 – x)’. Since f(x) = ex, we know f’(x) = ex too, so, 3 f’ (x3 – x) = e(x - 3) and thus 3 3 h’(x) = e(x - 3)(3x2 – 1) = (3x2 – 1) e(x - 3). English Equivalent <memorize this> “The derivative of e to any power is e to that very same power times the derivative of that power.” raj Mt020.02 Slide 4 on 4/14/00 Example: Differentiate y = x 5e 2x . Solution: Use the product rule first: y’ = [ x5 ]’ e2x + x5 [e2x ]’ = 5x4 e2x + x5 (e2x)(2) = x4e2x ( 5 + 2x ) Here is another variation: 5 Example: Find f’(x) if f(x) = (12e x + x ) . Solution: Here the power rule comes first: f’(x) = 5(12 e x + x ) 4 (12 e x + x )' 1 - x 4 x 1 2 =5(12 e + x ) (12 e + 2 ( x ) ) 1 = 5(12 e x + x ) 4 (12 e x + ) 2 x raj Mt020.02 Slide 5 on 4/14/00 Example: Find all the critical points of the e x graph of the equation y = . e x +1 Solution : Quotient rule: dy (e x +1)e x - e x (e x ) ex = 2 = 2 dx (ex + 1) (e x +1) Critical points happen when the derivative is undefined (no such place with this derivative) or where the derivative is equal to zero? When is this derivative zero? So where are the critical points of this function? Can you say anything about intervals where this function is increasing or decreasing? raj Mt020.02 Slide 6 on 4/14/00 Example: What are the extreme values (maximum and minimum) of this continuous function on the interval [ -2 , 2 ]? 2 f(x) = e(x - 4) Solution: raj Mt020.02 Slide 7 on 4/14/00 Example: The Lady Bug (a woman’s clothing chain) found that t days after the end of a sales promotion, the volume of sales was given by the function S(t) = 20,000(1+e-0.5t) in thousands of dollars for 0 £ t £ 5. Find the rate of change in sales volume when t = 1, 2, 3, 4. Solution : raj.