1. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 55 § 4.2 The Exponential Function e x

2. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 55  e  The Derivatives of 2 x, b x, and e x Section Outline

3. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 55 The Number e DefinitionExample e: An irrational number, approximately equal to 2.718281828, such that the function f (x) = b x has a slope of 1, at x = 0, when b = e

4. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 55 The Derivative of 2 x

5. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 55 Solving Exponential EquationsEXAMPLE SOLUTION Calculate.

6. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 55 The Derivatives of b x and e x

7. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 55 Solving Exponential EquationsEXAMPLE SOLUTION Find the equation of the tangent line to the curve at (0, 1). We must first find the derivative function and then find the value of the derivative at (0, 1). Then we can use the point-slope form of a line to find the desired tangent line equation. This is the given function. Differentiate. Use the quotient rule.

8. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 55 Solving Exponential Equations Simplify. CONTINUED Factor. Simplify the numerator. Now we evaluate the derivative at x = 0.

9. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 55 Solving Exponential EquationsCONTINUED Now we know a point on the tangent line, (0, 1), and the slope of that line, -1. We will now use the point-slope form of a line to determine the equation of the desired tangent line. This is the point-slope form of a line. (x 1, y 1 ) = (0, 1) and m = -1. Simplify.