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Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 50

Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 50. A major complaint of professors teaching calculus is that students don't have the appropriate background to work through the calculus course successfully. This text is targeted directly at this underprepared audience. This is a single-variable (2-semester) calculus text that incorporates a conceptual re-introduction to key precalculus ideas throughout the exposition as appropriate. This is the ideal resource for those schools dealing with poorly prepared students or for schools introducing a slower paced, integrated precalculus/calculus course | Exploratory Problem for Chapter 14 471 y logb x by x ln by In x y ln b In x ln x y Inb In x 1 log x ---- -------In x b In b In b inb is simply a constant. We know that the derivative of a constant times x is that constant times x so ln x ln x i. dx ln b ln b dx ln b x Conclusion d 1 1 logb x --------. dx ln b x The derivative oflogb x is just a constant times 1. For example log x jo because log x lnio ln x Notice that there is no escaping the natural logarithm. It pops its head3 right up into the derivative of the log base b of x no matter what the value of b EXAMPLE 14.1 Find the equation of the line tangent to x log3 27x2 at the point 1 3 . SOLUTION The slope of the tangent line at x 1 is given by 1 . The equation of the line is y - y1 m x - x1 or y - 3 1 x - 1 . x log3 27 log3 x2 3 log3 x2. For positive x x 3 2 log3 x. We need to differentiate this and evaluate the derivative at x 1. Recall 3 2 log3 x says take the derivative of 3 2 log3 x x x 1 and evaluate it at x 1. - 3 2 log3 x x x 1 1 1 ln 3 x _ 2 ln 3 x1 0 2 Therefore the equation of the tangent line is 2 2 2 y - 3 x - 1 or y x - 3. ln 3 ln 3 ln 3 This is a linear equation jny is a constant. PROBLEMS FOR SECTION 14.1 In Problems 1 through 7 find yf. 3 Of course we could write 7 in place of ln . In particular ln 10 og 10 So - log x 1 log e 1 but e log e log e log e x ln 10 x K x is involved however we present the derivative. 472 CHAPTER 14 Differentiating Logarithmic and Exponential Functions 1. y 2 ln 5x 2. y n ln x 3. y 4. y x In x 5. y 6. y 3 log x 7. y 8. Show that f x 1 1 23x 3 is invertible. Find f 1 x . 9. Show that g x n log2 nx - n2 is invertible. Find g-1 x . 10. Find and classify the critical points of f x x ln x. 11. What is the lowest value taken on by the function g x x2 ln x Is there a highest value Explain. 12. Use a tangent line approximation of ln x at x 1 to approximate a ln 0.9 . b ln 1.1 . 13. Graph f x x ln x indicating all local maxima minima andpoints of inflection. Do this without .