Quiz 9 in MATH 211: Derivatives of Exponential and Logarithmic Functions, Quizzes of Mathematics

Download Quiz 9 in MATH 211: Derivatives of Exponential and Logarithmic Functions and more Quizzes Mathematics in PDF only on Docsity! MATH 211 QUIZ 9 NAME Prof. J. Beachy Friday, 11/8/02 Circle C1(T10) C2(T11) C3(Th11) These are the new formulas you need to know, in addition to the product rule, quotient rule, and extended power rule. Also remember the chain rule: d dx f(u(x)) = f ′(u(x))u′(x) Exponentials: d dx ex = ex d dx eu(x) = eu(x) u′(x) or d dx eu(x) = u′(x)eu(x) Natural logs: d dx lnx = 1 x d dx lnu(x) = 1 u(x) u′(x) or d dx lnu(x) = u′(x) u(x) You should also remember these useful rules for exponents and natural logs: eu(x)ev(x) = eu(x)+v(x) ( eu(x) )k = eku(x) ln (u(x)v(x)) = lnu(x) + ln v(x) ln ( u(x) v(x) ) = lnu(x)− ln v(x) lnu(x)k = k lnu(x) 1. (14 pts) Find these derivatives: (a) (4.1 #57) d dx ( e3x + 1 )5 = 5(e3x + 1)4 · d dx e3x = 5(e3x + 1)4 · e3x · 3 = 15e3x (e3x + 1)4 (b) (4.1 #63) d dx ( e √ x + √ ex ) = d dx ( ex 1/2 + (ex)1/2 ) = d dx ex 1/2 + d dx e 1 2x = ex 1/2 · 1 2 x−1/2 + e 1 2x · 1 2 = e √ x 2 √ x + 1 2 e 1 2x (c) (4.2 #53) d dx (ln(ln 4x)) = 1 ln 4x · d dx ln 4x = 1 ln 4x · 1 4x · 4 = 1 x ln 4x (d) (4.2 #84) d dx ( ln √ 5 + x2 ) = d dx ( ln(5 + x2)1/2 ) = d dx ( 1 2 ln(5 + x2) ) = 1 2 · 1 5 + x2 · 2x = x 5 + x2 2. (6 pts; 4.3 #24) The cost of a first-class stamp in 1962 was 4 cents. In 1999, the cost was 33 cents. Assume that this increase can be described as exponential growth. (a) Write down the equation for exponential growth. P (t) = P0ekt (b) Find the constants in your equation for exponential growth. (Write an equation for the growth rate k in terms of the given information, and solve for k, but do not try to get a numerical answer, which would require a calculator.) Let t = 0 in 1962. Then P0 = 4, and we have the equation P (t) = 4ekt as the first step. In 1999 the cost was 33 cents, and t = 1999− 1962 = 37. Substitute these values into the equation: 33 = 4ek·37 33 4 = ek·37 ln ( 33 4 ) = ln ( ek·37 ) = k · 37 k = 1 37 ln ( 33 4 )