Math 10A, Summer, 2001, Lecture 12 (REVISED)Theorem 1 For any x,ddx(sin x) = cos x (1)ddx(cos x) = −sin x (2)Two facts:PRQOPQOR bisects and isperpendicular to QPlimQ→PP Q_P Q= 1Proof of (1)ddx(sin x) = lim∆x→0sin(x + ∆x) − sin x∆xsin(x + ∆x) − sin x = RQα = x +12∆xsin(x + ∆x) − sin x∆x=RQ_P Q≈RQP Q= cos α → cos x as ∆x → 0uvP = (cos x, sin x)Q = (cos(x + ∆x), sin(x + ∆x))R∆xαx1Math 10A, Summer, 2007 Lecture 12 (REVISED), p. 2Example 1 What is the derivative ofddx(5 sin x − 6 cos x)?Answer:ddx(5 sin x − 6 cos x) = 5 cos x + 6 sin xExample 2 (a) Find the linear function that best approximates f(x) = 5 sin x−6 cos x near x = 2.(b) Generate the graphs of the function and its linear approximation in the window−3 ≤ x ≤ 8, −12 ≤ y ≤ 12.Answer: (a) The linear function that best approximates y = f(x) near x = 2 is the function y = f(2)+f0(2)(x−2) whose graph is the tangent line to y = f(x) at x = 2. • y = [5 sin(2) − 6 cos(2)] + [5 cos(2) + 6 sin(2)](x − 2)(b) Figure A2x2 6y−10510y = 5 sin x − 6 cos xFigure A2Find t h e derivatives in Examples 3 through 5.Example 3 dy/dx for y = sin2x + 1Answer:dydx=ddx(sin2x + 1) =ddx[(sin x)2+ 1] = 2 sin xddx(sin x) + 0 = 2 sin x cos xExample 4ddx(x2cos x)Answer:ddx(x2cos x ) = x2ddx(cos x) + cos xddx(x2) = −x2sin x + 2x cos xExample 5ddxsin xxAnswer:ddxsin xx=xddx(sin x) − sin xddx(x)x2=x cos x − sin xx2Theorem 2 For any x where tan x is defined,ddx(tan x) =1cos2x= sec2x (3)Lecture 12 (REVISED), p. 3 Math 10A, Summer, 2007Proof:ddx(tan x) =ddxsin xcos x=cos xddx(sin x) − sin xddx(cos x)cos2x=cos2x + sin2xcos2x=1cos2x= sec2xExample 6 Find t h e derivative of y = tan x + x2at x = 1.Answer: y0= sec2x + 2x • y0(1) = sec2(1) + 2Example 7 Give an equation of the tangent line to y = tan x at x = 0. Then generate the curveand tangent line on a calculator.Answer: The tangent line to y = y(x) at x = 0 is y = y(0) + y0(0)x. • y(x) = tan x and y0(x) = sec2x •y(0) = tan(0) = 0 and y0(0) = sec2(0) = 1/ cos2(0) = 1 • Tangent line: y = x • Figure A7xy−2232π−32πy = tan xFigure A7Other derivativesSimilar calculations yield the other three of th e formulas,ddx(sin x) = cos xddx(cos x) = −sin xddx(tan x) = sec2xddx(cot x) = −csc2xddx(sec x) = sec x tan xddx(csc x) = −csc x cot xNotice that the formulas on the right can be obtained from those on the left by interchanging the functions(sine, tangent, secant) with the cofunctions (cosine, cotangent, cosecant) and adding minus signs.Math 10A, Summer, 2007 Lecture 12 (REVISED), p. 4Derivatives of y = sin–1x and y = tan–1xTheorem 3 (a) For any x with −1 < x < 1,ddx(sin–1x) =1p1 − x2(4)(b) For any x,ddx(tan–1x) =11 + x2(5)Proof: (a) Set y = sin–1x with −1 < x < 1. Then sin y = x, and by the general Chain Rule to becovered in Lecture 13,ddx(sin y) =dydx(x)cos ydydx= 1p1 − sin2ydydx= 1p1 − x2dydx= 1dydx=1p1 − x2(b) Set y = tan–1x with any x. Then tan y = x, and and by the general Chain Rule to be covered inLecture 13,ddx(tan y) =dydx(x)sec2ydydx= 1(1 + tan2y)dydx= 1(1 + x2)dydx= 1dydx=11 + x2Example 8 What is the derivative of y = sin–1x at x =12?Answer:ddx(sin–1x)x=0.5="1p1 − x2#x=1/2=1p1 − (1/2)2=23√3Lecture 12 (REVISED), p. 5 Math 10A, Summer, 2007Example 9 Give an equation of the tangent line to y = tan–1x at x = 2. Then generate the curveand the tangent line in the window −5 ≤ x ≤ 5, −2 ≤ y ≤ 2 on a calculator.Answer: y = tan–1x • y0=11 + x2• y(2) = tan–1(2) • y0(2) =11 + 22=15•Tangent line: y = tan–1(2) +15(x − 2) • Figure A9x2Y1tan–1xFigure A9Interactive examplesFor more practice, see Interac tive Examples 1, 2, and 5 in Section 3.5 and Example 1 of Section 3.6 onthe web page http//www.math.ucsd/∼ashenk/.Homework 12Find t h e derivatives in Exercises 1 through 8.1.ddx(x2+ 8 sin x − 2)2.ddx(√cos x)3. dy/dx for y = x2sin x4. y0(x) for y = 5 cos x − 3 sin x5. y0(0) for y(x) = 6 sin x + ex6. dy/dx for y = ln x − cos x7. y0(3) for y = x2tan x8.ddx(x tan–1x)9. Give an equation of the tangent line to y = tan2x at x =14π. Then generate the curve andtangent line on a calculator.More practiceFind t h e derivatives in Exercises 10 through 12.10.ddx[(sin–1x)2]11. y0(1) for y = cos4x12. y0(x) for y = 2 tan x − 3 sin x13. Give an equation of t h e tangent line to y = sin x + cos x at x =12π. The n generate the curve andtangent line on a calculator.Answers1ddx(x2+ 8 sin x − 2) =ddx(x2) + 8ddx(sin x) −ddx(2) = 2x + 8 cos x2.ddx√cos x =ddx[(cos x)1/2] =12(cos x)−1/2ddx(cos x) = −12(cos x)−1/2sin xMath 10A, Summer, 2007 Lecture 12 (REVISED), p. 63.dydx=ddx(x2sin x) = x2ddx(sin x) + sin xddx(x2) = 2x sin x + x2cos x4. y0(x) = 5ddx(cos x) − 3ddx(sin x) = −5 sin x − 3 cos x5. y0(x) =ddx(6 sin x + ex) = 6 cos x + ex• y0(0) = 6 cos(0) + e0= 76.dydx=ddx(ln x) −ddx(cosx) = x−1+ sin x7. y0= x2ddx(tan x) + tan xddx(x2) = x2sec2x + 2x tan x • y0(3) = 9 se c2(3) + 6 tan(3)8.ddx(x t an–1x) = xddx(tan–1x) + tan–1xddx(x) =x1 + x2+ tan–1x9. y(x) = tan2x • y0(x) = 2 tan xddx(tan x) = 2 tan x sec2x • y(14π) = tan (14π) = 1 •y0(14π) = 2 tan(14π) sec2(14π) = 4 • Tangent line: y = 1 + 4(x −14π)) • Figure A9xy24614πFigure A910.ddx(sin–1x)2= 2(sin–1x)ddx(sin–1x) =2 sin–1xp1 − x211. y0(x) =ddx(cos4x) = 4 cos3xddx(cos x) = −4 sin x cos3x • y0(1) = −4 sin(1) cos3(1)12. y0(x) = 2ddx(tan x) − 3ddx(sin x) = 2 sec2x − 3 cos x13. y(x) = sin x + cos x • y0(x) = cos x − sin x • y(12π) = 1 • y0(12π) = cos(12π) − sin(12π) = −1• Tangent line: y = 1 − (x −12π) • Figure A13xy212πFigure