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# Derivatives of Exponential and Trigonometric Functions

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Department
Mathematics
Course Code
MAT1320
Professor
Kirill Zaynullin

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Calculus and Vectors – How to get an A+ 5.1 5.2 Derivative of Exponential Function A Review of Exponential Functions Ex 1. Use the graph of the exponential function to The exponential function is defined as: evaluate each limit. x x y = f (x) = b ; b > 0,b ≠1 a) lim 2 x→∞ The graph of the exponential function is represented lim 2 = ∞ below: x→∞ b) lim 2 x x→−∞ x x→−∞ 2 = 0 c) lim 0.2x x→∞ x lim 0.2 = 0 x→∞ d) lim 0.2 x x→−∞ The x-axis ( y = 0) is a horizontal asymptote. lim 0.2 = ∞ x→−∞ B Number e Ex 2. Estimate the number e using formula (1) and by The number e is defined by: taking n =100000 . n e = lim 1 + 1 ⎞ (1) 100000 n→∞ ⎜ n⎟ ⎛ 1 ⎞ ⎝ ⎠ e ≅ ⎜+ ⎟ ≅ 2.7183 which can be written also as: ⎝ 100000 ⎠ 1 u e = lim(1+u) (2) u→ 0 C Exponential Function Ex 3. Estimate e using formula (2) and by taking x The exponential function e may be evaluate using n =100000 . the limit: 100000 n 1/2 ⎛ 1/2 ⎞ x ⎛ x ⎞ e = e ≅ ⎜+ ⎟ ≅1.6487 e = lim ⎜ + ⎟ (3) ⎝ 100000 ⎠ n→∞ ⎝ n ⎠ Proof: x x n ⎛ n⎞ ⎛ n⎞ ⎛ x⎞ ⎜⎛ x⎞ x⎟ ⎜ ⎛ x⎞ x⎟ lim⎜1 + ⎟ = lim⎜⎜1 + ⎟ ⎟ = ⎜lim ⎜ + ⎟ ⎟ = n→∞ ⎝ n⎠ n→∞ ⎜⎝ n⎠ ⎟ ⎜n→∞ ⎝ n⎠ ⎟ ⎝ ⎠ ⎝ ⎠ x ⎛ 1 ⎞ = ⎜ lim(1+u) u ⎟ = ex ⎜u→ 0 ⎟ ⎝ ⎠ D Derivative of e x Ex 4. Differentiate and simplify. 2 x (e )'= ex a) x e 2 x 2 x 2 x x 2 x x d x x (4) (x e )'= (x )'e + x (e )'= 2xe + x e = xe (2+ x) e = e dx x/2 Proof: b) e n n−1 1 1 ⎛ x ⎞ x ⎛ x⎞ ⎛1 ⎞ x (ex/2)' ( ex )' e = e x/2 ⎜1 + ⎟ → e ⇒ n⎜1+ ⎟ ⎜ ⎟ → ( )' x 2 ⎝ n ⎠ ⎝ n ⎠ ⎝ n⎠ 2 e n−1 n n n −x ⎜ ⎛ x ⎞ ⎟ ⎛ x ⎞ x x x c) e ⎜ ⎜ + ⎟ ⎟ → ⎜1+ ⎟ → e ⇒ ∴ (e )' e −x x −1 x −2 x −2x+x −x ⎝ ⎝ n ⎠ ⎠ ⎝ n ⎠ (e )'=[(e ) ]'= (−1)(e ) (e ) = −e = −e 5.1 5.2 Derivative of Exponential Function ©2010 Iulia & Teodoru Gugoiu - Page 1 of 3 Calculus and Vectors – How to get an A+ f (x) Ex 5. Differentiate. E Derivative of e Using (4) and the chain rule: a) e−3x −3x −3x −3x (e )'= e (−3x)'= −3e f( ) f( ) 2 (e )'=e f'( ) b) e−1/ x d (5) 2 ef (x)= ef (xf '(x) 2 2 2e −1/x dx (e−1/x )' e −1/ ( x −2 )' x3 x +1 c) e x +1 x 1 x +1 x 1 2 (2x)e xe (e )'= e ( x +1)'= 2 = 2 2 x +1 x +1 Ex 6. The hyperbolic functions are defined by: c) (cosh x)'= sinh x e −e −x e +e −x sinh x ' sinh x = , cosh x = , tanh x = . ⎛e x+ e−x ⎞ (e +e −x)' e +e −x( 1) 2 2 cosh x (cosh x)'= ⎜ ⎟ = = ⎝ 2 ⎠ 2 2 Prove that: e −e −x 2 2 = = sinh x a) cosh x −sinh x =1 2 2 2 2 2 ⎛e x+ e−x ⎞ ⎛e x− e −x⎞ d) (tanh x'= 1 cosh x − sinh x = ⎜ ⎟ − ⎜ ⎟ cosh x ⎝ 2 ⎠ ⎝ 2 ⎠ sinh x ' (sinh x)'cosh x −sinh x(cosh x)' e2x + e−2x + 2 e2x + e−2x −2 (tanh x)'=⎜ ⎟ = = − =1 ⎝ cosh x ⎠ cosh x2 4 4 cosh x sinh x 2 1 = = b) (sinh x)'= cosh x cosh x cosh x ' ⎛ e −e −x ⎞ ( −e −x)' e −e −x ( 1) (sinh x = ⎜ ⎟ = = ⎜ 2 ⎟ 2 2 ⎝ ⎠ e +e −x = = cosh x 2 x F Derivative of b , b > 0,b ≠1 Ex 7. Differentiate. a) 3x (b )'= (lnb)b x (3 )'= (ln3)3 x d x x (6) b = (lnb b dx 2 x Proof: b) x 2 x xlnb xlnb x 2 x 2 x 2 x x 2 x (b )'= (e )'= e (lnb) = (lnb)b (x 2 )'= (x )'(2 )+(x )(2 )'= 2x(2 )+ x (2 )(ln2) x = x(2 )(2+ xln2) c) (4 + x ) 3 x 4 3 x 4 2 x 4 [(4 + x ) ]'= 3(4 + x ) (4 + x )' x 4 2 x 3 = 3(4 + x ) [(ln4)4 + 4x ] f (x) Ex 8. Differentiate. G Derivative of b 3 Using (6) and the chain rule: a) 2 −x (2−x3)'= (ln2)(2 −x3)(−x )'= −3(ln2)(2 −x3)x2 f (x) f (x) b )'= (lnb b f x ) d (7) b f (x= (lnb)b f (xf '(x) b) 10 e −x dx 5.1 5.2 Derivative of Exponential Function ©2010 Iulia & Teodoru Gugoiu - Page 2 of 3 Calculus and Vectors – How to get an A+ x 2 ' x 2 ⎜10 e −x ⎟ = (ln10) ⎜0 e− x ⎟( ex − x2)' ⎝ ⎠ ⎝ ⎠ ⎛ ex−x2 ⎞ e − x2 = (ln10)⎜10 ⎟ ⎝ ⎠2 e − x 2 2 Ex 9. Find the equation of the tangent line to the Ex 10. Find the local extrema for y = f (x) = x e−x . graph of y = f (x) = x(2−x) at (0,0). −x2 2 − x2 −x2 2 −x −x f '(x) = 2xe + x e (−2x) = 2xe (1− x ) f '(x) = 2 − x(ln2)(2 ) f '(x) = 0 at x = 0,−1,+1 −0 −0 m = f '(0) = 2 −(0)(ln2)(2 ) =1 f (0) = 0, f (−1) = e , f (1) = e1 y −0 =1(x −0) ⇒ ∴ y = x x −1 0 1 f (x) _ 1/e ` 0 _ 1/e ` f '(x) + 0 - 0 + 0 - Local maximum points: (±1,1/e) . Local minimum point: (0,0) . −x2 3 −x Ex 11. Find the points of inflection for y = f (x) = e . Ex 12. Find the global extrema for f (x) = x 10 over −x2 [−1,2]. f '(x) = −2xe 2 2 f '(x= 3x 10 −x− x (ln10)(10 )−x = x 10 [3 (l−10)x] f ''(x) = −2ex −2x(−2xe −x ) f '(x) 0 at x = 0,3/ln10 1.303 = 2e −x2(2x −1) f (0) 0 3 −(3/ln10) f ''(x) = 0 at x = ±1/ 2 f (3/ln10) (3/ln10) 10 ≅ 0.110 −1/2 f (±1/ 2) = e =1/ e f ( 1)− 10 −2 f (2) =8(10 ) = 0.08 x −1/ 2 1/ 2 The global maximum point is: (1.303,0.110) . f (x) ∪ 1/ e ∩ 1/ e ∪ The global minimum point is (−1,−10). f ''(x) + 0 - 0 + The inflection points are (±1/ 2,1/ e) . Reading: Nelson Textbook, Pages 227-232 Homework: Nelson Textbook: Page 232 #2ef, 3cdf, 4abc, 5a, 8, 10a, 13, 16, 17 Reading: Nelson Textbook, Pages 235-239 Homework: Nelson Textbook: Page 240 #1bd, 2abcd, 3, 4, 6, 8, 9 5.1 5.2 Derivative of Exponential Function ©2010 Iulia & Teodoru Gugoiu - Page 3 of 3 Calculus and Vectors – How to get an A+ 5A Derivative of Logarithmic Function A Review of Logarithmic Function Ex 1. Use the graph of the logarithmic function to evaluate y = bx ⇔ x = log y each limit. b a) lim ln x y = f (x) = lbg x, b > 0,b ≠1,x > 0 x→0+ logb(xy) = logbx+log xb logbx = nlog xb li+ ln x = −∞ x→0 x logax logb y = logbx−log xb logbx = b) lim log x logba x→0+ 0.5 lnx = log x log 1= 0 e b lim+log 0.5x = ∞ logx = log x log b =1 x→0 10 b c) x→∞ logx lim logx = ∞ x→∞ d) x→∞ log0.1x lim log0.1x = −∞ x→∞ B Derivative of lnx Ex 2. Differentiate and simplify. 2 (ln x)'=1 a) x ln x x 2 2 2 21 (1) (x lnx)= (x )'lx + x (lnx)'=2 xn x+ x d lnx = 1 x dx x = 2xln x + x = x(2ln x +1) Proof: y= ln x ⇒ x= e y ⇒ ( )'= (y)' ⇒ ln x b) y 1 1 1 x 1= e y' ⇒ y'= y ⇒ y'= ⇒ ∴(ln )'=x 1 e x x ' x lnx ⎛ln x⎞ = (ln )'x− (ln )( )' = x = 1 lnx ⎜ x ⎟ 2 2 2 ⎝ ⎠ x x x x c) e ln x x x x x x1 x⎛ 1⎞ (e ln x = e )'lnx +e (lnx = e lnx +e = e ⎜lnx + ⎟ x ⎝ x⎠ C Derivative of ln f (x) Ex 3. Differentiate and simplify. Using (1) and the chain rule: a) ln(x + x ) [ln f x]'= f x ) 3 2 (x3+ x )' 3x2 + 2x 3x + 2 f x) [ln(x + x )]'= 3 2 = 3 2 = 2 (2) x + x x + x x + x d f '(x) ln f (x)= dx f (x) b) lnx −1 x +1 ⎡ x− ⎤ ' x + x⎛1 − ⎞' x + (1)(x +) (− 1)−1) 2 ⎢ln ⎥ = ⎜ ⎟ = 2 = 2 ⎣ x+1 ⎦ x − x⎝1 + ⎠ x − (x+ 1) x −1 c) ln(e + e−x) (e x+ e−x)' e x− e−x [lne x+ e−x)]'= = = tanh x e x+ e−x e x+ e−x 5A Derivative of Logarithmic Function ©2010 Iulia & Teodoru Gugoiu - Page 1 of 3 Calculus and Vectors – How to get an A+ D Derivative of logbx Ex 4. Differentiate. 1 a) logx (logb x)'= (lnb x (logx)'= 1 (3)
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