Course II Lesson 5 Derivatives of Logarithmic Functions and Exponential Functions 5A • Derivative of logarithmic functions 1 Review of the Logarithmic Function y x Exponential function y = a Fig.1 x Fig.1 y = a (a > 0, a ≠1) O x Logarithmic function y y = a x Fig.2 Fig.1 y = a x x y We replace the notation y = loga x y Fig.3 x = a y = loga x O x 2 Derivative of the Logarithmic Function From the definition ʹ loga (x + h)− loga x (loga x) = lim h→0 h 1 ⎛ h ⎞ ⎧1 x ⎛ h ⎞⎫ = lim loga ⎜1+ ⎟ = lim⎨ ⋅ loga ⎜1+ ⎟⎬ h→0 h ⎝ x ⎠ h→0⎩ x h ⎝ x ⎠⎭ h We put = k . When h → 0 , k → 0 . x ʹ ⎧1 1 ⎫ (loga x) = lim⎨ ⋅ loga (1+ k)⎬ k →0⎩ x k ⎭ ? 1 1 1 1 k ⎡ k ⎤ = limloga (1+ k) = loga lim(1+ k) x k →0 x ⎣⎢k →0 ⎦⎥ 3 Napier’s Constant Trial We expect that approaches one value as . Napier’s Constant n ⎛ 1 ⎞ e = lim⎜1+ ⎟ = n→∞⎝ n ⎠ Important Mathematical Constants “π=3.1415…” was known 4000 years ago th “e=2.7182…” was found in 17 century 4 Natural Logarithm Then If the base is e , we have Natural logarithm is the logarithm to the base e . Notation: loge x →ln x Summary d 1 d 1 (log x) = , (ln x) = dx a xlna dx x 5 1 Examples [ Example 5.1] Find the derivative of the following functions. (1) y = ln2x, (2) y = log2(3x + 2), (3) y = xln3x Ans. 1 3 (2) yʹ = ⋅(3x + 2)ʹ = (3x + 2) ln 2 (3x + 2) ln 2 3 (3) yʹ = xʹln3x + x(ln3x)ʹ = ln3x + x ⋅ = ln3x +1 3x 6 Examples [ Example 5.2] Calculate the money which you can receive one year later using various compound systems. The principal is 10000 yen. (1) Annual interest is 100%. (2) Half a year interest is 50%, (3) Monthly interest is 100/12%, (4) Daily interest is 100/360%. Ans. 1 (1) 10000×(1+1) = 20,000 yen (2) 10000×(1+1/ 2)2 = 22,500 yen (3) 10000×(1+1/12)12 = 26,130 yen (4) 10000×(1+1/ 365)365 = 27,148 yen [ Note ] Napier’s Constant n ⎛ 1 ⎞ lim⎜1+ ⎟ = e = 2.7182 n→∞ ⎝ n ⎠ 7 Exercise [Ex.5.1] Find the derivatives of the following functions. 2 (1) y = (ln x ) , (2) y = x ln x , (3) y = log10 x Pause the video and solve the problem by yourself. 8 Answer to the Exercise 2 3 [Ex.5.1] Find the derivatives of (1) y = (ln x ) and (2) y = ln( x + 1 ) Ans. (1) d d 2ln x (ln x)2 = 2ln x ⋅ ln x = dx dx x (2) d ʹ ʹ 1 (xln x)= (x) ⋅ln x + x⋅(ln x) = ln x + x⋅ = ln x +1 dx x ln x (3) log x = 10 ln10 d d ln x 1 d 1 log x = = ln x = dx 10 dx ln10 ln10 dx (ln10)x 9 Course II Lesson 5 Derivatives of Logarithmic Functions and Exponential Functions 5B • Derivative of exponential functions 10 Derivative of Inverse Functions Let f and g be inverse functions. Then y = f (x) x = g(y) Differentiate both sides of (1) by y and from the chain rule, we have df (x) df (x) dx df (x) dg(y) df (x) dg(y) 1 = = = 1 = dy dx dy dx dy dx dy Therefore df (x) 1 dy 1 = = dx ⎛ dg(y) ⎞ or dx ⎛ dx ⎞ ⎜ ⎟ ⎜ ⎟ dy ⎝ dy ⎠ ⎝ ⎠ 11 Derivative of the Exponential Function Exponential function of base e y = f (x) = ex x = g(y) = ln y Therefore, from the previous slide we have dy df (x) 1 1 d = = = = y (ex ) = ex dx dx ⎛ dg(y) ⎞ ⎛ 1 ⎞ dx ⎜ ⎟ ⎜ ⎟ ⎝ dy ⎠ ⎝ y ⎠ Case y = a x x If a = e , then x = ln a . Therefore x lna x x lna y = a = (e ) = e d (a x )= a x lna From the Chain Rule dt d d d (ax )= (ex lna )= ex lna (xlna) = ax lna 12 dt dt dt Examples [ Example 5.3] Find the derivative of the following functions. 2x −2x (1) y = e , (2) y = a Ans. (1) Chain rule yʹ = e2x ⋅(2x)ʹ = 2e2x (2) Chain rule yʹ = (a−2x loga)⋅(−2x)ʹ= −2a−2x loga 13 Exercise [Ex.5.2] Find the derivatives of the following functions. x ln x x −x (1) y = x a (2) y = 2 (3) e − e y = x −x e + e Pause the video and solve the problem by yourself. 14 Answer to the Exercise [Ex.5.2] Find the derivatives of the following functions. x ln x x −x (1) y = x a (2) y = 2 (3) e − e y = x −x e + e (1) Product rule x x x x x yʹ = (x)ʹa + x(a )ʹ = a + x ⋅ a loga = a (1+ xloga) (2) Chain rule 1 2ln x ln2 yʹ = 2ln x ln2⋅ ( )= x x (3) Quotient rule (ex + e−x )2 − (ex − e−x )2 4 yʹ = x −x 2 = x −x 2 (e + e ) (e + e ) 15 .