Today’s Agenda • Upcoming Homework • Section 3.3: Derivatives of logarithmic and exponential functions Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 1 / 10
Upcoming Homework • Written HW G: Section 3.2, #18,36,40,74,75. Due 10/16/2016. • WeBWorK HW #14: Section 3.3, due 10/19/2015 • WeBWorK HW #15: Section 3.5, due 10/21/2015 • Written HW H: Section 3.3, #48,70. Section 3.5, #8,26,36,38. Due 10/23/2015. • WeBWorK HW #16: Sections 3.7 and 4.1, due 10/26/2015 Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 2 / 10
Section 3.3 Recall the definition of the number e from Section 3.1: Definition 3.3.1 The number e is defined as x → 0 (1 + x ) 1 / x . e = lim We will use this definition to calculate the derivative of the general logarithmic function f ( x ) = log a x . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 3 / 10
Section 3.3 Recall the change of base formula from your previous math classes: log a x = ln x ln a . Therefore, if f ( x ) = log a x , we know that � d 1 � f ′ ( x ) = dx (ln x ) . ln a Now all that remains is to calculate d dx (ln x ). We will do this using the limit definition of the derivative. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 4 / 10
Section 3.3 Let f ( x ) = ln x . Then f ( x + h ) − f ( x ) ln( x + h ) − ln x f ′ ( x ) = lim = lim h h h → 0 h → 0 � x + h � � � 1 x · x 1 1 + h = lim h ln = lim h ln x x h → 0 h → 0 = 1 1 � 1 + h � x lim h / x ln x h → 0 � 1 / ( h / x ) = 1 � 1 + h x lim h → 0 ln x � � 1 / ( h / x ) � = 1 � 1 + h x ln lim x h → 0 = 1 x ln e = 1 x . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 5 / 10
Section 3.3 Therefore, Derivatives of logarithmic functions If f ( x ) = ln x , then f ′ ( x ) = 1 x , and if g ( x ) = log a x , then 1 g ′ ( x ) = x ln a . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 6 / 10
Section 3.3 The derivative of the general exponential function f ( x ) = a x is much easier to calculate. We write f ( x ) = y and use implicit differentiation: ⇒ ln y = ln a x = x ln a , y = a x ⇐ so dx (ln y ) = d d dx ( x ln a ) ⇒ 1 y · dy dx = ln a ⇐ ⇒ dy dx = y ln a , ⇐ and substituting, f ′ ( x ) = ln a · a x . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 7 / 10
Section 3.3 Derivatives of exponential functions If f ( x ) = a x , then f ′ ( x ) = ln a · a x . In particular, if g ( x ) = e x , then g ′ ( x ) = ln e · e x = e x . So, the function g ( x ) = e x is very special, because g ( x ) = g ′ ( x )!!! Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 8 / 10
Section 3.3 We will now do some examples of a technique called ”logarithmic differentiation.” One way in which this technique is useful is when you have a product or quotient of ”scary-looking” functions. In other words, it can be a shortcut to use in place of the product or quotient rule when the product or quotient rule would yield a royal mess. Example 3.3.2 (logarithmic differentiation) Find the derivatives of the following functions: 1 f ( x ) = x 3 / 4 √ x 2 + 1 (3 x + 2) 5 2 g ( x ) = e − x cos 2 x x 2 + x + 1 Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 9 / 10
Section 3.3 Practice Problems Differentiate each of the following functions. 1 f ( x ) = log 10 ( x 3 + 1) 2 g ( x ) = sin(ln x ) 3 f ( t ) = t ln t − t 4 f ( r ) = √ re r 5 g ( t ) = ( t 3 + 2 t ) e t � x − 1 6 h ( x ) = x 4 + 1 7 h ( t ) = √ te t 2 − t ( t + 1) 2 / 3 Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 10 / 10
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