Todays Agenda Upcoming Homework Section 3.3: Derivatives of - - PowerPoint PPT Presentation

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 e = lim

x→0(1 + x)1/x.

We will use this definition to calculate the derivative of the general logarithmic function f (x) = loga 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: loga x = ln x ln a . Therefore, if f (x) = loga x, we know that f ′(x) = 1 ln a d dx (ln x)

• .

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) = lim

h→0

f (x + h) − f (x) h = lim

h→0

ln(x + h) − ln x h = lim

h→0

1 h ln x + h x

• = lim

h→0

1 x · x h ln

• 1 + h

x

• = 1

x lim

h→0

1 h/x ln

• 1 + h

x

• = 1

x lim

h→0 ln

• 1 + h

x 1/(h/x) = 1 x ln

• lim

h→0

• 1 + h

x 1/(h/x) = 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) = loga x, then g′(x) = 1 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) = ax is much easier to calculate. We write f (x) = y and use implicit differentiation: y = ax ⇐ ⇒ ln y = ln ax = x ln a, so d dx (ln y) = d dx (x ln a) ⇐ ⇒ 1 y · dy dx = ln a ⇐ ⇒ dy dx = y ln a, and substituting, f ′(x) = ln a · ax.

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) = ax, then f ′(x) = ln a · ax. In particular, if g(x) = ex, then g′(x) = ln e · ex = ex. So, the function g(x) = ex 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) = x3/4√

x2 + 1 (3x + 2)5

2 g(x) = e−x cos2 x

x2 + 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) = log10(x3 + 1) 2 g(x) = sin(ln x) 3 f (t) = t ln t − t 4 f (r) = √rer 5 g(t) = (t3 + 2t)et 6 h(x) =

• x − 1

x4 + 1

7 h(t) = √tet2−t(t + 1)2/3 Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 14 October 2015 10 / 10