MATH 12002 - CALCULUS I 2.2: The Derivative as a Function Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.2: The Derivative as a Function

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 7

The Derivative Function

Definition

Let y = f (x) be a function. The derivative (function) of f is the function f ′(x) defined by f ′(x) = lim

h→0

f (x + h) − f (x) h . The domain of f ′ is the set of all numbers x in the domain of f such that this limit exists. Note: There are several other standard notations for the derivative of y = f (x) (or more precisely, the derivative of f with respect to x), including f ′(x) = y′ = dy dx = df dx = d dx f (x). The notation dy

dx is called Leibniz notation.

D.L. White (Kent State University) 2 / 7

Examples

Example

Find the derivative of f (x) = 5x2 + 3x.

Solution

By definition, f ′(x) = lim

h→0

f (x + h) − f (x) h = lim

h→0

[5(x + h)2 + 3(x + h)] − [5x2 + 3x] h = lim

h→0

[5x2 + 10xh + 5h2 + 3x + 3h] − [5x2 + 3x] h = lim

h→0

5x2 + 10xh + 5h2 + 3x + 3h − 5x2 − 3x h [Continued →]

D.L. White (Kent State University) 3 / 7

Examples

Solution [continued]

= lim

h→0

5x2 + 10xh + 5h2 + 3x + 3h − 5x2 − 3x h = lim

h→0

10xh + 5h2 + 3h h = lim

h→0

h(10x + h + 3) h = lim

h→0(10x + h + 3), since h = 0,

= 10x + 0 + 3 = 10x + 3. Therefore, f ′(x) = 10x + 3 or, in Leibniz notation, d dx (5x2 + 3x) = 10x + 3.

D.L. White (Kent State University) 4 / 7

Examples

Example

Find the derivative of f (x) =

1 x+2 and find the slope of the tangent line

to f at x = 3.

Solution

By definition, f ′(x) = lim

h→0

f (x + h) − f (x) h = lim

h→0 1 x+h+2 − 1 x+2

h = lim

h→0 x+2 (x+2)(x+h+2) − x+h+2 (x+h+2)(x+2)

h = lim

h→0

• (x+2)−(x+h+2)

(x+2)(x+h+2)

• h

[Continued →]

D.L. White (Kent State University) 5 / 7

Examples

Solution [continued]

= lim

h→0

• (x+2)−(x+h+2)

(x+2)(x+h+2)

• h

= lim

h→0

x + 2 − x − h − 2 (x + 2)(x + h + 2) · h = lim

h→0

−h (x + 2)(x + h + 2) · h = lim

h→0

−1 (x + 2)(x + h + 2), since h = 0, = −1 (x + 2)(x + 0 + 2) = − 1 (x + 2)2 . [Continued →]

D.L. White (Kent State University) 6 / 7

Examples

Solution [continued]

We have f ′(x) = − 1 (x + 2)2 , and so the slope of the tangent line at x = 3 is f ′(3) = − 1 (3 + 2)2 = − 1 52 = − 1 25.

D.L. White (Kent State University) 7 / 7