# note5 - Math021, week 5 Example 5.1 Evaluate the derivative...

This preview shows pages 1–3. Sign up to view the full content.

Math021, week 5 Example 5.1 Evaluate the derivative of the function f ( x ) = 1 x for all numbers x 6 = 0 solution: For h 6 = 0 and x 6 = 0, f ( x + h ) - f ( x ) h = 1 h [ 1 x + h - 1 x ] = x - ( x + h ) hx ( x + h ) = - 1 x ( x + h ) . So, f 0 ( x ) = lim h 0 - 1 x ( x + h ) = - 1 x 2 for all x 6 = 0. Example 5.2 Evaluate the derivative of the function f ( x ) = x for all numbers x > 0 . solution: For h 6 = 0 and x > 0, f ( x + h ) - f ( x ) h = x + h - x h = ( x + h - x )( x + h + x ) h ( x + h + x ) = ( x + h ) - x h ( x + h + x ) = 1 ( x + h + x ) . So, f 0 ( x ) = lim h 0 1 ( x + h + x ) = 1 2 x for all x > 0. Example 5.3 Evaluate the derivative of f at 0 if f ( x ) = | x | for all numbers x 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
soltion: For h 6 = 0, f (0+ h ) - f (0) h = | h | h . But lim h 0 | h | h does not exist (why?). So, the function f does not have a derivative at 0. Sometimes, we say that f is not “diﬀerentiable” at 0. Theorem 5.4 A diﬀerentiable function is continuous. proof: diﬃcult Standard Formula for Derivatives Theorem 5.5 If a is a number and f ( x ) = a for all x . Then, f 0 ( x ) = 0 . proof: For h 6 = 0 and all x , f ( x + h ) - f ( x ) h = a - a h = 0 So, f 0 ( x ) = lim h 0 0 = 0 . Theorem 5.6
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/13/2012 for the course MATH 021 taught by Professor Luxnu during the Fall '10 term at HKUST.

### Page1 / 6

note5 - Math021, week 5 Example 5.1 Evaluate the derivative...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online