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Unformatted text preview: MATH 137 WEEK 4 NOTES PROF. DOUG PARK 1. Definition of derivative L x ( x , f ( x )) ( t , f ( t )) x t h y = f ( x ) C Let t = x + h . Line L has slope = f ( t ) f ( x ) t x = f ( x + h ) f ( x ) h . As t → x , or equivalently as h → 0, line L approaches tangent line C at ( x,f ( x )). [slope of L ]→ [slope of tangent line C ] Def = f ( x ) Def . The derivative of f at x is the slope of tangent line to graph y = f ( x ) at the point ( x,f ( x )), and equals f ( x ) = lim t → x f ( t ) f ( x ) t x = lim h → f ( x + h ) f ( x ) h [Memorize!] If either limit exists as a finite number, then we say f is differentiable at x . Remark . All h ’s and t ’s will disappear after successfully evaluating the above limits. Notation . f ( x ) = ( f ( x )) = d dx f ( x ) = df dx = dy dx = y = ˙ y = ··· Physics . s ( t ) = position of particle at time t s ( t ) = velocity of particle at time t s 00 ( t ) = ( s ( t )) = acceleration of particle at time t Date : October 9, 2009. 1 2 DOUG PARK 2. Computing derivatives using limit definition Ex . Let f ( x ) = 3 √ x = x 1 / 3 . (i) If x 6 = 0, find f ( x ) using the definition. (ii) Show that f (0) DNE. (iii) Show that y = 3 √ x has a vertical tangent line at x = 0. Sol’n . (i) Recall A 3 B 3 = ( A B )( A 2 + AB + B 2 ). f ( x ) = lim t → x f ( t ) f ( x ) t x = lim t → x 3 √ t 3 √ x t x = lim t → x 3 √ t 3 √ x ( 3 √ t ) 3 ( 3 √ x ) 3 = lim t → x 1 ( 3 √ t ) 2 + 3 √ t · 3 √ x + ( 3 √ x ) 2 If we replace t by x , then the denominator is not 0, because x 6 = 0 by assumption. Hence f ( x ) = 1 3( 3 √ x ) 2 = 1 3 · 1 x 2 / 3 = 1 3 x 2 / 3 . (ii) Note t 2 / 3 = ( t 1 / 3 ) 2 = square ≥ 0. f (0) = lim t → f ( t ) f (0) t = lim t → t 1 / 3 t = lim t → 1 t 2 / 3 = 1 + = ∞ So f (0) DNE (as a real number). (iii) Theorem . Let f be a continuous function. Then y = f ( x ) has a vertical tangent line at x = a ⇐⇒ lim x → a  f ( x )  = ∞ . We check lim x →  f ( x )  = lim x → 1 3 · 1 x 2 / 3 = lim x → 1 3 · 1 x 2 / 3 = 1 3 · 1 + = ∞ Done! Ex . Decide whether f , g are differentiable at x = 0. f ( x ) = ( x sin π x if x 6 = 0 if x = 0 g ( x ) = ( x 2 sin π x if x 6 = 0 if x = 0 Sol’n . f (0) = lim t → f ( t ) f (0) t = lim t → t sin ( π t ) t = lim t → sin π t Note that lim t → ± π t = ±∞ , hence by letting u = π t f (0) = lim u →±∞ sin u = DNE, so f is not differentiable at x = 0. g (0) = lim t → g ( t ) g (0) t = lim t → t 2 sin ( π t ) t = lim t → t sin π t MATH 137 3 Since 1 ≤ sin π t ≤ 1 by multiplying by t , t ≤ t sin π t ≤ t if t > t ≥ t sin π t ≥ t if t < In both cases, lim t → ( ± t ) = 0, so by Squeeze Theorem, lim t → t sin π t = 0 = g (0) Thus...
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This note was uploaded on 12/18/2010 for the course ECONOMICS 120 taught by Professor Mesta during the Spring '10 term at Wilfred Laurier University .
 Spring '10
 Mesta

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