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Unformatted text preview: MATH 147 Assignment #4 Solutions 1) a) Find the derivative of each of the following functions. i) y = cos(3 x 2 ) + arcsin ( x ) ii) f ( x ) = tan( x ) e x iii) f ( x ) = x sin( x ) Solutions i) We have y = 6 x sin(3 x 2 ) + 1 √ 1 x 2 . ii) The Quotient Rule shows that f ( x ) = sec 2 ( x ) e x tan( x ) e x ( e x ) 2 = sec 2 tan( x ) e x iii) First note that if y = f ( x ) = x sin( x ) , then ln ( y ) = sin ( x )ln( x ) . Differentiating implicitly gives us that y y = cos( x ) ln ( x ) + sin( x ) x Therefore, y = y (cos( x ) ln ( x ) + sin( x ) x ) = x sin ( x ) (cos( x ) ln ( x ) + sin( x ) x ) Alternatively, we could write x sin( x ) = e sin ( x ) ln ( x ) The Chain Rule shows that 1 d dx ( x sin( x ) ) = d dx ( e sin( x ) ln( x ) ) = ( e sin ( x ) ln ( x ) ) d dx (sin( x )ln( x )) = ( e sin ( x ) ln ( x ) )(cos( x ) ln ( x ) + sin( x ) x ) = x sin ( x ) (cos( x ) ln ( x ) + sin( x ) x ) 2) a) Show that f ( x ) = x 2 sin( 1 x 2 ) if x 6 = 0 if x = 0 is differentiable every where but its derivative is not continuous at x = 0. Solution It follows from the standard rules of differentitation that f ( x ) is differentiable whenever x 6 = 0 with derivative f ( x ) = 2 x sin( 1 x 2 ) + ( x 2 cos( 1 x 2 ))( 2 x 3 ) = 2 x sin( 1 x 2 ) + 2cos( 1 x 2 ) x Consider lim h → f (0 + h ) f (0) h = lim h → f ( h ) f (0) h = lim h → h 2 sin( 1 h 2 ) h = lim h → h sin( 1 h 2 ) Since  sin( 1 h 2 ) ≤ 1 we have  h ≤ h sin( 1 h 2 ) ≤ h  and since lim h →  h  = 0 = lim h →  h  2 the Squeeze Theorem shows that f (0) = lim h → f (0 + h ) f (0) h = lim h → h sin( 1 h 2 ) = 0 . Finally, f ( x ) is not continuous at x = 0. To see, this first note that lim x → 2 x sin( 1 x 2 ) = 0 . From this it follows that f ( x ) will be continuous at x = 0 if and only if lim x → 2cos( 1 x 2 ) x = 0 . However, this latter limit does not exist since if x n = 1 √ 2 nπ then x n → 0 but lim n →∞ 2cos( 1 x 2 n ) x n = lim n →∞ 2(cos(2 nπ ))( √ 2 nπ ) = lim n →∞ 2( √ 2 nπ ) =∞ b) Let g ( x ) be such that  g ( x ) ≤ M for all x ∈ [ 1 , 1] . Let h ( x ) = x 2 g ( x ) if x 6 = 0 if x = 0 . Show that h ( x ) is differentiable at x = 0 and find h (0) . Solution Observe that h ( x ) h (0) x = xg ( x ) 3 for all x ∈ [ 1 , 1] with x 6 = 0. It follows from this that M  x ≤ h ( x ) h (0) x ≤ M  x  for all x ∈ [ 1 , 1] with x 6 = 0. The Squeeze Theorem clearly shows that h (0) = lim x → h ( x ) h (0) x = 0 Therefore, h ( x ) is differentiable at x = 0 with h (0) = 0. 3) Assume that f ( x ) is continuous on [3 , 5] ,f (3) = 2 and that f ( x ) = 1 1+ x 3 on (3 , 5) . Show that 127 63 ≤ f (5) ≤ 29 14 . Solution Note that f ( x ) is clearly decreasing on [3 , 5]. It follows that f (5) = 1 126 ≤ f ( x ) ≤ 1 28 = f (3) for any x ∈ [3 , 5] The Mean Value Theorem shows that there exists a c ∈ (3 , 5) such that f (5) f (3) 5 3 = f ( c ) But then 1 126 ≤ f (5) f (3) 5 3 ≤ 1 28 ....
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This note was uploaded on 10/21/2010 for the course MATH 147 taught by Professor Wolzcuk during the Fall '09 term at Waterloo.
 Fall '09
 Wolzcuk
 Derivative, Quotient Rule

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