Is called the derivative of f with respect to x the

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is called the derivative of f with respect to x . The domain of f consists of all x in the domain of f for which the limit above exists.// (b) Using only the definition of the derivative as a limit, show all steps of the computation of f ( x ) when f ( x ) = 1/ x . f ( x ) lim h 0 f( x h ) f( x ) h lim h o ( x h ) 1 x 1 h lim h 0 x ( x h ) hx ( x h ) lim h 0 1 x ( x h ) 1 x 2 for every real number x 0.
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TEST2/MAC2311 Page 4 of 4 ______________________________________________________________________ 8. (5 pts.) Determine whether the following function is differentiable at x = 1. Since lim x 1 f ( x ) lim x 1 3 x 3 and , lim x 1 f ( x ) lim x 1 ( x 2 x 2) 4 f is not continuous at x = 1. Consequently, f is cannot be differentiable there either. [Check continuity first, folks. There was a red herring here!] ______________________________________________________________________ 9. (5 pts.) Compute f (x) when f( x ) sin(2 x 3 ). f ( x ) cos(2 x 3 ) (6 x 2 ) 6 x 2 cos(2 x 3 ). f ( x ) 12 x cos(2 x 3 ) 6 x 2 sin(2 x 3 ) (6 x 2 ) 12 x cos(2 x 3 ) 36 x 4 sin(2 x 3 ). ______________________________________________________________________ 10. (10 pts.) A spherical balloon is to be deflated so that its radius decreases at a constant rate of 15 cm/min. At what rate must air be removed when the radius is 9 cm.? [ V = (4/3) π r 3 ??] Let r ( t ) denote the radius of the sphere at time t in minutes. Then the volume is given by V ( t ) = (4/3) π ( r ( t )) 3 . We are told that r ( t ) = -15 (cm/min) for any t . The question is, what is V ( t 0 ) at the moment t 0 when r ( t 0 ) = 9 cm? Evidently, V ( t ) = 4 π ( r ( t )) 2 r ( t ) for any t . Substituting t 0 into this last equation, using the know values of r and r at t 0 , and doing the boring multiplication by hand yields V ( t 0 ) = -4860 π cubic cm/min.
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