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Let us nd a formula for the second derivative y , where y stands for either of the two functions. Dx (x 2 + y2 ) = Dx (9) 2x + 2yy = 0 x + yy = 0 Next, differentiate both sides of (1) with respect to x, Dx (x + yy ) = Dx (0) 1 + yDx (y ) + y Dx y = 0 1 + yy + y y = 0 1 + yy + (y )2 = 0 Solve (1) for y in terms of x and y, y = Substitute (x/y) for y in (2), x2 1 + yy + 2 = 0 y y2 + y3 y + x 2 = 0 Solve for y , y = From (0), we may substitute 9 for x 2 + y2 , 9 y = 3 y x 2 + y2 y3 [multiplying by y2 ] x y (2) [Dx y2 = 2yy by the power chain rule] (1)
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Acceleration Let an object move along a coordinate axis according to the equation s = f (t), where s is the coordinate of the object and t is the time. From 18, the object s velocity is given by v= ds = f (t) dt
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The rate at which the velocity changes is called the acceleration a. De nition: a = EXAMPLES
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(a) For an object in free fall, s = s0 + v0 t 16t 2 , where s, measured in feet, is positive in the upward direction and t is measured in seconds. Recall that s0 and v0 denote the initial position and velocity; that is, the values of s and v when t = 0. Hence, v= ds = v0 32t dt dv = 32 a= dt
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d2s dv = 2 = f (t) dt dt
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Thus, the velocity decreases by 32 feet per second every second. This is sometimes expressed by saying that the (downward) acceleration due to gravity is 32 feet per second per second, which is abbreviated as 32 ft/sec2 .
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(b) An object moves along a straight line according to the equation s = 2t 3 3t 2 + t 1. Then, v= ds = 6t 2 6t + 1 dt dv a= = 12t 6 dt
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In this case, the acceleration is not constant. Notice that a > 0 when 12t 6 > 0, or t > 1 . This implies (by Theorem 17.3) 2 that the velocity is increasing for t > 1 . 2
Solved Problems
22.1 Describe all the derivatives ( rst, second, etc.) of the following functions: x 1 (b) f (x) = (a) f (x) = x 4 5x 4 x+1
(a) f (x) = x 3 5 (b) f (x) = = f (x) = 3x 2 f (x) = 6x f (4) (x) = 6 f (n) (x) = 0 for n 5 [by the quotient rule] (x + 1)Dx x xDx (x + 1) (x + 1)2 x+1 x 1 (x + 1)(1) x(1) = = = (x + 1) 2 (x + 1)2 (x + 1)2 (x + 1)2 [by the power chain rule]
f (x) = 2(x + 1) 3 Dx (x + 1) 2 = 2(x + 1) 3 (1) = (x + 1)3 +6 f (x) = 2( 3)(x + 1) 4 = (x + 1)4 24 f (4) (x) = 2( 3)( 4)(x + 1) 5 = (x + 1)5 . . . f (n) (x) = 2( 3)( 4)( 5) (n)(x + 1) (n+1) = ( 1)n 1 n! (x + 1)n+1
algebra ( 1)n 1 will be 1 when n is odd and 1 when n is even. n! stands for the product 1 2 3 n of the rst n positive integers.
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