sol12-6 - + 3) ]( x 2 + 4)-( x 2 + 4) 3 (2 x ) ( x 2 + 4) 2...

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SOLUTIONS CHAPTER 12.6 MATH 132 WI01 4. Proof. dy dx = dy dz dz dx but dy dz = ( z 1 3 ) 0 = 1 3 z 1 3 - 1 = 1 3 z - 2 3 dz dx = ( x 6 - x 2 + 1) 0 = 6 x 5 - 2 x We have to multiply these two, but since we don’t want to have z in the final formula, we’ll replace every occurence of z by its formula, z = x 6 - x 2 + 1 (BEWARE! not the formula for the DERIVATIVE of z !!!!): dy dx = 1 3 z - 2 3 (6 x 5 - 2 x ) = 1 3 ( x 6 - x 2 + 1) - 2 3 (6 x 5 - 2 x ) ± 8. Proof. First find dy dx , using the above method: dy du = 9 u 2 - 2 u + 7 du dx = 3 and hence dy dx = (9 u 2 - 2 u + 7)(3) = [9(3 x - 2) 2 - 2(3 x - 2) + 7](3) Now plug in x = 1; we get dy dx (1) = [9(3 - 2) 2 - 2(3 - 2) + 7](3) = 14 · 3 = 42 ± Date : 01/21/2000. 1
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2 MATH 132 WI01 22. Proof. y = 3 p 8 x 2 - 1 = (8 x 2 - 1) 1 3 so then, using Chain Rule , we get y 0 = 1 3 (8 x 2 - 1) 1 3 - 1 (8 x 2 - 1) 0 = 1 3 (8 x 2 - 1) - 2 3 · 16 x ± 28. Proof. y = 1 (1 - x ) 3 = (1 - x ) - 3 so then, using Chain Rule , we get y 0 = ( - 3)(1 - x ) - 3 - 1 (1 - x ) 0 = - 3(1 - x ) - 4 ( - 1) ± 46. Proof. Use Quotient Rule : y 0 = [(2 x + 3) 3 ] 0 ( x 2 + 4) - (2 x + 3) 3 ( x 2 + 4) 0 ( x 2 + 4) 2 = = [3(2 x + 3) 2 (2 x
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Unformatted text preview: + 3) ]( x 2 + 4)-( x 2 + 4) 3 (2 x ) ( x 2 + 4) 2 = = [3(2 x + 3) 2 (2)]( x 2 + 4)-( x 2 + 4) 3 (2 x ) ( x 2 + 4) 2 ± 62. Proof. First we need y : y =-3 (3 x 2 + 1) 3 = (-3) · (3 x 2 + 1)-3 y = (-3)[(-3)(3 x 2 + 1)-3-1 (3 x 2 + 1) ] = (-3)[(-3)(3 x 2 + 1)-4 (6 x )] Simplify: y = 54 (3 x 2 + 1) 4 Now we can find the slope of the tangent line by plugging in x = 0: slope = 54 (3 · 2 + 1) 4 = 54 1 4 = 54 SOLUTIONS CHAPTER 12.6 3 Having the slope and the point (0 ,-3) we get for the equation of the tangent line using point-slope formula y-(-3) = 54( x-0) ±...
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This note was uploaded on 07/26/2011 for the course MATH 132 taught by Professor Staff during the Spring '08 term at Ohio State.

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sol12-6 - + 3) ]( x 2 + 4)-( x 2 + 4) 3 (2 x ) ( x 2 + 4) 2...

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