MATH 112 Midterm: MATH 112 UW derivative Solutions

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31 Jan 2019
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9. dy dx dy dx dy dx dy dt dy dx dy dx dy dx dy dx dy dx. 3 (9x + 7)(x 4) ( 1 (9x + 7)2. 8(3x 7)7(3) + (3x 7)8. 2(cid:17)(cid:16)2x 3x2(cid:17: f (r) = (cid:0)3r + r3(cid:1)(cid:0) 6r + 20r4(cid:1) (cid:0)1 3r2 + 4r5(cid:1)(cid:0)3 + 3r2(cid:1, f (t) = (cid:20)t2. 3(cid:16)t4 + t2 + 1(cid:17)2(cid:16)4t3 + 2t(cid:17) +(cid:16)t4 + t2 + 1(cid:17)3 (3r + r3)2 (2t)(cid:21) + 6 (2: g (v) = 2 5 . 3 (cid:16)7v3 + 10(cid:17) 1/3 (21v2: h (v) = (2v 5) . # + 6x (x2 + 4x + 2)2 (2u + 3) = "(cid:0)x2 + 4x + 2(cid:1) (2x + 2) (cid:0)x2 + 2x(cid:1) (2x + 4) 3: f (x) = (cid:16)x4, g (x) = 2h(cid:16)x4, r (x) =

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