Basic Differentiation Rules For Derivatives
Basic Differentiation Rules For Derivatives

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Use the properties of the derivative to find the following:
(a) r(t) = ti + 4tj + t^2k, u(t) = ti + t^2j + t^3k
r'(t) = i + 4j + 2tk
(b) [r(t) – u(t)] = 3i + 8tj + 4t^2k
(c) [(2t)u(t)] = 20ti + 6t^2j + 8t^3k
(d) a[t * (r(t) – u(t))] = 10ti + 12t^2j + 5t^4k
(e) d/dt[r(t) * u(t)] = 10t^2i + 32t^3j + 2t^2k
-r(5t)

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01:23

Use the properties of the derivative to find the following.

r(t) = 9ti + (t − 1)j, u(t) = ti + t^2j + (4/3)t^3k

(a) r'(t)(b) d/dt [u(t) − 2r(t)](c) d/dt [(2t)r(t)](d) d/dt [r(t) · u(t)](e) d/dt [r(t) ✕ u(t)](f) d/dt [u(2t)]

05:29

Use the properties of the derivative to find the following:

r(t) = 3ti + (t – 1)j, u(t) = ti + t^2j + 2t^3k

(a) r(t)3i + j

(b)[u(t) – 2r(t)]5i + (2t – 2)j + 2/k

(c)[(2t)r(t)]12ti + (4t – 2)j

(d)[r(t) u(t)]32t^2 + 7t

(e)[r(t) x u(t)][u(2t)]

04:14

(a) r(t) = 9ti + (t – 1)j(b) [u(t) – 2r(t)] = ~17i + (2t – 2)j + 2/2k(c) [(4t)r(t)] = 72ti + (8t – 4)j(d) -[r(t) u(t)] = 16t + 372(e) -[r(t) x u(t)] = 24 – 23 – 1(64 – 0) + (9? – ? + 4) X-[u(2t)] = 18i + 4j

01:01

Use the properties of the derivative to find the following:(a) r'(t) = ti + 5tj + t^2k, u(t) = 4ti + t^2j + t^3k(b) (i)n – (1)az1p dt = -2i + (10 – 2t)j + (4t – 32)k(c) d[(4t)u(t)] = 32ti + 12e^2j + 16t^3k(d) dt(e) dtX(f) d r(4t) ap

07:42

Use the properties of the derivative to find the following: r(t) = ti + 4tj + tZk, u(t) = 4ti + t^2j + 3k

(a) r(t) + u(t) = i + 4j + 2tk (b) 2r(t) – u(t) = 2i + 8j – 2tj (c) 4t – 3u(t) = 12t – 3t^2j (d) -[(3t)u(t)] = -3t^3j (e) [r(t) u(t)] = 4t^2j (f) [r(t) u(t)] = 4t^2j (g) r(3t) = 3ti + 12tj + 3tk

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You are watching: SOLVED: Use the properties of the derivative to find the following: (a) r(t) = ti + 4tj + t^2k, u(t) = ti + t^2j + t^3k r'(t) = i + 4j + 2tk (b) [r(t). Info created by Bút Chì Xanh selection and synthesis along with other related topics.