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Differentiate following functions.
(a) f ( x ) = 2 x5 cosh x
(b) h (t ) = sinh t / t + 1
Solution
(a) f ′ ( x ) = 10x4 cosh x + 2x5 sinh x
(b) h′ (t ) = (t + 1) cosh t - sinh t /(t + 1)2
A triangle has vertices A (-1, 3, 4) B (3, -1, 1) and C (5, 1, 1). The area of ABC is a) 30.1 b) 82.1 c) 9.1 d) 52.1
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