AMA1110 Tutorial - 4s.pdf - AMA1110 2019/20 Semester 1 Exercise 4 AMA1110 Basic Mathematics I Calculus and Probability Statistics Exercise 4 Keywords

# AMA1110 Tutorial - 4s.pdf - AMA1110 2019/20 Semester 1...

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AMA1110 2019/20 Semester 1 Exercise 4 AMA1110 Basic Mathematics I - Calculus and Probability & Statistics Exercise 4 Keywords : Continuity. The First Principles. Differentiability. L’Hˆopital’s Rule. Linear Approximations. Note : If the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form. The indeterminate forms include 0 0 , , 0 × ∞ , ∞ - ∞ , 1 , 0 0 and 0 . A Revision Definition A.1 (The First Principles) The derivative of f ( x ) is defined by the for- mula f 0 ( x ) = lim Δ x 0 f ( x + Δ x ) - f ( x ) Δ x . (A.1) Note that, the function f ( x ) is differentiable on the interval J if f 0 ( x ) exist at every x in J . Theorem A.2 (L’Hˆopital’s Rule) Let f ( x ) and g ( x ) be two differentiable functions at all the points near a , but not at a . Suppose g 0 ( x ) 6 = 0 for all x , with x 6 = a . If (1) lim x a f ( x ) = lim x a g ( x ) = 0 or ±∞ . (2) lim x a f 0 ( x ) g 0 ( x ) exists or tends to infinity, then lim x a f ( x ) g ( x ) = lim x a f 0 ( x ) g 0 ( x ) (A.2) B Exercise 1. Differentiate the following functions y with respect to x from the first principle. (a) y = cos x (b) y = x 3 (c) y = x 1 - x 2 (d) y = 1 x + 2 2. A function f is defined as f ( x ) = ( 4 cos x + 1 if x 0 , ax + b if x > 0 . If f is differentiable everywhere, find the values of a and b . 3. Find the values of a and b such that f 0 (1) exists, where f ( x ) = 1 x if | x | > 1 , ax + b if | x | ≤ 1 . 4. Find the derivatives of the following functions. 1

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AMA1110 2019/20 Semester 1 Exercise 4 (a) 3 5 x 3 + 4 x 2 - 5 2 x + 7 + 9 x - 1 - 7 x - 2 (b) (2 x + 3) - 2 (4 x - 1) 2 (c) 3 + x 2 - x (d) sin 2 x cos 5 x (e) sin m x cos n x (f) ( x 4 - 3 x 2 + 5) 3 (g) x + 1 3 x 4 (h) 2 x x 2 + 1 (i) e sin 2 x (j) x 1 - x 2 (k) xe - 1 /x (l) tan( 1 - x ) (m) sec 2 x 1 + tan 2 x (n) e e x (o) (1 - x - 1 ) - 1 (p) ln(sin x ) - 1 2 sin 2 x (q) x tan - 1 (4 x ) (r) ln(sec 5 x + tan 5 x ) (s) cot(3 x 2 + 5) (t) sin(tan( 1 + x 3 )) (u) tan 2 (sin x ) (v) x + 1(2 - x ) 4 ( x + 3) 7 (w) ln x x + 1 (x) x 3 sin 5 x ( x 2 + x ) 3 (y) sin 2 x sin 3 x sin 4 x sin 5 x (z) (sin mx ) n (cos nx ) m 5. Consider the function f ( x ) = x 2 ( x + 1) 2 3
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