Barrons AP Calculus - David Bock
(E) As x → 0+, and therefore y → 0. As x → 0−, → −∞, so and therefore Because the two one-sided limits are not equal, the limit ...
(A) but f (−1) = 2. The limit does not exist at a = 1 and f (2) does not exist. ...
41. (B) ...
(D) but since these two limits are not the same, does not exist. ...
Answers Explained Many of the explanations provided include intermediate steps that would normally be reached on the way to a fi ...
(A) By the Quotient Rule, (6), ...
(B) Since y = (3 − 2x)1/2, by the Power Rule, (3), ...
(B) Since y = 2(5x + 1)−3, y ′ = −6(5x + 1)−4 (5). ...
5. (E) ...
(D) Rewrite: ...
(A) Rewrite: y = (x^2 + 2x − 1)1/2; then (x^2 + 2x − 1)−1/2 (2x + 2). ...
(D) Use the Quotient Rule: ...
(C) Since ...
(E) Use formula (18): ...
(A) Use formulas (13), (11), and (9): ...
(D) By the Quotient Rule, ...
(D) Since ln (x^2 + 1) ...
14. (C) ...
(A) Since (−csc 2x cot 2x · 2). ...
(A) y ′ = e−x (−2 sin 2x) + cos 2x(−e−x). ...
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