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MATH 1271 Calculus I October 20, 2011 Fall 2011 Lecturer: Prof. Scot Adams Print name: Quiz 2 Time: 30 minutes (Show all your work for credit.) (1) Use logarithmic differentiation to find the derivative of √ 2 f (x) = 3 x · ex +x · (x − 1)1/4 . Solution. Recall that f 0 (x) d (ln f (x)) = . dx f (x) Thus d √ x2 +x 1/4 3 x·e · (x − 1) f (x) · ln dx d 1 1 2 f (x) · ln(x) + (x + x) ln(e) + ln(x − 1) dx 3 4 d 1 1 2 f (x) · ln(x) + x + x + ln(x − 1) dx 3 4 1 1 f (x) · + 2x + 1 + . 3x 4(x − 1) 0 f (x) = = = = 1 2 (2) Evaluate the limit lim x→1 ln x . 11 sin(xπ/2)(x2 − x) (a) 1/18 (b) 1 11 (c) 1 21 (d) 1/7 (e) undefined Solution. Naively plugging in 1 gives us an indeterminate of the form 00 . Thus we may apply L’Hospital’s rule after first applying the product rule of limits to simplify: ln x 1 ln(x) = lim · lim 2 2 x→1 11 sin(xπ/2)(x − x) x→1 11 sin(xπ/2) x→1 x − x (1/x) 1 = ( ) · lim 11 x→1 2x − 1 1 = . 11 lim (3) Differentiate f (x) = (4 − x)sin(x) . Solution. We use logarithmic differentiation: d f (x) = f (x) · (sin(x) ln(4 − x)) dx sin(x) = f (x) · cos(x) ln(4 − x) − . 4−x 0 3 (4) Evaluate lim (2 − x)(tan(πx/2)) . x→1 (a) e2/π (b) e−1 (c) e1 (d) eπ/2 (e) undefined Proof. Plugging in 1 gives us an indeterminate of the form 1∞ . Let L = lim (2 − x)(tan(πx/2)) . x→1 Then sin(πx/2) · ln(2 − x) x→1 cos(πx/2) ln 2 − x = lim sin(πx/2) · lim x→1 x→1 cos(πx/2) (−1/(2 − x)) = (1) · lim x→1 (π/2) sin(πx/2) 1 = 2/π 2 . = π Thus as ln L = π2 , it follows that L = eπ/2 . ln(L) = lim