1S1 Tutorial Sheet 6: Solutions1 3. (2) Find x2 − 1 x→1 x2 − x both by factorizing and by l’Hò‚pital’s rule. lim 7 February 2007 Questions (7 Solution: First by factorizing: lim 1. (4) Find the first four terms in the Taylor expansion of ln(x) about x = 1. Solution: So we are interest in the expansion of ln (1 + h) in h: 1 2 d2 ln x d ln x + ln (1 + h) = ln (1) + h h 2 dx x=1 2 4 dx x=1 1 3 d3 ln x 1 4 d ln x + h + h +... 6 dx3 x=1 24 dx4 x=1 x→1 x2 − 1 (x − 1)(x + 1) x+1 = lim = lim =2 x→1 x2 − x x→1 x(x − 1) x (8 and, we can see from the above that both numerator and denominator vanish at x = 1 s we can use l’Hopital’s rule: x2 − 1 2x = lim =1 x→1 x2 − x x→1 2x − 1 lim (1) (9 Extra Questions Now d ln x dx d2 ln x dx2 d3 ln x dx3 d4 ln x dx4 so = 1 x = − = 1. Find df /dx where f (x) = cos−1 x. Solution: So, say y = cos−1 x and hence x = cos y. Now differentiate 1 x2 2 x3 = − 6 x3 1 1 1 ln (1 + h) = h − h2 + h3 − h4 + . . . 2 3 4 1 Conor Houghton, houghton@maths.tcd.ie, see also http://www.maths.tcd.ie/~houghton/1S1 1 (11 (3) p and now it would be good to rewrite this in terms of x; sin y = ± 1 − cos2 y and choosing the positive square root, making the sine positive, as is appropriate for th interval zero to π. dy 1 = −√ (12 dx 1 − x2 (4) where we have used cos y = cos (cos−1 x) = x. 2. Find dy/dx where ln y + ln x = exp xy. (5) Solution: Differentiating so, solving for dy/dx we get 1/x + 2y 1 + 2xy dy =− =− dx 1 + 2y x + 2xy 1 dy =− dx sin y and hence Solution: Differenciating across we get 1 dy dy + 2y + 2x + 2y x dx dx (10 (2) 2. (2) Use implicite differentiation to find dy/dx where ln x + 2xy + y 2 = 0 dy dx 1 = − sin y (6) 1 dy 1 dy + = yexy + x exp xy y dx x dx (13 yexy − x1 xy 2 exy − y dy = 1 = xy dx x − xyexy − xe y (14 so, solving for dy/dx 2 3. Differentiate ln cos x. 8. Find lim Solution: This is done using the chain chain rule sin x d ln cos x = − = − tan x dx cos x x→1 (15) x→1 x2 Solution: x2 − 1 − 2x + 1 (24 Solution: Again, and bottom are zero at x = 1, so differentiate lim 4. Differentiate ln x3 both by the chain rule and by ln x3 = 3 ln x. x2 x2 − 1 2x = lim − 2x + 1 x→1 2x − 2 (25 so now the bottom is zero but the top isn’t and the sign is different for x approachin from above and x approaching from below, so the limit is not defined. So, by the chain rule we get and 3x2 3 d ln x3 = 3 = dx x x (16) 3 d 3 ln x = dx x (17) 5. Find lim x→0 ex − 1 sin x (18) Solution: So the top and bottom are both zero at the limit point x = 0, so, differentiating top and bottom ex − 1 ex = lim =1 (19) lim x→0 sin x x→0 cos x 6. Find lim x→1 ln x 1−x (20) Solution: Again the top and bottom are both zero at x = 1 so lim x→1 ln x 1 = − lim = −1 x→1 x 1−x (21) ln x exp (x − 1) (22) 7. Find lim x→1 Solution: Trick question, the denominator isn’t zero at x = 1, in fact you kiust substitute ln x lim =0 (23) x→1 exp (x − 1) 3 4