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