Page 291 - Calculus Demystified

Solutions to Exercises
2
2
278 = lim [−1/(1 + h) ]−[−1/1 ]
h→0 h
−1 −[−(1 + h) ]
2
= lim h(1 + h) 2
h→0
2h + h 2
= lim 2 3
h→0 h + 2h + h
2 + h
= lim 2
h→0 1 + 2h + h
= 2

The derivative is therefore equal to 2.
d x (x + 1) · 1 − x · 2x 1 − x 2
2
4. (a) dx x + 1 = (x + 1) 2 = (x + 1) 2 .
TEAMFLY
2
2
2
d 2 d 2 d 2 2
(b) dx sin(x ) = dx sin (x ) · dx x =[cos(x )]· 2x.
d d d
3
2
2
3
2
2
3
3
2
(c) dt tan(t −t ) = dt tan (t −t )· dt (t −t ) = sec (t − t ) ·
(3t − 2t).
2
d x − 1 (x + 1) · (2x) − (x − 1) · (2x) 4x
2
2
2
(d) 2 = 2 2 = 2 2 .
dx x + 1 (x + 1) (x + 1)
(e) d [x · ln(sin x)] = 1·ln(sin x)+x · cos x = ln(sin x)+ x · cos x .
dx sin x sin x
d
(f) ds e s(s+2) = e s(s+2) ·[1 · (s + 2) + s · 1]= e s(s+2) ·[2s + 2].
d 2 2 d 2
2
2
(g) dx e sin(x ) = e sin(x ) · dx [sin(x )]= e sin(x ) · cos(x ) · 2x.

x 1 x e + 1
x
(h) ln(e + x) = e + x · (e + 1) = e + x .
x
x
5. (a) Since the ball is dropped, v 0 = 0. The initial height is h 0 = 100.
Therefore the position of the body at time t is given by
2
p(t) =−16t + 0 · t + 100.
The body hits the ground when
2
0 = p(t) =−16t + 100
or t = 2.5 seconds.
(b) Since the ball has initial velocity 10 feet/second straight down, we
know that v 0 =−10. The initial height is h 0 = 100. Therefore the

®
Team-Fly