Page 314 - Calculus Demystified

Chapter 6
5. We do (a) and (b). 301
2
x + 1
3
(a) Let A = x · . Then
3
x − x
2
3
ln A = 3ln x + ln(x + 1) − ln(x − x)
hence
2
dA/dx d 3 2x 3x − 1
= ln A = + − .
2
3
A dx x x + 1 x − x
Multiplying through by A gives
2 2
dA 3 x + 1 3 2x 3x − 1
= x · · + − .
3
3
2
dx x − x x x + 1 x − x
3
sin x · (x + x)
(b) Let A = . Then
2
x (x + 1)
3 2
ln A = ln sin x + ln(x + x) − ln x − ln(x + 1)
hence
2
dA/dx d cos x 3x + 1 2x 1
= ln A = + − − .
3
A dx sin x x + x x 2 x + 1
Multiplying through by A gives
3 2
dA sin x · (x + x) cos x 3x + 1 2x 1
= · + − − .
3
2
dx x (x + 1) sin x x + x x 2 x + 1
6. Let R(t) denote the amount of substance present at time t. Let noon on
January 10 correspond to t = 0 and noon on February 10 correspond to
t = 1. Then R(0) = 5 and R(1) = 3. We know that
R(t) = P · e Kt .
Since
5 = R(0) = P · e K·0 ,

we see that P = 5. Since
3 = R(1) = 5 · e K·1 ,

we find that K = ln 3/5. Thus
t
3
t ln(3/5)
R(t) = 5 · e = 5 · .
5