Chapter 6: The Calculus of Exponential and Logarithmic Functions

if g(x) = ∫^x_a f(t)dt where a stands for a constant, and f is continuos in the neighbourhood of a, then g’(x) = f(x)

ln x = ∫^x₁ 1/t dt

where x is a positive number

d/dx (lnx) = 1/x

∫ 1/u du = ln |u| + C

If: 1. f’(x) = g’(x) for all values of x in the domain, and

2. f(a) = g(a) for one value, x = a, in the domain,

then f(x) = g(x) for all values of x in the domain

if two functions have the same derivative everywhere and they also have a point in common, then they are the same function

ln (ab) = lna + lnb

ln(a/b) = lna - lnb

ln(a^r) = r lna

ln 1 = 0

a = log_bc if and only if b^a = c where b > 0, b ≠ 1, and c > 0

log_bx = log_ax/log_ab

in particular: log_bx = log_ex/log_eb = lnx/lnb = 1/lnb * lnx

e = lim n→0 (1+n)^1/n = lim n→∞ (1 + 1/n)ⁿ

e = 2.7182818284…

0/0, 0⁰, 1^∞, ∞/∞, 0(∞), ∞⁰, ∞-∞

∞+∞=∞, -∞-∞=-∞, 0^∞=0, 0^-∞=∞, ∞*∞=∞

like e or π, numbers that are not only irrational, but that are not able to be expressed using only a finite number of the operations of algebra performed on integers, it goes beyond these operations

ln(e^x) = x, and e^lnx = x

b^x = e^xlnb

for any positive constant b≠1 (to avoid division by zero)

d/dx (b^x) = b^xlnb ∫b^udu = b^u/lnb + C

1. take ln of both sides

2. ln of a power

3. differentiate implicitly with respect to x

4. solve for y’

5. substitute for y

if f(x) = g(x)/h(x) and if lim g(x0 as x—>c = lim h(x) as x—>c = 0

then lim f(x) as x—>c = lim g’(x)/h’(x) as x—>c provided the latter limit exists

for the function f(x) = g(h(x)), if h(x) has a limit, L, as x approaches c and if g is continuous at L, then lim g(h(x)) as x—>c = g(lim h(x) as x—>c)

= -cosx + c

= sinx + c

= -ln|cosx| + C = ln|secx| + C

= ln|sinx| + C = -ln|cscx| + C

= ln|secx + tanx| + C

= -ln|cscx + cotx| + C = ln|cscx - cotx| + C