differentiable at x = c
if lim f(x)-f(c)/x-c exists
or lim f(c+h)-f(c)/h where h = x-c
derivative
if the limit exists say lim f(x)-f(c)/x-c is the derivative of f at c = f’(c)
first order taylor
f is diff at c if and only if there exits a constant M and function r(x) on X with
f(x) = f(c) +M(x-c) + r(x)(x-c) such that
- r(x) is cont at x = c
- lim r(x) = r(c) = 0 as x goes to c
if so M = f’(c)
another way of writing first order taylor
f(x) = f(c) + f1(x)(x-c)
f1(x) = M + r(x)
f1 cont at x = c and f1(c) = m = f’(c)
diff and cont theorem
f diff at x = c
then f also cont at x = c
diff implies cont
properties of the derivative
(f+g)’(c) = f’(c) + g’(c)
(µf)’(c) = µf’(c)
(fg)’(c) = f(c)g’(c) + f’(c)g(c)
(fºg)’(c) = f’(g(c))g’(c)
(1/f)’(c) = -f’(c)/f^2(c)
assuming f and g are differentiable at x = c. then the above are also differentiable at x = c with derivative as above
inverse functions (if f is differentiable does inverse exist)
f is differentiable and 1-1 so inverse f^-1 = g exists and is continuous on Y = f(I) (image of f)
derivative of inverse
f differentiable and 1-1 with f’(d) ≠ 0 and c = f(d) then the inverse f^-1 is differentiable at y = c with
f^-1(c) = 1/f’(d) = 1/f’(f^-1(c))
lipschitz continuous
|f(x) - f(y)| <= L|x-y|
max/ min and derivative
if f differentiable at c and f has a local max/min at x = c then f’(c) = 0
rolles theorem
f : [a,b] -> R continuous and differentiable on (a,b) with f(a) = f(b) then there exists a c in (a,b) such that f’(c) = 0
mean value theorem
f [a,b] -> R cont and diff on (a,b) then there exists a c in (a,b) such that f’(c) = f(b)-f(a)/b-a
growth theorem
f: I ->R cont on I and differentiable at interior points
(i) if f’(x) = 0 for all x in I then f constant on I
(ii) f’(x) >= (>) 0 for all x in I then f (strictly) increasing on I
(iii) f’(x) <= (<) 0 for all x in I then f (strictly) decreasing on I
warning about growth theorem
only works on an interval I
generalised MVT
f,g [a,b]-> R cont and diff on (a,b)
assume g’(x) ≠ 0 for all x in [a,b]
then there exists a c in (a,b) st
f’(c)/g’(c) = f(b)-f(a) / b-a
if g(x) = x get MVT
lhopitals rule
f,g (a,b)-> R diff
f(x), g(x) ≠ 0 for all x in (a,b)
but lim f(x) = lim g(x) = 0 as x -> a
if L = lim f’(x)/g’(x) exists as x -> a
lim f(x)/g(x) exists and = L
for ‘0/0’ form
lhopitals rule remarks
a can be ∞
- lim f(x)/g(x) = lim f(1/x)/g(1/x) as x -> ∞ then use lhopitals
- ∞/∞ also works as lim f(x)/g(x) = lim 1/g(x) / 1/f(x)
- 0.∞ also workds f(x).g(x) = f(x)/ 1/g(x)
- check hypothesis - has to be 0/0 to use
powers beat log
⍺>0 lim logx/x^⍺ = 0
⍺> 0 lim logx . x^⍺ = 0
both by lhopitals
exponentials beat powers
⍺ >0 lim x^⍺/e^x = 0
taylor using lhopitals
apply lhopitals then find limit
original function = limit + remainder
rearrange to find taylor
taylors theorem A
f: I -> R n times diff and c in I
then there exists a function r(x) (depending on c) st f(x) = Tn(x) + rn(x)(-c)^n
with lim rn(x) = 0 as x-> 0
taylor theorem B - lagrange
everything as in A assume f is n+1 times differentiable on I
then f(x) = Tn(x) + f^(n+1)(ø)/(n+1)! (x-c)^n+1
for some ø in between x and c
