geometrical representation of f’(a)
f’(a) of f(x) at x = a is equal to the slope of the tangent to the graph of f at a
definition of f’(a)
the limit of the slope of a line segment joining 2 nearby points on the graph of f
= lim h-> 0 ( f(a+h) - f(a) ) /h
if the limit exists we say f is differentiable at x = a
differentiable
if f’(a) exists for all a in the domain of f we say f is differentiable and the function f’(x) is the derivative
equation of tangent to f at (a,f(a)
y = f(a) + f’(a) (x-a)
necessary condition for f’(a) to exists
is for f to be continuous at a but this is not sufficient (can be continuous and not differentiable)
ways function can fail to be differentiable at a point where its continuous
twice differentiable
f(x) is differentiable then f’(x) may be differentiable if it is we say its twice differentiable
(smooth means infinitely differentiable)
(f’(x))’ = f’‘(x)
how to write nth derivative
f^(n)(x)
d^nf/dx^n
D^n f(x)
if its a variable of time we dot it to show differentiation
leibniz rule
if f(x) and g(x) are differentiable n times so is there product fg(x)
D(fg) = (Df)g + f(Dg)
D^n(fg) = ∑ (nCk) (D^k f)(D^n-k g) from k=0 to n
chain rule
for differentiating the composition of functions
fºg(x)
if g(x) is differentiable at x and f(x) is differentiable at g(x)
fºg’(x) = f’(g(x))g’(x)
or d/dx(f(g(x)) = df/dg dg/dx
L’Hopitals rule
applies to limits in the form 0/0 or ♾️/♾️
if f(x) and g(x) are differentiable on I = (a-h,a) U (a,a+h) for some small h-> 0 with lim f(x) to a = lim g(x) to a = 0
and lim f’(x)/lim g’(x) exists
then lim f(x)/g(x) = lim f’(x)/g’(x)
indeterminate form of the limit
lim x->a f(x) /g(x) doesnt exists if L=/= 0 and M = 0
if L= M = 0 then its indeterminate form
might or might not have a well defined value - we cant tell by knowing L and M
eg 0/0, ♾️/♾️, 0*♾️, ♾️-♾️, 0^0 , 1^♾️, ♾️^0
proof of L’Hopitals rule
where f’(a) and g’(a) exist and the derivatives are continuous and g’(a) =/= 0
lim f(x)/g(x) = lim f(x) - 0/ g(x) -0 = lim f(x) - f(a) / g(x) -g(a) = lim (f(x)-f(a))/(x-a) / (g(x)-g(a))/(x-a) = f’(a)/g’(a) = lim f’(x)/g’(x)
bounded above
f(x) is a function on an interval I
is bounded above by an upper bound k1 if f(x) <= k1 for all x in I
f(x) need not take the value k1 in I
global maximal value
if there exists an x in I such that f(x) = k1 the upper bound is attained and k1 is the global max
bounded
f(x) is bounded in I if its bounded above and below in I
there exists a k in R such that |f(x)| <= k for all x in I
extreme value theorem
if f(x) is continuous on a closed interval [a,b] then its bounded on that interval and has upper/ lower bounds which are attained
(has global max/min)
there exists x1,x2 in [a,b] such that f(x2) <= f(x) <= f(x1) for all x in [a,b]
monotonic increasing
if in [a,b] f(x1) <= f(x2) for all x1,x2 such that a<= x1 < x2 <= b
similar for decreasing
strictly monotonic increasing
if in [a,b] f(x1) < f(x2) for all x1,x2 such that a <= x1 < x2 <= b
similar for decreasing
local maximum (or minimum)
f(x) has a local max (or min) at x = a if there exists h>0 such that f(a) >= f(x) (or f(a)<= f(x))
on (a-h,a+h)
local extremum
either a local max or a local min
max/ min and derivative
if f(x) has a maximum (or minimum) at x = a and is differentiable at x = a
f’(a) = 0
stationary point
f(x) has one at x = a if its differentiable at x = a and f’(a) = 0
critical point
an interior point x = a of the domain of f(x) is a critical point if either f’(a) = 0 or f’(a) does not exist