Chapter 5 differentiation Flashcards
Differentiation definition
Let f be a real-valued function with domain D_f. We say that f is differentiable at a ∈D_f if Lim( x->a) [ (f(x)-f(a))/(x-a)] exists and is finite/real.
Then we have
f’(a) =
Lim( x->a) [ (f(x)-f(a))/(x-a)] = Lim( h->0) [ (f(a+h)-f(a))/(h)]
And we call f’(a) in R the derivative of f at a.
We say that f is differentiable on S subset of D_f is it is differentiable at every point a in S
Derivative function
From the limit,
Real valued function f’ with domain
D_f’ = {x∈D_f| f’(x) exists}
Notation for n∈N with n>2 we define the nth derivative of f at a by f^(n) (a) = ( f^(n-1)’)(a)
Whenever the limit on the right hand side exists we sat f us infinitely differentiable or smooth at a if f’(a) exists
Theorem 5.2.4 differentiable and continuous?
If a real valued function f is differentiable at a∈D_f then f is continuous at a
Not true that every function that is continuous is differentiable
Proof:
By showing that f(x) -f(a) =( [f(x)-f(a) ]/ [x-a] )•(x-a) and as differentiable by taking limits = f’(a)•0 =0 .
Definition: left derivative and right derivative
We say that the REAL VALUED function f has a left derivative at a∈D_f if f’_(a) = lim h↑0 ( f(a+h) -f(a))/h exists and is finite that it has a right derivative at a∈D_f if f’+(a) =lim h↓0 (f(a+h)-f(a))/h exists and is finite
Theorem 5.2.7 when is a real valued function is differentiable
A real valued function f is differentiable at a∈D_f
IF AND ONLY IF
Both the left and right derivatives at a exist and are equal. In this case f’(a) = f’_(a) = f’+(a)
(Left and right equal)
Proof:
Exercise
Theorem 5.3.1 Rules for differentiation
Sum
Muliplied
Quotient
Of functions
Let f and g be real-valued functions that are differentiable at a∈(D_f ∩ D_g). Then the following hold:
1) for each α,β∈R, the function of αf+βg is differentiable at a and (αf+βg)’(a) = αf’(a) +βg’(a)
2) the product rule: the function fg is differentiable at a and (fg)’(a)=f’(a)g(a)+f(a)g’(a).
3. The quotient rule: if g(a)≠0 then f/g is differentiable at a and (f/g)’(a) = (g(a)f’(a)-f(a)g’(a))/(g(a)²)
Proof:
1) algebra of limits
2) relevant limits
3) …
Theorem 5.3.2
Differentiation for composites
The chain rule:
Let f,g be real valued functions such that the range of g is contained in the domain of f. Suppose that g is differentiable at a and that f is differentiable at g(a). Then (f∘g)’(a) = f’(g(a))g’(a)
Definition 5.4.1: for functions local minimum and local maximum and turning points
A REAL-valued function f has a local minimum at a∈D_f if there exists δ>0 such that (a-δ,a+δ) ⊂D_f and f(x) ≥ f(a) for all x∈(a-δ,a+δ).
A real-valued function f has a local maximum at a ∈D_f if there exists δ>0 such that (a-δ,a+δ) ⊂D_f and f(x) ≤ f(a) for all x∈(a-δ,a+δ).
A turning point/extreme point for f is a point in its domain that is either a local minimum or local maximum.
Difference between local and global maxima minima
Eg if f: [a,b] -> R us continuous then by theorem 4.3.4 it attains its supremum and infimum on [a,b]. So these are the global maximum and minimum
But aren’t necessarily turning points as they might be at the end points of the interval [a,b].
Local need not be global.
Theorem 5.4.3 differentiable turning points values
If f is DIFFERENTIABLE at a∈D_f and a is a turning point for f then f’(a) =0.
These are sometimes called stationary points
Turning points are stationary points but not all stationary points are turning points: some are inflection points
Theorem 5.4.4 Rolle’s theorem
Let f be a real-values function that is continuous on [a,b] and differentiable on (a,b) with f(a) =f(b) then there exists ∈(a,b) such that f’(c) =0
Diagram
Mean value theorem 5.5.1
**
If a real-valued function f is continuous on [a,b] and differentiable on (a,b), then there exists c∈(a,b) such that
f’(c) = (f(b) -f(a))/(b-a)
So when you know it’s differentiable …
Does not say unique c!
Example- let a and b in R and consider limit as n tends to infinity of ( |a|^n + |b|^n) ^(1/n)
Does the limit exist? Prove your guess.
By considering cases of a and b eg 0s we guess that converges to max{|a|,|b|}
Proving this:
Let M equal this guess. Then we have |a| and |b| less than or equal to M.
So raising to the power of n.
|a|^n + |b|^n less than or equal to 2M^n
And so
(|a|^n + |b|^n)^(1/n) less than or equal to (2M^n) = 2^(1/n) M
Using the algebra of limits limit as n tends to infinity of this gives M. Hence it follows by the sandwich our original limit converges to M.
Example:
f(x)=c where c in R us constant
Checking if differentiable
f(x) = x^n
1) directly from definition of differentiability
2) f(x) = x^n for x in R where n in N fixed. By the binomial theorem
f is differentiable at all x in R:
( f(a+h) - f(a)) /h
= ((a+h)^n - a^n)/h
= ( a^n + na^{n-1} h + 0.5•n•(n-1)•a^(n-2)•h²+…+ nah^{n-1} + h^n -a^n) /h
= na^{n-1} + + 0.5•n•(n-1)•a^(n-2)•h+…+ nah^{n-2} + h^{n-1}
Thus the limit as h tends to 0 is na^{n-1}
So f’(x) =nx^n-1. For all x in R and D_f’ = D_f =R
SUMMARY: find the limit defined in the differentiable definition, for any a in the domain. (This was found using the BINOMIAL THEOREM)
Example: consider a function f(x) =|x| with D_f =R which is continuous at every point in R and clearly differentiable at every x not equal to 0.
WE SHOW THAT ITS NOT DIFFERENTIAVLE AT 0 by showing the Left and RIGHT limits are DIFFERENT
Lim_ h↑0 [ (f(0+h - f(0) )/h] = lim_h↑0[ |h| /h] = lim_h↑0[-h/h] =-1
Lim_ h↓0 [ (f(0+h - f(0) )/h] = lim_h↓0[ |h| /h] = lim_h↑0[h/h] =1
Thus the left and right limits are different and so limit as h tends to 0 of f(0+h) -f(0) /h does not exist.
Example: consider function f; [-3,2] to R defined by
f(x) =
{ x+2 if x in [-3,-1]
{x^2 if x in [-1,2]
Identify any turning points and global maximum and minimum
The function is continuous: at x=-1 we can check this.
Global max 4 at x=2
Global min -1 x=-3
These aren’t turning points as they’re end points. That is we can’t find a delta st all points within distance delta are contained in the domain.
Local minimum at x=0.
Stationary points and turning points
Tps are stationary points
Not all stationary points are tps
f(x)=x^3. The. f’(0) =0 but 0 is neither local max nor local min but is an INFLECTION POINT
Corollary 5.5.2
Monotonicity revisited
f’(x) for monotone functions
Suppose that a real valued function f is continuous on [a,b] and differentiable on (a,b)
If for all x in (a,b) we have
f’(x) ≥ 0 : f is monotone increasing on [a,b]
f’(x) > 0 : f is strictly monotone increasing on [a,b]
f’(x) ≤ 0 : f is monotone decreasing on [a,b]
f’(x) < 0 : f is strictly monotone decreasing on [a,b]
By using the MVT showing existence and bigger than or equal to 0 etc
•corollary useful tool to study inverses of functions when used in conjunction with the inverse function theorem
Theorem 5.5.3
Inverses revisited.
Continuity of inverses and delta
Suppose that f:R to R is continuous on [a,b] and differentiable on (a,b) and that f’ is continuous at c∈(a,b). If f’(c) ≉ 0 then the following hold:
1) there exists δ>0 so that f is invertible on ]c-δ, c+δ]
And f^- is continuous on ( f(c-δ), f(c+δ)) if f’(c) > 0, and on ( f(c+δ), f(c-δ)) if f’(c) < 0.
2) the mapping f^-1 is differentiable at f(c) and
(f^-1)’(f(c)) = 1/ [f’(c)]
• in calculus du/dx = 1/ (du/dx)
Theorem 5.5.4
Cauchys mean value theorem
Let f and g each be continuous on [a,b] and differentiable on (a,b) with
g’(x) ≠ 0 for all x∈(a,b).
Then there exists c∈(a,b) so that
f’(c)/g’(c)
= ( f(b) -f(a)) / ( g(b) - g(a))
Proof:
By rolled theorem…
Corollary 5.5.5 l’Hôpitals rule
Suppose that f and g are each differentiable on (a,b) with g’(x) ≠ 0 for all x∈(a,b).
1) if c∈(a,b) with f(c) = g(c)=0 then
Limit as x tends to c of (f(x)/g(x))
= limit as x tends to c of (f’(x)/g’(x))
Whenever the limit on RHS exists and is finite.
2) and 3) similarly if limit as x tends to a-= limit as x tends to b- =0
And if limit as x tends to b+ = limit as x tends to a+ =0
Works for infinity also
Example 5.5.6 evaluate
Limit as x↓0 of x^x
Define f(x) =a^x as f(x) = e^{xlnx} for x in the reals.
By l’Hôpitals
Limit as x↓0 of xlnx = Limit as x↓0 of (ln(x)/(1/x))
= -Limit as x↓0 of ((1/x)/(1/x²)) = -Limit as x↓0 of x =0
So by the CONTINUITY of the exponential function
Limit as x↓0 of x^x = Limit as x↓0 of e^xln(x) = e^ ( Limit as x↓0 of xlnx) = e^0 = 1
Definition 5.6.1:
For each n in N real vector space for
For each n in N we define the real vector space C^n (a,b) to consist of functions f([a,b] to R for which
- the nth derivative f^(n) of f exists for all points (a,b)
- f^(n) is continuous on (a,b)
We define vector space with n= infinity as that of functions that are infinitely differentiable on (a,b)
Clearly this is a subset of n=n which is a subset of n= n-1 … and of C(a,b) (which is a space of continuous functions on (a,b))
Definition 5.6.2
TAYLOR COEFFICIENTS
AND. TAYLOR POLYNOMIAL..
Let f∈C^n (a,b) for some n∈N. fix x₀∈ (a,b). The real numbers f^(k) (x₀)/ k! For k=0,1,..,n are TAYLOR COEFFICIENTS of f at x₀.
We define function T^(n) _f ∈ C^n (a,b) by
T^(n) _f (x) = Σ from k=0 to n
Of [( f^(k) (x₀)) / k!] • (x-x₀)^k
This function is the TAYLOR POLYNOMIAL of f of degree n around x₀
Theorem 5.6.3 Taylor’s theorem
Let f∈C^{n+1} (a,b) for some n∈N. fix x₀∈ (a,b).
Then for all x∈ (a,b),
f(x) = Σ from k=0 to n
Of
[( f^(k) (x₀)) / k!] • (x-x₀)^k + (R^{n+1}_f) (x)
Where R^{n+1}_f (x) + (f^{n+1}(c)) /( (n+1)!) • (x- x₀)^{n+1}, for some c depending on x with c ∈ ( x₀,x) if x > x₀ and c ∈ ( x,x₀) if x< x₀
The R term measures the error in approximating f by it’s Taylor polynomial degree n at f (remainder term of degree n+1) = f(x) - T^(n) _f (x)
If 0 in (a,b) can take x_0 as 0 Maclaurins
f is represented by its Taylor series
For f in C^ infinity (a,b) and if series terms of Taylor’s converges for all x in (a,b) then we can write f(x) = this limit as n tends to infinity
Examples we can check differentiability of functions
By the definition of the limit of differentiation. Showing it exists for all a in the domain of f.
