MA 105 : Calculus
Division 1, Lecture 04
Prof. Sudhir R. Ghorpade IIT Bombay
Recap of the previous lecture
Equivalent definition of continuity (in terms of -condition)
Intermediate Value Property (IVP). Examples Intermediate Value Theorem. Application
Extreme Value Property (of continuous functions defined on closed and bounded intervals), Examples
Limits of functions of a real variable. Examples Continuity and Limit
Limit theorems for functions of a real variable Right hand and left hand limits
Equivalent condition for the existence of a limit Differentiability: Definition and Examples Right hand and left hand derivatives
Differentiability
Definition
LetD ⊂R, and let c be an interior point of D. A function f :D →R is said to be differentiable at c if
lim h→0
f(c+h)−f(c) h
exists. In this case, this limit is denoted byf0(c), and it is called the derivativeof f at c.
Geometrically speaking,f0(c) is equal to the slope of the tangentto the curve y =f(x) at x =c, and so the curve is ‘smooth’ at c; there is no corner or cuspat c.
Also,f0(c) can be interpreted as the rate of change in f atc.
Examples:
(i) If f :R→Ris a constant function, then f is differentiable and f0(c) = 0 for all c ∈R.
(ii) If f :R→Ris the identity function f(x) := x for x ∈R, then f is differentiable andf0(c) = 1 for all c ∈R. (iii) Letf(x) :=x2/3 forx ∈R. Then f is not differentiable
at 0 since f(0 +h)−f(0)
h =
1
h1/3 for h6= 0.
(iv) Letf(x) := sinx forx ∈R. Then f is differentiable at 0 since lim h→0 f(0 +h)−f(0) h = limh→0 sinh−sin 0 h−0 = 1. (v) Letf(0) := 0 and f(x) :=xsin(1/x) for x ∈R\ {0}.
Thenf is not differentiable at 0 since f(0 +h)−f(0) h = sin 1 h forh 6= 0.
Right hand and left hand derivatives
LetD ⊂R, and let f :D →R.Supposec ∈D is such that [c,c +r)⊂D for some r >0. If the limit
lim h→0+
f(c+h)−f(c) h
exists, then it is called theright hand derivative of f at c, and we denote it byf+0(c).
Theleft hand derivative of f atc is defined similarly, and we denote it byf−0(c).
If c is an interior point of D, then f is differentiable at c ⇐⇒ f+0(c) and f−0(c) both exist and are equal. Example: Letf(x) :=|x| forx ∈R. Then f−0(0) =−1 and f+0(0) = 1. Hence f is not differentiable at 0.
If I is a nonempty open interval inR, that is, if I := (a,b), (a,∞), (−∞,b) or R, then we say that a function f :I →R is differentiable onI if f is differentiable at everyc ∈I.
If I := [a,b], then we say that f :I →R is
differentiable onI if f is differentiable on (a,b), and if bothf+0(a) and f−0(b) exist.
Similarly, we define the differentiability of a function on I := [a,b), (a,b], [a,∞) or (−∞,b].
In these cases, there is a newfunction f0 :I →R, called the
derivative (function) of f defined onI.
Letc ∈I. If g :=f0 is differentiable at c, then we say that f istwice differentiable atc, and write f00(c) := g0(c). f00(c) is called thesecond derivative of f atc. Also, for n∈N, we denote thenth derivativeof f at c byf(n)(c).
Equivalent condition for differentiability
LetD ⊂R, letf :D →R, and letc be an interior point of D. Recall thatf is said to be differentiable at c if the limit
f0(c) := lim h→0
f(c +h)−f(c)
h exists.
(Carath´eodory Lemma: C-lemma)
Letf :D →R, and c be an interior point of D. Then f is differentiable at c ⇐⇒ there is a function f1 :D →R which is continuous atc and satisfies
f(x)−f(c) = (x −c)f1(x) for all x ∈D. In this case, the functionf1 is unique and f0(c) =f1(c). The functionf1 :D →Ris called the increment function associated withf andc.
Proof: Supposef is differentiable at c. Define f1(x) := f(x)−f(c) x −c if x ∈D\ {c}, f0(c) if x =c.
Thenf1 is continuous at c since limx→cf1(x) =f0(c) =f1(c). Conversely, suppose there isf1 :D →R as stated. Let
h:=x −c for x ∈D, and so c+h=x. Then lim h→0 f(c +h)−f(c) h = limx→c f(x)−f(c) x −c = limh→0f1(x) =f1(c), since f1 is continuous at c. Hence f is differentiable at c. The uniqueness of the increment functionf1 is obvious. If D is an interval and c is an end-point of D, then the C-lemma is valid atc.
The C-lemma relates the concepts of differentiability and continuity. It enables us to prove the following results neatly.
Differentiability =
⇒
Continuity
Theorem
Supposef :D →R is differentiable at an interior point c of D. Then f is continuous at c.
Proof.
Letf1 be the increment function associated with f and c. Thenf(x) =f(c) + (x−c)f1(x) for x ∈I. Since f1 :I →R is continuous at c, so is f.
Remarks:
1. If a functionf is not continuous at an interior pointc of D, then it cannot be differentiable atc.
Example: Letf(x) := [x] for x ∈R, and c := 1. 2. The converse of the above theorem is false. Example: f(x) =|x| forx ∈R, and c := 0.
Differentiability: Algebraic rules
Letf and g be differentiable atc. Then f ±g and f ·g are differentiable atc, and
(f ±g)0(c) = f0(c)±g0(c) resp.,
(f ·g)0(c) =f0(c)g(c) +f(c)g0(c).
Further, ifg(c)6= 0, then f/g is differentiable at c, and f g 0 (c) = f 0(c)g(c)−f(c)g0(c) g(c)2 .
Note: If f is differentiable at c, then fn is differentiable atc for each n∈N, and (fn)0(c) =n f(c)n−1f0(c). This also holds for a negative integern if f(c)6= 0. Examples: Every polynomial function p is differentiable on R. A rational functionp/q is differentiable at c if q(c)6= 0.
Differentiability: Chain Rule
Theorem
LetD ⊂R and f :D →R. Also, let E ⊂R and g :E →R be such thatf(D)⊂E. Suppose c is an interior point of D and f(c) is an interior point of E. Iff is differentiable at c, and ifg is differentiable at f(c), then g◦f :D →R is differentiable at c, and (g ◦f)0(c) =g0(f(c))f0(c).
Proof. Letf1 be the increment functions associated with f and c, and let g1 be the increment functions associated with g and d :=f(c). Then f(x)−f(c) = (x −c)f1(x) for all x ∈D, wheref1 is continuous at c, and g(y)−g(d) = (y −d)g1(y) for all y ∈E, where g1 is continuous at d.
Defineh:=g ◦f :D →R. Then for x ∈D, h(x)−h(c) = g(f(x))−g(f(c))
= f(x)−f(c)g1(f(x)) = (x−c)f1(x)(g1◦f)(x).
Leth1 :=f1·(g1◦f) :D →R. Then h1 is continuous atc since f and f1 are continuous atc and g1 is continuous at d =f(c). It follows that h1 is the increment function associated withg ◦f and c, and
(g ◦f)0(c) = h1(c) =g0(f(c))f0(c). Conclusion of the Chain Rule in the Leibnitz notation: If y :=f(x) and z :=g(y), thenz = (g ◦f)(x) and
dz dx x=c = dz dy y=f(c)· dy dx x=c Warning:
Cancelling out dy in the RHS above to obtain the LHS does notyield a proof of the Chain Rule!
Example: Consider f(x) := ( x2sin1 x if x ∈R\ {0}, 0 if x := 0. Let f1(x) := ( xsin1 x if x ∈R\ {0}, 0 if x := 0.
Thenf1 is continuous at 0, and f(x)−f(0) = (x−0)f1(x) for allx ∈R. By the C-lemma, f is differentiable at 0, and
f0(0) =f1(0) = 0.
Also, if c 6= 0, then f is differentiable at c, and f0(c) = 2csin1
c −cos 1
c. (Multiplication rule and chain rule.) Note: The derivative f0 of f is not continuous at 0. Why?
Differentiability of the inverse function
Theorem
LetI be an interval, and letc be an interior point ofI. If f :I →Ris a one-one continuous function such that f is differentiable at c and f0(c)6= 0, then f−1 :f(I)→
R is differentiable at f(c) and f−10 f(c) = 1 f0(c).
Proof: Skipped. You may try, if interested. Or look up [GL-1]. Note: Since f :I →R is continuous, it has the IVP, and so J =f(I) is an interval. Hence f(c) is either an interior point of J or an end-point of J. Since f is also one-one, it can be shown to be strictly monotonic (to be defined later), and so f(c) is in fact an interior point ofJ.
Examples of derivatives of inverse functions
(i) Let n∈N, and let f(x) :=xn for x ∈(0,∞). Then f is one-one and continuous. Considerc ∈(0,∞). Nowf0(c) = n cn−1 6= 0. Further,f (0,∞)
= (0,∞). If d ∈(0,∞) and f(c) =cn=d, thenc =d1/n and
f−10(d) = 1 f0(c) = 1 n cn−1 = 1 n d(n−1)/n = 1 nd (1/n)−1.
(ii) Let f(x) := sinx for x ∈(−π
2,
π
2). Then f is one-one and continuous. Considerc ∈(−π 2, π 2). Now f 0(c) = cosc 6= 0. Further, f (−π 2, π 2) = (−1,1). If d ∈(−1,1) and f(c) = sinc =d, then f−10(d) = 1 f0(c) = 1 cosc = 1 p 1−sin2c = √ 1 1−d2.
Local Extremum
LetD ⊆R, and let c be an interior point of D. We say that a functionf :D →R has
(i) a local minimum at c if there is δ >0 such that (c −δ,c+δ)⊆D and
f(x)≥f(c) for all x ∈(c−δ,c+δ).
(ii) a local maximum atc if there is δ >0 such that (c −δ,c+δ)⊆D and
f(x)≤f(c) for all x ∈(c−δ,c+δ).
(iii) a local extremumat c if it has a local maximum or a local minimum at c.
Proposition
LetD ⊆R, and let c be an interior point of D. Iff :D →R has a local extremum at c, and iff is differentiable at c, then f0(c) = 0.
Proof.
Letf be differentiable atc. By the C-lemma, there is f1 :D →R such that f1 is continuous at c and
f(x)−f(c) = (x−c)f1(x) for all x ∈D.
Letf have a local minimum at c. Then there is δ >0 such thatf(x)≥f(c) for all x ∈(c −δ,c+δ). Hence f1(x)≥0 for all x ∈(c,c+δ) and f1(x)≤0 for all x ∈(c−δ,c). Now 0≤ lim
x→c+f1(x) =f1(c) = limx→c−f1(x)≤0, since f1 is continuous at c. Hence f0(c) =f1(c) = 0.
A similar argument holds iff has a local maximum at c.
Examples:
(i) LetD := [0,1], and f(x) :=x forx ∈D. Then
f(0) = min{f(x) :x ∈D}, f(1) = max{f(x) :x ∈D}. Butf+0(0) = 16= 0 and f−0(1) = 16= 0. Note: Neither 0 nor 1 is an interior point ofD.
(ii) LetD := [−1,1], and f(x) =|x| forx ∈D. Then f has a local minimum at the interior point 0 of D. But f+0(0) = 16= 0 and f−0(1) =−16= 0. Note: f is not differentiable at 0.
(iii) LetD:= [−1,1], andf(x) :=x2 forx ∈D. Thenf has a local minimum at the interior point 0 ofD, and f0(0) = 0. (iv) LetD := [−1,1], and f(x) :=x3 forx ∈D. Then 0 is an interior point of D and f0(0) = 0. But f does not have a local extremum at 0. Thus the converse of the above proposition is false.
Rolle’s Theorem
Theorem
Leta <b and f : [a,b]→R be a function such that (i) f is continuous on [a,b],
(ii) f is differentiable on (a,b), and (iii) f(a) = f(b).
Then there exists c ∈(a,b) such that f0(c) = 0.
Thus there exists a point c ∈(a,b) such that the tangent to the graph at (c,f(c)) is parallel to the x-axis.
