MA 105 : Calculus Division 3, Lecture 23srg/courses/autumn2019/MA105... · MA 105 : Calculus...

MA 105 : Calculus

Division 3, Lecture 23

Prof. Sudhir R. GhorpadeIIT Bombay

Prof. Sudhir R. Ghorpade, IIT Bombay MA 105 Calculus: Division 3, Lecture 23

Recap of the previous lecture

Directional derivatives. Examples

Differentiation of a function of two real variables

Differentiability and Total Derivative

Linear Maps and Derivatives

Increment Lemma. Application: Differentiability impliesContinuity

Necessary condition for differentiability

Sufficient condition for differentiability


Differentiability and Total DerivativeSuppose D ⊂ R2 and (x0, y0) is an interior point of D.Consider a funtion f : D → R.

Definition

We say that f is differentiable at (x0, y0) if there is(α, β) ∈ R2 such that

lim(h,k)→(0,0)

f (x0 + h, y0 + k)− f (x0, y0)− αh − βk√h2 + k2

= 0.

In this case, the pair (α, β) ∈ R2 is called the total derivativeof f at (x0, y0). We have seen that if f is differentiable at(x0, y0), then both the partial derivatives at (x0, y0) exist andin fact, (α, β) = (∇f )(x0, y0). Moreover, directionalderivatives at (x0, y0) exists along any unit vector uuu exist, and

(DDDuuuf )(x0, y0) = (∇f )(x0, y0) · uuu.Prof. Sudhir R. Ghorpade, IIT Bombay MA 105 Calculus: Division 3, Lecture 23

Increment Lemma

We have the following two-dimensional analogue of theCaratheodory Lemma.

Proposition (Increment Lemma)

Let D ⊂ R2, and let (x0, y0) be an interior point of D. Thenf : D → R is differentiable at (x0, y0) if and only if there existfunctions f1, f2 : D → R such that f1 and f2 are continuous at(x0, y0), and for all (x , y) ∈ D,

f (x , y)− f (x0, y0) = (x − x0)f1(x , y) + (y − y0)f2(x , y).

In this case, ∇f (x0, y0) =(f1(x0, y0), f2(x0, y0)

).

A pair (f1, f2) of functions stated in the Increment Lemma iscalled a pair of increment functions associated with thefunction f and the point (x0, y0).


We now consider an important consequences of the IncrementLemma.

Proposition ( Differentiability =⇒ Continuity)

Let D ⊂ R2, and let (x0, y0) be an interior point of D.Suppose f : D → R is differentiable at (x0, y0). Then f iscontinuous at (x0, y0).

Proof:If (f1, f2) is a pair of increment functions associated with f and(x0, y0), then

f (x , y) = f (x0, y0) + (x − x0)f1(x , y) + (y − y0)f2(x , y)

for all (x , y) ∈ D. Consequently, the continuity of f at (x0, y0)follows from the continuity of f1 and f2 at (x0, y0).


Differentiability: Necessary Conditions

Let (x0, y0) be an interior point of a subset D of R2, and letf : D → R. Suppose f is differentiable at (x0, y0). Then thefollowing conditions necessarily hold, that is, they arenecessary conditions.

Both partial derivatives fx(x0, y0) and fy (x0, y0) exist.

The directional derivative (DDDuuuf )(x0, y0) exists for everyunit vector uuu ∈ R2.

(DDDuuuf )(x0, y0) =(fx(x0, y0), fy (x0, y0)

)·uuu = (∇f )(x0, y0) ·uuu

for every unit vector uuu ∈ R2.

f is continuous at (x0, y0).

None of the above conditions are sufficient to conclude that afunction is differentiable. In fact, we shall give an example of afunction for which all the above conditions hold at (x0, y0), butthe function is not differentiable at (x0, y0) (Example (vii)).


Proposition (Sufficient condition for differentiability)

Let (x0, y0) ∈ R2, r > 0, and let f : S → R, whereS := {(x , y) ∈ R2 : |x − x0| < r and |y − y0| < r}. Supposeone of the partial derivatives of f exists on S and iscontinuous at (x0, y0), while the other exists at (x0, y0). Thenf is differentiable at (x0, y0).

Proof. Without loss of generality, suppose fx exists on S and iscontinuous at (x0, y0), while fy exists at (x0, y0).For (x , y)∈S , f (x , y)− f (x0, y0) = A(x , y) + B(y), where

A(x , y) := f (x , y)−f (x0, y) and B(y) := f (x0, y)−f (x0, y0).

Define f1, f2 : S → R as follows.

f1(x , y) :=

A(x , y)

x − x0if x 6= x0,

fx(x0, y) if x = x0,

f2(x , y) :=

B(y)

y − y0if y 6= y0,

fy (x0, y0) if y = y0.


x

y

b

b b

x0 − r x0 + rx0

y0

y0 − r

y0 + r

(x0, y0)

(x0, y)(x, y)

Figure: The dotted lines linking (x , y) to (x0, y), and (x0, y) to(x0, y0).


By considering the cases x 6= x0, x = x0, y 6= y0 and y = y0,it follows that for all (x , y) ∈ S ,

f (x , y)− f (x0, y0) = (x − x0)f1(x , y) + (y − y0)f2(x , y).

To show f1 is continuous at (x0, y0), considerx ∈ (x0− r , x0)∪ (x0, x0 + r) and y ∈ (y0− r , y0 + r). Since fxexists on S , there is cy ∈ R between x0 and x such that

A(x , y) = f (x , y)− f (x0, y) = (x − x0)fx(cy , y).

by the MVT. Hence f1(x , y) = fx(cy , y). Also, by definition,f1(x0, y) = fx(x0, y) for all y ∈ (y0 − r , y0 + r). Since fx iscontinuous at (x0, y0), we see that f1 is continuous at (x0, y0).

Also, f2 is continuous at (x0, y0) since fy (x0, y0) exists.

By the Increment Lemma, f is differentiable at (x0, y0).


How to check differentiability?

To check the differentiability of a function of two variables, wemay see whether f violates one of the necessary conditions fordifferentiability. Also, we may see whether f satisfies asufficient condition for differentiability. Alternatively, we mayturn to the definition of differentiability itself.

Examples:

(i) Let f (x , y) :=√

x2 + y 2 for (x , y) ∈ R2.Then f is continuous at (0, 0). But (DDDuuuf )(0, 0) does not existfor any unit vector uuu. Hence f is not differentiable at (0, 0).

(ii) Let f (x , y) := 1 if 0 < y < x2 and f (x , y) := 0 otherwise.If uuu := (u1, u2), then f (tu1, tu2) = 0 for all t in an intervalabout 0, and so (DDDuuuf )(0, 0) = 0 = (∇f )(0, 0) ·uuu. But f is notcontinuous at (0, 0) since f (0, 0) = 0 and f (1/n, 1/2n2) = 1for all n ∈ N. Hence f is not differentiable at (0, 0).


(iii) Let f (x , y) := x2y/(x2 + y 2) for (x , y) 6= (0, 0), andf (0, 0) := 0. Then f is continuous at (0, 0), and(DDDuuuf )(0, 0) = u2

1u2 for every unit vector uuu := (u1, u2). But(DDDuuuf )(0, 0) 6= (∇f )(0, 0) · uuu unless u1 = 0 or u2 = 0.Hence f is not differentiable at (0, 0).

(iv) Let f (x , y) := x2 + y 2 for (x , y) ∈ R2. Let (x0, y0) ∈ R2.Then f is continuous at (x0, y0) and for every unit vector uuu,(DDDuuuf )(x0, y0) = (∇f )(x0, y0) · uuu. Since fx and fy exist on R2

and fx is continuous at (x0, y0), f is differentiable at (x0, y0).

(v) Let f (x , y) := x2y 2/(x4 + y 2) for (x , y) 6= (0, 0), andf (0, 0) := 0. We check that f is continuous at (0, 0), and(DDDuuuf )(0, 0) = 0 = (∇f )(0, 0) · uuu for every unit vector uuu.

Let us find fx(x0, y0) and fy (x0, y0) for (x0, y0) 6= (0, 0).


For (x0, y0) 6= (0, 0), it can be checked that

fx(x0, y0) =2x0y

20 (y 2

0 − x40 )

(x40 + y 20 )2

and fy (x0, y0) =2x60y0

(x40 + y 20 )2

.

Now fx is continuous at (0, 0) as y 40 , 2x

40y

20 ≤ (x40 + y 2

0 )2.Hence f is differentiable at (0, 0). (We note that fy is notcontinuous at (0, 0) since fy (1/n, 1/n2) = 1/2 for n ∈ N.)(Aliter: Here is a direct proof. For (h, k) 6= (0, 0),

0 ≤ h2k2 − 0− 0·h − 0·k(h4 + k2)

√h2 + k2

= hk2

(h4 + k2)

h√h2 + k2

≤ |h| → 0

as (h, k)→ (0, 0), and so f is differentiable at (0, 0).)


(vi) Let f (x , y) := |xy | for (x , y) ∈ R2. Then f is continuousat (0, 0), and (DDDuuuf )(0, 0) = 0 = (∇f )(0, 0) · uuu for every unitvector uuu. On the other hand, fx(0, y0) does not exist if y0 6= 0and fy (x0, 0) does not exist if x0 6= 0. For (h, k) 6= (0, 0),

Q(h, k) :=f (h, k)√h2 + k2

=|hk |√h2 + k2

≤ |h| → 0

as (h, k)→ (0, 0). Hence f is differentiable at (0, 0).

(vii) Let f (x , y) := x3y/(x4 + y 2) for (x , y) 6= (0, 0), andf (0, 0) := 0. Then (DDDuuuf )(0, 0) = 0 = (∇f )(0, 0) · uuu for everyunit vector uuu. Also, f is continuous at (0, 0). Further, fx andfy are not continuous at (0, 0). (Check!) For (h, k) 6= (0, 0),

Q(h, k) :=f (h, k)√h2 + k2

=h3k

(h4 + k2)√h2 + k2

6→ 0

as (h, k)→ (0, 0) since Q(1/n, 1/n2)→ 1/2. Hence f is notdifferentiable at (0, 0).


We obtain basic properties of differentiable functions, and alsothe gradients of a sum, of a difference, of a product and of aquotient of functions from the Increment Lemma as follows.

Proposition

Let D ⊂ R2 and let (x0, y0) be an interior point of D. Supposef , g : D → R are functions that are differentiable at (x0, y0).Then f ± g and f g are differentiable at (x0, y0), and moreover,

∇(f ± g)(x0, y0) = (∇f )(x0, y0)± (∇g)(x0, y0) resp.,

∇(f g)(x0, y0) = g(x0, y0)(∇f )(x0, y0) + f (x0, y0)(∇g)(x0, y0).

If g(x0, y0) 6= 0, then f /g is differentiable at (x0, y0) and

∇( fg

)(x0, y0) =

g(x0, y0)(∇f )(x0, y0)− f (x0, y0)(∇g)(x0, y0)

g(x0, y0)2.


Proposition (Chain Rules)

Let D ⊂ R2 and let (x0, y0) be an interior point of D, and letf : D → R be differentiable at (x0, y0).

(i) Let E ⊂ R be such that f (D) ⊂ E and z0 := f (x0, y0) is aninterior point of E . If g : E → R is differentiable at f (x0, y0),then the function h := g◦f :D → R is differentiable at (x0, y0),

hx(x0, y0) = g ′(z0) fx(x0, y0) and hy (x0, y0) = g ′(z0) fy (x0, y0).

(ii) Let E ⊂ R, and let t0 be an interior point of E . Ifx , y : E → R are differentiable at t0, and if (x(t), y(t)) ∈ Dfor all t ∈ E and (x(t0), y(t0)) := (x0, y0), then the functionφ : E → R defined by φ(t) := f (x(t), y(t)) for t ∈ E isdifferentiable at t0, and

φ′(t0) = fx(x0, y0) x ′(t0) + fy (x0, y0) y ′(t0).


Proposition (Chain Rules: continued)

(iii) Let E ⊂ R2, and let (u0, v0) be an interior point of E . Ifx , y : E → R are differentiable at (u0, v0), and if(x(u, v), y(u, v)) ∈ D for all (u, v) ∈ E and(x(u0, v0), y(u0, v0)) := (x0, y0), then the function F : E → Rdefined by F (u, v) := f (x(u, v), y(u, v)) for (u, v) ∈ E isdifferentiable at (u0, v0),

Fu(u0, v0) = fx(x0, y0) xu(u0, v0) + fy (x0, y0) yu(u0, v0),and

Fv (u0, v0) = fx(x0, y0) xv (u0, v0) + fy (x0, y0) yv (u0, v0).

It is often helpful to write the identities given by the ChainRules in an informal but suggestive notation as follows.

(i) If z = f (x , y) and w = g(z), then w is a function of (x , y),

and∂w

∂x=

dw

dz

∂z

∂x,

∂w

∂y=

dw

dz

∂z

∂y.


(ii) If z = f (x , y) and if x = x(t), y = y(t), then z is afunction of t, and

dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt.

(iii) If z = f (x , y) and if x = x(u, v), y = y(u, v), then z is afunction of u and v , and

∂z

∂u=∂z

∂x

∂x

∂u+∂z

∂y

∂y

∂uand

∂z

∂v=∂z

∂x

∂x

∂v+∂z

∂y

∂y

∂v.

It should be noted that the identities in (i), (ii), and (iii) aboveare valid when the concerned (partial) derivatives areevaluated at appropriate points and when, for instance, eachof the (partial) derivatives exists in a neighbourhood of therelevant point and is continuous at that point.


The conclusions of Chain Rules (i), (ii) and (iii) can be writtenin terms of the gradient (∇f )(x0, y0) of f at (x0, y0) as follows:

(i) (∇h)(x0, y0) = g ′(z0)(∇f )(x0, y0)

(ii) φ′(t0) = (∇f )(x0, y0) ·(x ′(t0), y ′(t0)

)(iii) (∇F )(u0, v0) =

((∇f )(x0, y0) ·

(xu(u0, v0), yu(u0, v0)

),

(∇f )(x0, y0) ·(xv (u0, v0), yv (u0, v0)

))The Chain Rules given above can be proved by using theIncrement Lemma.


Examples illustrating the Chain Rules:

(i) Define f : R2 → R and g : R→ R by f (x , y) := xy for(x , y) ∈ R2, and g(z) := sin z for z ∈ R. By the Chain Rule,the composite function (g◦f )(x , y) := sin(xy) is differentiableat every point of R2, and

∂(g◦f )

∂x=

dg

dz

∂z

∂x= (cos xy)y and

∂(g◦f )

∂y=

dg

dz

∂z

∂y= (cos xy)x .

(ii) Define f : R2 → R by f (x , y) := x2 + y 2 for (x , y) ∈ R2,and x , y : R→ R by x(t) := et and y(t) := sin t for t ∈ R.By the Chain Rule, the composite function φ : R→ R givenby φ(t) := f (x(t), y(t)) = e2t + sin2 t is differentiable at everypoint of R, and

dφ

dt=

∂f

∂x

dx

dt+∂f

∂y

dy

dt= (2x(t))et + (2y(t)) cos t

= 2e2t + 2 sin t cos t.


(iii) As in (ii) above, define f : R2 → R by f (x , y) := x2 + y 2

for (x , y) ∈ R2, and x , y : R2 → R by x(u, v) := u2 − v 2 andy(u, v) := 2uv for (u, v) ∈ R2. By the Chain Rule, thecomposite function F : R2 → R given by

F (u, v) := f (x(u, v), y(u, v)) = (u2 − v 2)2 + (2uv)2

= u4 + 2u2v 2 + v 4

is differentiable at every point of R2, and

∂F

∂u=

∂f

∂x

∂x

∂u+∂f

∂y

∂y

∂u= 2x(2u) + 2y(2v)

= 2(u2 − v 2)(2u) + 2(2uv)(2v) = 4u(u2 + v 2),

∂F

∂v=

∂f

∂x

∂x

∂v+∂f

∂y

∂y

∂v= 2x(−2v) + 2y(2u)

= 2(u2 − v 2)(−2v) + 2(2uv)(2u) = 4v(u2 + v 2).


Geometric Interpretation of the Gradient

Suppose f : D ⊂ R2 → R is differentiable at an interior point(x0, y0) of D, and suppose (∇f )(x0, y0) 6= (0, 0). Then forevery unit vector uuu,

(DDDuuuf )(x0, y0) = (∇f )(x0, y0) · uuu = ‖(∇f )(x0, y0)‖ cos θ,

where θ ∈ [0, π] is the angle between (∇f )(x0, y0) and uuu.Hence

(DDDuuuf )(x0, y0) is maximum when cos θ = 1, that is, θ = 0,

so that uuu =(∇f )(x0, y0)

‖(∇f )(x0, y0)‖ .

(DDDuuuf )(x0, y0) is minimum when cos θ=−1, that is, θ = π,

so that uuu = − (∇f )(x0, y0)

‖(∇f )(x0, y0)‖ .

(DDDuuuf )(x0, y0) = 0 when cos θ = 0, that is, θ = π/2,so that uuu is perpendicular to (∇f )(x0, y0).


Example:Let f (x , y) := 4− x2 − y 2 for (x , y) ∈ R2. Then f isdifferentiable on R2, and (∇f )(x0, y0) = (−2x0,−2y0) for all(x0, y0) ∈ R2.

Let (x0, y0) := (1, 1). Then (∇f )(1, 1) = (−2,−2).

Consider the surface z = f (x , y). At (1, 1),

the direction of steepest ascent is(−2,−2)/‖(−2,−2)‖ = (−1/

√2,−1/

√2),

the direction of steepest descent is−(−2,−2)/‖(−2,−2)‖ = (1/

√2, 1/√

2),

the directions (u,u2) of no change (in height) satisfy(−2,−2) · (u1, u2) = 0, and so, they are (1/

√2,−1/

√2)

and (−1/√

2, 1/√

2).

See the picture on the next slide.


b

b (1, 1)

2

(−1√2,−1√2) (

1√2,

1√2)

(−1√2,

1√2)

(1√2,−1√2)

Figure: Directions of steepest ascent, of steepest decent and of nochange in height


Tangent Line

Recall the one variable situation:Let a < b and x0 ∈ (a, b). Let a function f : (a, b)→ R bedifferentiable at x0. Then the equation of the tangent line tothe curve y = f (x) in R2 at (x0, f (x0)) is given by

y − f (x0) = f ′(x0)(x − x0).

More generally, suppose D ⊂ R2 and P0 := (x0, y0) is aninterior point of D. Let a function F : D → R have partialderivatives at P0, and let (∇F )(P0) 6= (0, 0). Suppose Fdefines a curve C in R2 (implicitly) by the equationF (x , y) = 0 for (x , y) ∈ D, and P0 lies on C . Then theequation of the tangent line to C at P0 is given by

Fx(P0)(x − x0) + Fy (P0)(y − y0) = 0.

Note: If D := (a, b)× R, f : (a, b)→ R, and for (x , y) ∈ D,F (x , y) := y − f (x), then y0 = f (x0), Fx(P0) = −f ′(x0) andFy (P0) = 1. We recover the earlier equation of the tangent line.


MA 105 : Calculus Division 3, Lecture 23srg/courses/autumn2019/MA105... · MA 105 : Calculus...

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Transcript of MA 105 : Calculus Division 3, Lecture 23srg/courses/autumn2019/MA105... · MA 105 : Calculus...