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More on Partial Derivatives

Module by: Lawrence Baggett. E-mail the author

Summary: We close the chapter with a little more concerning partial derivatives. Thus far, we have discussed functions of a single variable, either real or complex. However, it is difficult not to think of a function of one complex variable z=x+iyz=x+iy as equally well being a function of the two real variables xx and y.y. We will write (a,b)(a,b) and a+bia+bi to mean the same point in C≡R2,C≡R2, and we will write |(a,b)||(a,b)| and |a+bi||a+bi| to indicate the same quantity, i.e., the absolute value of the complex number a+bi≡(a,b).a+bi≡(a,b). We have seen before that the only real-valued, differentiable functions of a complex variable are the constant functions. However, this is far from the case if we consider real-valued functions of two real variables, as is indicated previously. Consequently, we make the following definition of differentiability of a real-valued function of two real variables. Note that it is clearly different from the definition of differentiability of a function of a single complex variable, and though the various notations for these two kinds of differentiability are clearly ambiguous, we will leave it to the context to indicate which kind we are using.

We close the chapter with a little more concerning partial derivatives. Thus far, we have discussed functions of a single variable, either real or complex. However, it is difficult not to think of a function of one complex variable z=x+iyz=x+iy as equally well being a function of the two real variables xx and y.y. We will write (a,b)(a,b) and a+bia+bi to mean the same point in C≡R2,C≡R2, and we will write |(a,b)||(a,b)| and |a+bi||a+bi| to indicate the same quantity, i.e., the absolute value of the complex number a+bi≡(a,b).a+bi≡(a,b). We have seen in (Reference) that the only real-valued, differentiable functions of a complex variable are the constant functions. However, this is far from the case if we consider real-valued functions of two real variables, as is indicated in (Reference). Consequently, we make the following definition of differentiability of a real-valued function of two real variables. Note that it is clearly different from the definition of differentiability of a function of a single complex variable, and though the various notations for these two kinds of differentiability are clearly ambiguous, we will leave it to the context to indicate which kind we are using.

Definition 1:

Let f:S→Rf:S→R be a function whose domain is a subset SS of R2,R2, and let c=(a,b)c=(a,b) be a point in the interior S0S0 of S.S. We say that ff is differentiable, as a function of two real variables, at the point (a,b)(a,b) if there exists a pair of real numbers L1L1 and L2L2 and a function θθ such that

f ( a + h 1 , b + h 2 ) - f ( a , b ) = L 1 h 1 + L 2 h 2 + θ ( h 1 , h 2 ) f ( a + h 1 , b + h 2 ) - f ( a , b ) = L 1 h 1 + L 2 h 2 + θ ( h 1 , h 2 )
(1)

and

lim | ( h 1 , h 2 ) | → 0 θ ( h 1 , h 2 ) | ( h 1 , h 2 ) | = 0 . lim | ( h 1 , h 2 ) | → 0 θ ( h 1 , h 2 ) | ( h 1 , h 2 ) | = 0 .
(2)

One should compare this definition with part (3) of (Reference).

Each partial derivative of a function ff is again a real-valued function of two real variables, and so it can have partial derivatives of its own. We use simplifying notation like fxyxxfxyxx and fyyyxyy...fyyyxyy... to indicate “higher order” mixed partial derivatives. For instance, fxxyxfxxyx denotes the fourth partial derivative of f,f, first with respect to x,x, second with respect to xx again, third with respect to y,y, and finally fourth with respect to x.x. These higher order partial derivatives are called mixed partial derivatives.

Definition 2:

Suppose SS is a subset of R2,R2, and that ff is a continuous real-valued function on S.S. If both partial derivatives of ff exist at each point of the interior S0S0 of S,S, and both are continuous on S0,S0, then ff is said to belong to C1(S).C1(S). If all kkth order mixed partial derivatives exist at each point of S0,S0, and all of them are continuous on S0,S0, then ff is said to belong to Ck(S).Ck(S). Finally, if all mixed partial derivatives, of arbitrary orders, exist and are continuous on S0,S0, then ff is said to belong to C∞(S).C∞(S).

Exercise 1

  1. Suppose ff is a real-valued function of two real variables and that it is differentiable, as a function of two real variables, at the point (a,b).(a,b). Show that the numbers L1L1 and L2L2 in the definition are exactly the partial derivatives of ff at (a,b).(a,b). That is,
    L1=tialftialx(a,b)=limh→0f(a+h,b)-f(a,b)hL1=tialftialx(a,b)=limh→0f(a+h,b)-f(a,b)h
    (3)
    and
    L2=tialftialy(a,b)=limh→0f(a,b+h)-f(a,b)h.L2=tialftialy(a,b)=limh→0f(a,b+h)-f(a,b)h.
    (4)
  2. Define ff on R2R2 as follows: f(0,0)=0,f(0,0)=0, and if (x,y)≠(0,0),(x,y)≠(0,0), then f(x,y)=xy/(x2+y2).f(x,y)=xy/(x2+y2). Show that both partial derivatives of ff at (0,0)(0,0) exist and are 0. Show also that ff is not, as a function of two real variables, differentiable at (0,0).(0,0). HINT: Let hh and kk run through the numbers 1/n.1/n.
  3. What do parts (a) and (b) tell about the relationship between a function of two real variables being differentiable at a point (a,b)(a,b) and its having both partial derivatives exist at (a,b)?(a,b)?
  4. Suppose f=u+ivf=u+iv is a complex-valued function of a complex variable, and assume that ff is differentiable, as a function of a complex variable, at a point c=a+bi≡(a,b).c=a+bi≡(a,b). Prove that the real and imaginary parts uu and vv of ff are differentiable, as functions of two real variables. Relate the five quantities
    tialutialx(a,b),tialutialy(a,b),tialvtialx(a,b),tialvtialy(a,b),andf'(c).tialutialx(a,b),tialutialy(a,b),tialvtialx(a,b),tialvtialy(a,b),andf'(c).
    (5)

Perhaps the most interesting theorem about partial derivatives is the “mixed partials are equal” theorem. That is, fxy=fyx.fxy=fyx. The point is that this is not always the case. An extra hypothesis is necessary.

Theorem 1: Theorem on mixed partials

Let f:S→Rf:S→R be such that both second order partials derivatives fxyfxy and fyxfyx exist at a point (a,b)(a,b) of the interior of S,S, and assume in addition that one of these second order partials exists at every point in a disk Br(a,b)Br(a,b) around (a,b)(a,b) and that it is continuous at the point (a,b).(a,b). Then fxy(a,b)=fyx(a,b).fxy(a,b)=fyx(a,b).

Proof

Suppose that it is fyxfyx that is continuous at (a,b).(a,b). Let ϵ>0ϵ>0 be given, and let δ1>0δ1>0 be such that if |(c,d)-(a,b)|<δ1|(c,d)-(a,b)|<δ1 then |fyx(c,d)-fyx(a,b)|<ϵ.|fyx(c,d)-fyx(a,b)|<ϵ. Next, choose a δ2δ2 such that if 0<|k|<δ2,0<|k|<δ2, then

| f x y ( a , b ) - f x ( a , b + k ) - f x ( a , b ) k | < ϵ , | f x y ( a , b ) - f x ( a , b + k ) - f x ( a , b ) k | < ϵ ,
(6)

and fix such a k.k. We may also assume that |k|<δ1/2.|k|<δ1/2. Finally, choose a δ3>0δ3>0 such that if 0<|h|<δ3,0<|h|<δ3, then

| f x ( a , b + k ) - f ( a + h , b + k ) - f ( a , b + k ) h | < | k | ϵ , | f x ( a , b + k ) - f ( a + h , b + k ) - f ( a , b + k ) h | < | k | ϵ ,
(7)

and

| f x ( a , b ) - f ( a + h , b ) - f ( a , b ) h | < | k | ϵ , | f x ( a , b ) - f ( a + h , b ) - f ( a , b ) h | < | k | ϵ ,
(8)

and fix such an h.h. Again, we may also assume that |h|<δ1/2.|h|<δ1/2.

In the following calculation we will use the Mean Value Theorem twice.

0 ≤ | f x y ( a , b ) - f y x ( a , b ) | ≤ | f x y ( a , b ) - f x ( a , b + k ) - f x ( a , b ) k | + | f x ( a , b + k ) - f x ( a , b ) k - f y x ( a , b ) | ≤ ϵ + | f x ( a , b + k ) - f ( a + h , b + k ) - f ( a , b + k ) h k | + | f ( a + h , b ) - f ( a , b ) h - f x ( a , b ) k | + | f ( a + h , b + k ) - f ( a , b + k ) + ( f ( a + h , b ) - f ( a , b ) ) h k - f y x ( a , b ) | < 3 ϵ + | f ( a + h , b + k ) - f ( a , b + k ) + ( f ( a + h , b ) - f ( a , b ) ) h k - f y x ( a , b ) | = 3 ϵ + | f y ( a + h , b ' ) - f y ( a , b ' ) h - f y x ( a , b ) | = 3 ϵ + | f y x ( a ' , b ' ) - f y x ( a , b ) | < 4 ϵ , 0 ≤ | f x y ( a , b ) - f y x ( a , b ) | ≤ | f x y ( a , b ) - f x ( a , b + k ) - f x ( a , b ) k | + | f x ( a , b + k ) - f x ( a , b ) k - f y x ( a , b ) | ≤ ϵ + | f x ( a , b + k ) - f ( a + h , b + k ) - f ( a , b + k ) h k | + | f ( a + h , b ) - f ( a , b ) h - f x ( a , b ) k | + | f ( a + h , b + k ) - f ( a , b + k ) + ( f ( a + h , b ) - f ( a , b ) ) h k - f y x ( a , b ) | < 3 ϵ + | f ( a + h , b + k ) - f ( a , b + k ) + ( f ( a + h , b ) - f ( a , b ) ) h k - f y x ( a , b ) | = 3 ϵ + | f y ( a + h , b ' ) - f y ( a , b ' ) h - f y x ( a , b ) | = 3 ϵ + | f y x ( a ' , b ' ) - f y x ( a , b ) | < 4 ϵ ,
(9)

because b'b' is between bb and b+k,b+k, and a'a' is between aa and a+h,a+h, so that |(a',b')-(a,b)|<δ1/2<δ1.|(a',b')-(a,b)|<δ1/2<δ1. Hence, |fxy(a,b)-fyx(a,b)<4ϵ,|fxy(a,b)-fyx(a,b)<4ϵ, for an arbitrary ϵ,ϵ, and so the theorem is proved.

Exercise 2

Let ff be defined on R2R2 by f(0,0)=0f(0,0)=0 and, for (x,y)≠(0,0),(x,y)≠(0,0),f(x,y)=x3y/(x2+y2).f(x,y)=x3y/(x2+y2).

  1. Prove that both partial derivatives fxfx and fyfy exist at each point in the plane.
  2. Show that fyx(0,0)=1fyx(0,0)=1 and fxy(0,0)=0.fxy(0,0)=0.
  3. Show that fxyfxy exists at each point in the plane, but that it is not continuous at (0,0).(0,0).

The following exercise is an obvious generalization of the First Derivative Test for Extreme Values, (Reference), to real-valued functions of two real variables.

Exercise 3

Let f:S→Rf:S→R be a real-valued function of two real variables, and let c=(a,b)∈S0c=(a,b)∈S0 be a point at which ff attains a local maximum or a local minimum. Show that if either of the partial derivatives tialf/tialxtialf/tialx or tialf/tialytialf/tialy exists at c,c, then it must be equal to 0.

HINT: Just consider real-valued functions of a real variable like x→f(x,b)x→f(x,b) or y→f(a,y),y→f(a,y), and use (Reference).

Whenever we make a new definition about functions, the question arises of how the definition fits with algebraic combinations of functions and how it fits with the operation of composition. In that light, the next theorem is an expected one.

Theorem 2

(Chain Rule again) Suppose SS is a subset of R2,R2, that (a,b)(a,b) is a point in the interior of S,S, and that f:S→Rf:S→R is a real-valued function that is differentiable, as a function of two real variables, at the point (a,b).(a,b). Suppose that TT is a subset of R,R, that cc belongs to the interior of T,T, and that φ:T→R2φ:T→R2 is differentiable at the point cc and φ(c)=(a,b).φ(c)=(a,b). Write φ(t)=(x(t),y(t)).φ(t)=(x(t),y(t)). Then the composition f∘φf∘φ is differentiable at cc and

f ∘ φ ' ( c ) = t i a l f t i a l x ( a , b ) x ' ( c ) + t i a l f t i a l y ( a , b ) y ' ( c ) = t i a l f t i a l x ( φ ( c ) ) x ' ( c ) + t i a l f t i a l y ( φ ( c ) ) y ' ( c ) . f ∘ φ ' ( c ) = t i a l f t i a l x ( a , b ) x ' ( c ) + t i a l f t i a l y ( a , b ) y ' ( c ) = t i a l f t i a l x ( φ ( c ) ) x ' ( c ) + t i a l f t i a l y ( φ ( c ) ) y ' ( c ) .
(10)

Proof

From the definition of differentiability of a real-valued function of two real variables, write

f ( a + h 1 , b + h 2 ) - f ( a , b ) = L 1 h 1 + L 2 h 2 + θ f ( H 1 , h 2 ) . f ( a + h 1 , b + h 2 ) - f ( a , b ) = L 1 h 1 + L 2 h 2 + θ f ( H 1 , h 2 ) .
(11)

and from part (3) of (Reference), write

φ ( c + h ) - φ ( c ) = φ ' ( c ) h + θ φ ( h ) , φ ( c + h ) - φ ( c ) = φ ' ( c ) h + θ φ ( h ) ,
(12)

or, in component form,

x ( c + h ) - x ( c ) = x ( c + h ) - a = x ' ( c ) h + θ x ( h ) x ( c + h ) - x ( c ) = x ( c + h ) - a = x ' ( c ) h + θ x ( h )
(13)

and

y ( c + h ) - y ( c ) = y ( c + h ) - b = y ' ( c ) h + θ y ( h ) . y ( c + h ) - y ( c ) = y ( c + h ) - b = y ' ( c ) h + θ y ( h ) .
(14)

We also have that

lim | ( h 1 , h 2 ) | → 0 θ f ( ( h 1 , h 2 ) ) | ( h 1 , h 2 ) | = 0 , lim | ( h 1 , h 2 ) | → 0 θ f ( ( h 1 , h 2 ) ) | ( h 1 , h 2 ) | = 0 ,
(15)
lim h → 0 θ x ( h ) h = 0 , lim h → 0 θ x ( h ) h = 0 ,
(16)

and

lim h → 0 θ y ( h ) h = 0 . lim h → 0 θ y ( h ) h = 0 .
(17)

We will show that f∘φf∘φ is differentiable at cc by showing that there exists a number LL and a function θθ satisfying the two conditions of part (3) of (Reference).

Define

k 1 ( h ) , k 2 ( h ) ) = φ ( c + h ) - φ ( c ) = ( x ( c + h ) - x ( c ) , y ( c + h ) - y ( c ) ) . k 1 ( h ) , k 2 ( h ) ) = φ ( c + h ) - φ ( c ) = ( x ( c + h ) - x ( c ) , y ( c + h ) - y ( c ) ) .
(18)

Thus, we have that

f ∘ φ ( c + h ) - f ∘ φ ( c ) = f ( φ ( c + h ) ) - f ( φ ( c ) ) = f ( x ( c + h ) , y ( c + h ) ) - f ( x ( c ) , y ( c ) ) = f ( a + k 1 ( h ) , b + k 2 ( h ) ) - f ( a , b ) = L 1 k 1 ( h ) + L 2 k 2 ( h ) + θ f ( k 1 ( h ) , k 2 ( h ) ) = l 1 ( x ( c + h ) - x ( c ) ) + L 2 ( y ( c + h ) - y ( c ) ) + θ f ( k 1 ( h ) , k 2 ( h ) ) = L 1 ( x ' ( c ) h + θ x ( h ) ) + L 2 ( y ' ( c ) h + θ y ( h ) ) + θ f ( k 1 ( h ) , k 2 ( h ) ) = ( L 1 x ' ( c ) + L 2 y ' ( c ) ) h + L 1 θ x ( h ) + L 2 θ y ( h ) + θ f ( k 1 ( h ) , k 2 ( h ) ) . f ∘ φ ( c + h ) - f ∘ φ ( c ) = f ( φ ( c + h ) ) - f ( φ ( c ) ) = f ( x ( c + h ) , y ( c + h ) ) - f ( x ( c ) , y ( c ) ) = f ( a + k 1 ( h ) , b + k 2 ( h ) ) - f ( a , b ) = L 1 k 1 ( h ) + L 2 k 2 ( h ) + θ f ( k 1 ( h ) , k 2 ( h ) ) = l 1 ( x ( c + h ) - x ( c ) ) + L 2 ( y ( c + h ) - y ( c ) ) + θ f ( k 1 ( h ) , k 2 ( h ) ) = L 1 ( x ' ( c ) h + θ x ( h ) ) + L 2 ( y ' ( c ) h + θ y ( h ) ) + θ f ( k 1 ( h ) , k 2 ( h ) ) = ( L 1 x ' ( c ) + L 2 y ' ( c ) ) h + L 1 θ x ( h ) + L 2 θ y ( h ) + θ f ( k 1 ( h ) , k 2 ( h ) ) .
(19)

We define L=(L1x'(c)+L2y'(c))L=(L1x'(c)+L2y'(c)) and θ(h)=l1θx(h)+L2θy(h)+θf(k1(h),k2(h)).θ(h)=l1θx(h)+L2θy(h)+θf(k1(h),k2(h)). By these definitions and the calculation above we have Equation (4.1)

f ∘ φ ( c + h ) - f ∘ φ ( c ) = L h + θ ( h ) , f ∘ φ ( c + h ) - f ∘ φ ( c ) = L h + θ ( h ) ,
(20)

so that it only remains to verify (Reference) for the function θ.θ. We have seen above that the first two parts of θθ satisfy the desired limit condition, so that it is just the third part of θθ that requires some proof. The required argument is analogous to the last part of the proof of the Chain Rule ((Reference)), and we leave it as an exercise.

Exercise 4

  1. Finish the proof to the preceding theorem by showing that
    limh→0θf(k1(h),k2(h))h=0.limh→0θf(k1(h),k2(h))h=0.
    (21)
    HINT: Review the corresponding part of the proof to (Reference).
  2. Suppose f:S→Rf:S→R is as in the preceding theorem and that φφ is a real-valued function of a real variable. Suppose ff is differentiable, as a function of two real variables, at the point (a,b),(a,b), and that φφ is differentiable at the point c=f(a,b).c=f(a,b). Let g=φ∘f.g=φ∘f. Find a formula for the partial derivatives of the real-valued function gg of two real variables.
  3. (A generalized Mean Value Theorem) Suppose uu is a real-valued function of two real variables, both of whose partial derivatives exist at each point in a disk Br(a,b).Br(a,b). Show that, for any two points (x,y)(x,y) and (x',y')(x',y') in Br(a,b),Br(a,b), there exists a point (x^,y^)(x^,y^) on the line segment joining (x,y)(x,y) to (x',y')(x',y') such that
    u(x,y)-u(x',y')=tialutialx(x^,y^)(x-x')+tialutialy(x^,y^)(y-y').u(x,y)-u(x',y')=tialutialx(x^,y^)(x-x')+tialutialy(x^,y^)(y-y').
    (22)
    HINT: Let φ:[0,1]→R2φ:[0,1]→R2 be defined by φ(t)=(1-t)(x',y')+t(x,y).φ(t)=(1-t)(x',y')+t(x,y). Now use the preceding theorem.
  4. Verify that the assignment f→tialf/tialxf→tialf/tialx is linear; i.e., that
    tial(f+g)tialx=tialftialx+tialgtialx.tial(f+g)tialx=tialftialx+tialgtialx.
    (23)
    Check that the same is true for partial derivatives with respect to y.y.

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