Functions of Several VariablesA function of several variables is just what it sounds like. It may be

viewed in at least three different ways. We will use a function of twovariables as an example.

• z = f(x, y) may be viewed as a function of the two independentvariables x, y.

• It may be viewed as a function defined at different points (x, y)in the plane.

• It may be viewed as a function whose domain is the set ofvectors < x, y > or xi + yj.

Limits of Functions of Several VariablesWe define a limit of a function of several variables essentially the

same way we define a limit for an ordinary function:

Definition 1 (Limit). limx→c f(x) = L if ∀ε > 0, ∃δ > 0 such that|f(x)− L| < ε whenever 0 < |x− c| < δ.

Definition 2 (Limit). limx→c f(x) = L if ∀ε > 0, ∃δ > 0 such that|f(x)− L| < ε whenever 0 < |x− c| < δ.

Properties of LimitsRule of Thumb: If a property of limits makes sense when translated

to refer to a limit of a function of several variables, then it is valid fora function of several variables.

For example, the limit of a sum will be the sum of the limits, the limitof a difference will be the difference of the limits, the limit of a productwill be the product of the limits and the limit of a quotient will be thequotient of the limits, provided the latter limit exists.

ContinuityThe definition of continuity for a function of several variables is es-

sentially the same as the definition for an ordinary function.

Definition 3 (Continuity). A function f is continuous at c iflimx→c f(x) = f(c).

Definition 4 (Continuity for a Function of Several Variables). A func-tion f is continuous at c if limx→c f(x) = f(c).

As with ordinary functions, functions of several variables will generallybe continuous except where there’s an obvious reason for them not tobe.

1

2

Partial DerivativesFor a function of several variables, we have partial derivatives with

respect to each of its variables. The definition is based on the definitionof an ordinary derivative.

Definition 5 (Derivative). Let f : R → R.df

dx(x) = limh→0

f(x + h)− f(x)

h.

Definition 6 (Partial Derivative). Let f : R2 → R.∂f

∂x(x, y) =

limh→0f(x + h, y)− f(x, y)

h,

∂f

∂y(x, y) = limh→0

f(x, y + h)− f(x, y)

h.

The obvious generalizations hold for functions with more than twoindependent variables.

Calculation of Partial DerivativesEffectively, we calculate the partial derivative of a function with

respect to one of its independent variables by acting as if the otherindependent variables were actually constants.

NotationThe following notations for the partial derivatives of a function z =

f(x, y) are equivalent.

fx =∂f

∂x=

∂z

∂x= f1 = D1f = Dxf

fy =∂f

∂y=

∂z

∂y= f2 = D2f = Dyf

Higher Order DerivativesSince a partial derivative is itself a function of several variables, it

has its own partial derivatives.

(fx)y = fxy = f12 =∂

∂y

(∂f

∂x

)=

∂2f

∂y∂x=

∂2z

∂y∂x

(fy)x = fyx = f21 =∂

∂x

(∂f

∂y

)=

∂2f

∂x∂y=

∂2z

∂x∂y

Changing the Order of Differentiation

Theorem 1 (Clairaut’s Theorem). If fxy and fyx are both continuouson a disk containing (a, b), then fxy(a, b) = fyx(a, b).

3

Proof.Let φ(h) = f(x + h, y + h)− f(x, y + h)− f(x + h, y) + f(x, y). The

motivation comes from writing either fxy or fyx as a limit.

We may write φ(h) = α(y+h)−α(y), where α(t) = f(x+h, t)−f(x, t).The Mean Value Theorem implies α(y + h) − α(y) = α′(t)h for somet between y and y + h. Since α′(t) = f2(x + h, t) − f2(x, t), we haveφ(h) = [f2(x + h, t)− f2(x, t)]h.

If we write β(s) = f2(s, t), then f2(x+h, t)−f2(x, t) = β(x+h)−β(x).

Clairault’s Theorem

β(s) = f2(s, t), f2(x + h, t)− f2(x, t) = β(x + h)− β(x).

By the Mean Value Theorem, β(x + h) − β(x) = β′(s)h for some sbetween x and x + h. Since β′(s) = f21(s, t), we get f2(x + h, t) −f2(x, t) = f21(s, t)h, so φ(h) = f21(s, t)h

2.

Thusφ(h)

h2= f21(s, t) → f21(x, y) as h → 0, since f21 is continuous at

(x, y).

A similar calculation showsφ(h)

h2= f12(s, t) → f12(x, y) as h → 0,

showing f12(x, y) = f21(x, y).�

Tangent Planes

Consider a surface z = f(x, y) and suppose we are interested in theplane tangent to the surface at the point (a, b, c), where c = f(a, b).

Since∂z

∂xrepresents about how much z will change if x changes by

1 and y is fixed, here, and elsewhere as we look at tangent planes,tangent plane approximations and differentials, the partial derivativeshown really means the partial derivative’s value at the relevant point,

in this case (a, b), it seems reasonable to expect the vector < 1, 0,∂z

∂x>

to be tangent to the surface.

Similarly, it is reasonable to expect the vector < 0, 1,∂z

∂y> to be

tangent to the surface.

Tangent Planes

4

We thus expect n =

∣∣∣∣∣∣∣∣∣i j k

1 0∂z

∂x

0 1∂z

∂y

∣∣∣∣∣∣∣∣∣ = −∂z

∂xi − ∂z

∂yj + k to be a normal

vector to the tangent plane.

We thus take n =< −∂z

∂x,−∂z

∂y, 1 >.

We thus get < −∂z

∂x,−∂z

∂y, 1 > · < x−a, y−b, z−c >= 0 as an equation

for the tangent plane, or −∂z

∂x(x − a) − ∂z

∂y(y − b) + (z − c) = 0, or

z − c =∂z

∂x(x− a) +

∂z

∂y(y − b).

This should be reminiscent of the Point-Slope Formula for the equationof a line.

Tangent Hyperplanes

It generalizes to

y − b =∑n

i=1

∂y

∂xi

(xi − ai)

as an equation for the hyperplane tangent to the hypersurface y =f(x1, x2, . . . , xn) at the point (a1, a2, . . . , an, b).

Tangent Plane Approximations and Differentials

If we take z − c =∂z

∂x(x− a) +

∂z

∂y(y − b) and solve for z, we get

z = c +∂z

∂x(x− a) +

∂z

∂y(y − b)

This should be reminiscent of the Tangent Line Approximation for or-dinary functions.

We may use this formula to approximate f(x, y) at a point (x, y) closeto a point (a, b).

Definition 7 (Differentials). dx = ∆x = x− ady = ∆y = y − b

dz =∂z

∂x(x− a) +

∂z

∂y(y − b)

Differentials

5

We may use the differential dz to approximate the change ∆z = ∆fof a function f(x, y) if the independent variables x and y change byamounts dx and dy.

This generalizes in the obvious way to functions of more than twovariables.

DifferentiabilityRecall that for an ordinary function y = f(x) which was differen-

tiable at a point, we founddy −∆y

∆x→ 0 as ∆x → 0.

We take the analogue of this as a definition of differentiability for func-tions of several variables. We state the definition for the case of afunction of two variables; the variation for more variables should beobvious.

Definition 8 (Differentiable). We say a function f(x, y) is differen-

tiable at a point ifdz −∆z√

(∆x)2 + (∆y)2→ 0 as

√(∆x)2 + (∆y)2 → 0.

Differentiability

Recall√

(∆x)2 + (∆y)2 is the distance between (x, y) and the pointin question.

Effectively, we are defining a function of several variables to be differ-entialbe when an approximation using differentials is reasonable.

We still need a reasonable way of determining whether a function isdifferentiable. This is given by the following theorem.

Differentiability

Theorem 2. If both partial derivatives of a function z = f(x, y) arecontinuous in some open disc {(x, y) : (x− a)2 +(y− b)2 < r} centeredat (a, b), then f(x, y) is differentiable at (a, b).

Proof. We need to showdz −∆z√

(∆x)2 + (∆y)2→ 0 as

√(∆x)2 + (∆y)2 →

0.

We may write ∆z−dz = f(x, y)−f(a, b)−(

∂z

∂x(x− a) +

∂z

∂y(y − b)

)=

f(x, y)− f(a, y)− ∂z

∂x(x− a) + f(a, y)− f(a, b)− ∂z

∂y(y − b).

Proof

6

By the Mean Value Theorem, f(x, y) − f(a, y) =∂z

∂x(x∗, y)(x − a)

for some x∗ between a and x if x is close enough to a.

Similarly, f(a, y) − f(a, b) =∂z

∂y(a, y∗)(y − b) for some y∗ between b

and y if y is close enough to b.

We thus get ∆z − dz =∂z

∂x(x∗, y)(x− a)− ∂z

∂x(x− a) +

∂z

∂y(a, y∗)(y −

b)− ∂z

∂y(y−b) =

(∂z

∂x(x∗, y)− ∂z

∂x

)(x−a)+

(∂z

∂x(a, y∗)− ∂z

∂y

)(y−b).

Proof

Since both|x− a|√

(x− a)2 + (y − b)2≤ 1 and

|y − b|√(x− a)2 + (y − b)2

≤ 1,

we have

|(

∂z

∂x(x∗, y)− ∂z

∂x

)(x− a)|√

(x− a)2 + (y − b)2≤

∣∣∣∣∂z

∂x(x∗, y)− ∂z

∂x

∣∣∣∣ → 0

and

|(

∂z

∂y(a, y∗)− ∂z

∂y

)(y − b)|√

(x− a)2 + (y − b)2≤

∣∣∣∣∂z

∂y(a, y∗)− ∂z

∂y

∣∣∣∣ → 0

since both partial derivatives are continuous near (a, b).�

The Chain RuleFor an ordinary function, if y = f(u) and u = g(x), making y =

f ◦ g(x) a composite function, we can differentiate with respect to x

using the Chain Rule:dy

dx=

dy

du

du

dx.

Suppose we have a function z = f(x, y), but x = g(t) and y = h(t),making z = f(g(t), h(t)) a composite function of t. We can come upwith a variation of the Chain Rule, which holds under appropriateconditions. The conditions we will assume are that all the relevantderivatives exist and are continuous near t and all the relevant partialderivatives exist and are continuous near (f(t), g(t)).

By the definition of a derivative,

dz

dt= limk→0

f(g(t + k), h(t + k))− f(g(t), h(t))

k.

The Chain Rule

7

We can rewrite the numerator as f(g(t+k), h(t+k))−f(g(t), h(t)) =[f(g(t+k), h(t+k))−f(g(t), h(t+k))]+[f(g(t), h(t+k))−f(g(t), h(t))].

Using the Mean Value Theorem, the first difference may be written:

f(g(t+k), h(t+k))− f(g(t), h(t+k)) = f1(u, h(t+k))[g(t+k)− g(t)],where u is between g(t + k) and g(t).

But, also by the Mean Value Theorem, g(t+ k)− g(t) = g′(t∗)k, wheret∗ is between t and t + k.

We thus have f(g(t + k), h(t + k)) − f(g(t), h(t + k)) = f1(u, h(t +k))g′(t∗)k

Similarly, f(g(t), h(t + k))− f(g(t), h(t)) = f2(g(t), v)h′(t∗∗)k, where vis between h(t) and h(t + k) and t∗∗ is between t and t + k.

The Chain Rule

We thus getdz

dt= limk→0

f1(u, h(t + k))g′(t∗)k + f2(g(t), v)h′(t∗∗)k

k=

limk→0 f1(u, h(t + k))g′(t∗) + f2(g(t), v)h′(t∗∗) = f1(g(t), h(t))g′(t) +f2(g(t), h(t))h′(t).

Using Leibniz’ Notation, this may be written as:

dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt.

This is one variation of the Chain Rule.

Partial Derivatives Via the Chain RuleSuppose z = f(x, y), while x = g(s, t) and y = h(s, t). Then z =

f(g(s, t), h(s, t)) can be thought of as a function of s and t. We might

then want to calculate the partial derivatives∂z

∂sand

∂z

∂t.

By the nature of partial differentiation, the Chain Rule we just derivedcan be adjusted to give formulas for these partial derivatives.

∂z

∂s=

∂z

∂x

∂x

∂s+

∂z

∂y

∂y

∂s

∂z

∂t=

∂z

∂x

∂x

∂t+

∂z

∂y

∂y

∂t

If we have functions involving more than two variables, this may beadjusted in the hopefully obvious way.

Implicit DifferentiationThe Chain Rule may be used to derive a formula for implicit differ-

entiation.

8

Theorem 3 (Implicit Differentiation). If a differentiable function y =

f(x) is defined implicitly by an equation F (x, y) = 0, thendy

dx= −Fx

Fy

=

−

∂F

∂x∂F

∂y

.

Note: We have assumed y = f(x) is differentiable. We are not heredealing with how one knows whether such a function is differentiable.In general, if such a function is not differentiable, it will be relativelyobvious.

Implicit Differentiation

Proof. Using the Chain Rule,dF

dx=

∂F

∂x

dx

dx+

∂F

∂y

dy

dx=

∂F

∂x+

∂F

∂y

dy

dx.

Since F (x, y) = 0, it follows thatdF

dx= 0, so

∂F

∂x+

∂F

∂y

dy

dx= 0.

Solving fordy

dx, we get

∂F

∂y

dy

dx= −∂F

∂x, so

dy

dx= −

∂F

∂x∂F

∂y

. �

Directional DerivativesConsider a function z = f(x, y) and its graph, which will be a surface.

The partial derivative∂z

∂xmay be thought of as representing how fast

the surface is rising above one’s head if one is walking on the xy-planein the direction of the x-axis.

Similarly, the partial derivative∂z

∂ymay be thought of as representing

how fast the surface is rising above one’s head if one is walking on thexy-plane in the direction of the y-axis.

Directional DerivativeFor a given unit vector u, we define the directional derivative Duz

to represent how fast the surface is rising above one’s head if one iswalking on the xy-plane in the direction of u.

Definition 9 (Directional Derivative). Let f : Rn → R and let u ∈ Rn

be a unit vector. Let g(t) = f(x + ut). Duf(x) = g′(0) is called thedirectional derivative of f at x in the direction of u.

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Note that if n = 1, then the directional derivative is the same as theordinary derivative, while the directional derivatives in the directionsof the coordinate axes are the same as the partial derivatives.

The Del Operator and the Gradient

Definition 10 (Del Operator). 5 =

(∂

∂x,

∂

∂y

)Note this is really just a symbolic entity. By itself, it is meaningless,but we use it as a mneumonic device.

Definition 11 (Gradient). grad f = 5f =

(∂f

∂x,∂f

∂y

)The gradient turns out to be convenient when calculating directionalderivatives. It also generalizes to higher dimensions.

Calculating Directional Derivatives

Theorem 4. If all the partial derivatives of z = f(x) are continuousis some open ball centered at x, then Duf(x) = (5f) · u.

This theorem gives us a convenient way to calculate any directionalderivative of a function and also shows that it is sufficient to be ableto calculate all the partial derivatives.

ProofWe will prove the theorem for R2, but a similar proof will work for

higher dimensions; only the notation would get messier.

Proof. Consider a function f(x, y) and a unit vector u =< a, b >. Letz = g(t) be defined by letting z = f(x, y), where x = x0+at, y = y0+bt.

By definition, Duf(x0, y0) = g′(0).

By the Chain Rule, g′(t) =dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt= (5z) · u.

Evaluating this at 0 gives the result. �

Maximum Value of the Directional DerivativeDuf = (5f) · u = |5f | |u| cos θ, where θ is the angle between 5f

and u.

Since −1 ≤ cos θ ≤ 1, the maximal value obviously occurs when θ = 0and cos θ = 1, in other words, when u is in the same direction as 5f .

There’s a catch: This depends on the property u · v = |u||v| cos θ,which we’ve seen for R2 and R3, but whose very meaning is unclear forhigher dimensions.

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Cauchy-Schwarz InequalityWe can give u·v = |u||v| cos θ meaning through the Cauchy-Schwarz

Inequality u · v ≤ |u||v|. We will show the Cauchy-Schwarz Inequalityholds in any dimension, with equality holding if and only if one vectoris a multiple of the other.

Consider a vector u−tv. Certainly (u−tv)·(u−tv) ≥ 0, with equalityholding if and only if u is a multiple t of v or v = 0.

Since (u− tv) · (u− tv) = u ·u−2tu ·v+ t2vv = |v|2t2−2u ·vt+ |u|2,we get |v|2t2 − 2u · vt + |u|2 ≥ 0.

Cauchy-Schwarz InequalityIt follows that the quadratic equation |v|2t2 − 2u · vt + |u|2 = 0 in

t can’t have more than one solution, so the discriminant (−2u · v)2 −4|v|2|u|2 can’t be positive.

In other words, (−2u · v)2 − 4|v|2|u|2 ≤ 0, so 4(u · v)2 − 4|v|2|u|2 ≤ 0,so (u · v)2 − |v|2|u|2 ≤ 0, so (u · v)2 ≤ |v|2|u|2, so u · v ≤ |u||v|.Equality clearly holds if and only if either u − tv = 0 or if v = 0, inother words, if and only if either u is a scalar multiple of v or if v = 0.

Cauchy-Schwarz and Directional Derivatives

Since |u · v| ≤ |u||v|, it follows that −1 ≤ u · v|u||v|

≤ 1.

We may thus define the angle θ between u and v by θ = arccos

(u · v|u||v|

).

It follows that u · v = |u||v| cos θ, so the argument we used beforeabout the directional derivative being maximal in the direction of thegradient can legitimately be used.

Tangent Planes and GradientsRecall the formula for the plane tangent to the surface z = f(x, y)

at a point (a, b):

z − c =∂z

∂x(x− a) +

∂z

∂y(y − b).

Using the language of gradients, this could be written in the formz − c = (5f)· < x− a, y − b > or z − c = (5f) · (x− x0),where x =< x, y > and x0 =< a, b >.

Since one standard form for the equation of a plane is z− z0 = n · (x−x0), with n being a normal to the plane, it follows that 5f is normalto the tangent plane.

Tangent Planes for Surfaces Defined Implicitly

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Suppose a surface is the graph of an equation φ(x, y, z) = 0. Atmost points (where there is a tangent plane and the tangent plane isn’tvertical), a portion of the surface near the point can be consideredthe graph of a function z = f(x, y) defined implicitly by the equationφ(x, y, z) = 0 along with some side conditions.

By the formula for implicit differentiation,∂z

∂x= −

∂φ

∂x∂φ

∂z

and∂z

∂y=

−

∂φ

∂y∂φ

∂z

, so the equation of the tangent plane may be written

z − c = −

∂φ

∂x∂φ

∂z

(x− a)−

∂φ

∂y∂φ

∂z

(y − b).

Simplifying:∂φ

∂z(z − c) = −∂φ

∂x(x− a)− ∂φ

∂y(y − b),

∂φ

∂x(x− a) +

∂φ

∂y(y − b) +

∂φ

∂z(z − c) = 0.

This can also be written in the form (5φ)· < x− a, y − b, z − c >= 0.