Elements of the Differential and Integral Calculus/Chapter XV

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Elements of the Differential and Integral Calculus by William Anthony Granville
Chapter XV, §122-127

[edit] CHAPTER XV

PARTIAL DIFFERENTIATION

122. Continuous functions of two or more independent variables. A function $f (x, y)$ of two independent variables x and y is defined as continuous for the values $(a, b)$ of $(x, y)$ when

$\lim_{x \to a, y \to b} f(x, y) = f(a, b)$

no matter in what way x and y approach their respective limits a and b. This definition is sometimes roughly summed up in the statement that a very small change in one or both of the independent variables shall produce a very small change in the value of the junction.^[1]

We may illustrate this geometrically by considering the surface represented by the equation

z = f (x, y).

Consider a fixed point P on the surface where x = a and y = b.

Denote by $Δ x$ and $Δ y$ the increments of the independent variables x and y, and by $Δ z$ the corresponding increment of the dependent variable z, the coordinates of P' being

(x + Δ x, y + Δ y, z + Δ z).

At P the value of the function is

z = f (a, b) = M P .

If the function is continuous at P, then, however $Δ x$ and $Δ y$ may approach the limit zero, $Δ z$ will also approach the limit zero. That is, $M' P'$ will approach coincidence with MP, the point $P'$ approaching the point P on the surface from any direction whatever.

A similar definition holds for a continuous function of more than two independent variables.

In what follows, only values of the independent variables are considered for which a function is continuous.

123. Partial derivatives. Since x and y are independent in

z = f (x, y),

x may be supposed to vary while y remains constant, or the reverse.

The derivative of z with respect to x when x varies and y remains constant^[2] is called the partial derivative of z with respect to x, and is denoted by the symbol $\frac{\partial z}{\partial z}$ We may then write

(A)

$\frac{\partial z}{\partial x} = \lim_{\Delta x \to 0} \left[ \frac{f(x + \Delta x, y) - f(x, y)}{\Delta x} \right]$

Similarly, when x remains constant has been generally adopted to indicate partial differentiation. Other notations; however, which are in use are

$\left( \frac{dz}{dx} \right), \left( \frac{dz}{dy} \right); f'_x(x, y), f'_y(x, y); f_x(x, y), f_y(x, y); D_x f, D_y f; z_x, z_y.$

Our notation may be extended to a function of any number of independent variables. Thus, if

u = F (x, y, z),

then we have the three partial derivatives

$\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z}$ ; or, $\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}.$

ILLUSTRATIVE EXAMPLE 1. Find the partial derivatives of $z = a x 2 + 2 b x y + c y 2 .$

Solution.	$\frac{\partial z}{\partial x} = 2 ax + 2 by$ , treating y as a constant,
	$\frac{\partial z}{\partial y} = 2 bx + cy$ , treating x as a constant.

ILLUSTRATIVE EXAMPLE 2. Find the partial derivatives of $u = sin(a x + b y + c z).$

Solution.	$\frac{\partial u}{\partial x} = a \cos (ax + by + cz)$ , treating y and z as constants,
	$\frac{\partial u}{\partial y} = b \cos (ax + by + cz)$ , treating x and z as constants,
	$\frac{\partial u}{\partial z} = c \cos (ax + by + cz)$ , treating y and x as constants.

Again turning to the function

z = f (x, y),

we have, by (A), p. 191 [§123], defined $\frac{\partial z}{\partial x}$ as the limit of the ratio of the increment of the function (y being constant) to the increment of x, as the increment of x approaches the limit zero. Similarly, (B), p. 191, has defined $\frac{\partial z}{\partial y}$ . It is evident, however, that if we look upon these partial derivatives from the point of view of § 94, p. 141, then

$\frac{\partial z}{\partial x}$

may be considered as the ratio of the time rates of change of z and x when y is constant, and

$\frac{\partial z}{\partial y}$

as the ratio of the time rates of change of z and y when x is constant.

124. Partial derivatives interpreted geometrically. Let the equation of the surface shown in the figure be

z = f (x, y).

Pass a plane EFGH through the point P (where x = a and y = b) on the surface parallel to the XOZ-plane. Since the equation of this plane is

y = b,

the equation of the section JPK cut out of the surface is

z = f (x, b),

if we consider EF as the axis of Z and EH as the axis of x: In this plane $\frac{\partial z}{\partial x}$ means the same as $\frac{dz}{dx}$ and we have

$\frac{\partial z}{\partial x}$	$= tan M T P$
	= slope of section JK at P.

Similarly, if we pass the plane BCD through P parallel to the YOZ-plane, its equation is x = a, and for the section $DPI, \frac{\partial z}{\partial y}$ means the same as $\frac{dz}{dy}$ . Hence

$\frac{\partial z}{\partial y} = \frac{dz}{dy} = -\tan MT'P =$ slope of section DI at P.

ILLUSTRATIVE EXAMPLE 1. Given the ellipsoid $\frac{x^2}{24} + \frac{y^2}{12} + \frac{z^2}{6} = 1$ ; find the slope of the section of the ellipsoid made (a) by the plane y = 1 at the point where x = 4 and z is positive; (b) by the plane x = 2 at the point where y = 3 and z is positive.

Solution.	Considering y as constant,
	$\frac{2x}{24} + \frac{2z}{6} \frac{\partial z}{\partial x}$	= 0, or $\frac{\partial z}{\partial x} = -\frac{x}{4z}.$
When x is a constant	$\frac{2y}{12} + \frac{2z}{6} \frac{\partial z}{\partial y}$	= 0, or $\frac{\partial z}{\partial y} = -\frac{y}{2z}.$

(a) When y = 1 and x = 4, z = $\sqrt{\frac{3}{2}}$ . ∴ $\frac{\partial z}{\partial x} = -\frac{x}{4z}.$

(b) When x = 2 and y = 3, z = $\frac{1}{\sqrt{2}}$ . ∴ $\frac{\partial z}{\partial y} = -\frac{3}{2} \sqrt{2}.$ Ans.

EXAMPLES

1. $u = x 3 + 3 x 2 y - y 3 .$	Ans.	$\frac{\partial u}{\partial x} = 3 x^2 + 6xy;$
		$\frac{\partial u}{\partial y} = 3 x^2 + 3 y^2.$
2. $u = A x 2 + B x y + C y 2 + D x + E y + F .$		$\frac{\partial u}{\partial x} = 2 Ax + By + d;$
		$\frac{\partial u}{\partial y} = Bx + 2Cy + E.$
3. $u = (a x 2 + b y 2 + c z 2) n .$		$\frac{\partial u}{\partial x} = \frac{2anxu}{ax^2 + by^2 + cz^2};$
		$\frac{\partial u}{\partial y} = \frac{2bnxu}{ax^2 + by^2 + cz^2}$
4. $u = \arcsin \frac{x}{y}.$		$\frac{\partial u}{\partial x} = \frac{1}{\sqrt{y^2 - x^2}};$
		$\frac{\partial u}{\partial y} = -\frac{x}{y\sqrt{y^2 - x^2}}.$
5. $u = x y .$		$\frac{\partial u}{\partial x} = yx^{y - 1};$
		$\frac{\partial u}{\partial y} = x^y \log x.$
6. $u = a x 3 y 2 z + b x y 3 z 4 + c y 6 + d x z 3 .$		$\frac{\partial u}{\partial x} = 3ax^2y^2z + by^3z^4 + dz^3.$
		$\frac{\partial u}{\partial y} = 2ax^3yz + 3bxy^2z^4 + 6cy^5.$
		$\frac{\partial u}{\partial z} = ax^3y^2 + 4bxy^3z^3 + 3dxz^2.$

7. $u = x 3 y 2 - 2 x y 4 + 3 x 2 y 3$ ; show that $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 5u.$

8. $u = \frac{xy}{x + y};$ show that $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = u.$

9. $u = (y - z)(z - x)(x - y)$ ; show that $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0.$

10. $u = l o g (e x + e y);$ show that $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 1.$

11. $u = \frac{e^{xy}}{e^x + e^y};$ show that $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = (x + y - 1)u.$

12. $u = = x y y x;$ show that $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = (x + y + \log u)u.$

13. $u = log(x 3 + y 3 + z 3 - 3 x y z);$ show that $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3}{x + y + z}.$

14. $u = e x sin y + e y sin x;$ show that

$\left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial u}{\partial y} \right)^2 = e^{2x} + e^{2y} + 2 e^{x + y} \sin(x + y).$

15. $u = log(tan x + tan y + tan z);$ show that

$\sin 2x \frac{\partial u}{\partial x} + \sin 2y \frac{\partial u}{\partial y} + \sin \frac{\partial u}{\partial z} = 2.$

16. Let y be the altitude of a right circular cone and x the radius of its base. Show (a) that if the base remains constant, the volume changes $\frac{1}{3} \pi x^2$ times as fast as the altitude; (b) that if the altitude remains constant, the volume changes $\frac{2}{3} \pi xy$ times as fast as the radius of the base.

17. A point moves on the elliptic paraboloid $z = \frac{x^2}{9} + \frac{y^2}{4}$ and also in a plane parallel to the XOZ-plane. When x = 3 ft. and is increasing at the rate of 9 ft. per second, find (a) the time rate of change of z; (b) the magnitude of the velocity of the point; (c) the direction of its motion.

Ans. (a) v_z = 6 ft. per sec.; (b) v = $3\sqrt{13}$ ft. per sec.; (c) $\tau = \arctan \frac{2}{3}$ , the angle made with the XOY-plane.

18. If, on the surface of Ex. 17, the point moves in a plane parallel to the plane YOZ, find, when y = 2 and increases at the rate of 5 ft. per sec., (a) the time rate of change of z; (b) the magnitude of the velocity of the point; (c) the direction of its motion.

Ans. (a) 5 ft. per sec.; (b) $5\sqrt{2}$ ft. per sec.; (c) $\tau = \frac{\pi}{4},$ the angle made with the plane XOY.

125. Total derivatives. We have already considered the differentiation of a function of one function of a single independent variable. Thus, if

y = f (v)

and

v = ψ(x),

it was shown that

$\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}.$

We shall next consider a function of two variables, both of which depend on a single independent variable. Consider the function

u = f (x, y),

where x and y are functions of a third variable t.

Let t take on the increment Δt, and let Δx, Δy, Δu be the corresponding increments of x, y, u respectively. Then the quantity

Δ u = f (x + Δ x, y + Δ y) - f (x, y)

is called the total increment of u.

Adding and subtracting $f (x, y + Δ y)$ in the second member,

(A) $Δ u = [f (x + Δ x, y + Δ y) - f (x, y + Δ y)] + [f (x, y + Δ y) - f (x, y)].$

Applying the Theorem of Mean Value (46), p. 166 [§106], to each of the two differences on the right-hand side of (A), we get, for the first difference,

(B) $f(x + \Delta x, y + \Delta y) - f(x, y + \Delta y) = f_x'(x + \theta_1 \cdot \Delta x, y + \Delta y) \Delta x.$

[ $a = x,Δ a = Δ x$ , and since x varies while y + Δy remains constant, we get the partial derivative with respect to x.]

For the second difference we get

(C) $f(x, y + \Delta y) - f(x, y) = f_y'(x, y + \theta_2 \cdot \Delta y) \Delta y.$

[ $a = y,Δ a = Δ y,$ and since y varies while x remains constant, we get the partial derivative with respect to y.]

Substituting (B) and (C) in (A) gives

(D) $\Delta u = f_x' ( x + \theta_1 \cdot \Delta x, y + \Delta y) \Delta x + f_y' (x, y + \theta_2 \cdot \Delta y) \Delta y,$

where θ₁ and θ₂ are positive proper fractions. Dividing (D) by Δt,

(E) $\frac{\Delta u}{\Delta t} = f_x'(x + \theta_1 \cdot \Delta x, y + \Delta y) \frac{\Delta x}{\Delta t} + f_y'(x, y + \theta_2 \cdot \Delta y) \frac{\Delta y}{\Delta t}.$

Now let Δt approach zero as a limit, then

(F) $\frac{du}{dt} = f_x'(x, y) \frac{dx}{dt} + f_y'(x, y)\frac{dy}{dt}.$

[Since Δx and Δy converge to zero with Δt, we get $\lim_{\Delta t \to 0} f_x'(x + \theta_1 \cdot x, y + \Delta y) = f_x'(x, y),$ and $\lim_{\Delta t \to 0} f_y'(x, y + \theta_2 \cdot \Delta y) = f_y'(x, y), f_x'(x, y) \text{and} f_y'(x, y)$ being assumed continuous.]

Replacing $f (x, y)$ by u in (F), we get the total derivative

(51) $\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y}\frac{dy}{dt}.$

In the same way, if

u = f (x, y, z),

and x, y, z are all functions of t, we get

(52) $\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} + \frac{\partial u}{\partial z} \frac{dz}{dt},$

and so on for any number of variables.^[3]

In (51) we may suppose t = x; then y is a function of x, and u is really a function of the one variable x, giving

(53) $\frac{du}{dx} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \cdot \frac{dy}{dx}.$

In the same way, from (52) we have

(54) $\frac{du}{dx} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \cdot \frac{dy}{dx} + \frac{\partial u}{\partial z} \cdot \frac{dz}{dx}.$

The student should observe that $\frac{\partial u}{\partial x}$ and $\frac{du}{dx}$ have quite different meamngs. The partial derivative $\frac{\partial u}{\partial x}$ is formed on the supposition that the particular variable x alone varies, while

$\frac{du}{dx} = \lim_{\Delta x \to 0} \left( \frac{\Delta u}{\Delta x} \right),$

where $Δ u$ is the total increment of u caused by changes in all the variables, these increments being due to the change $Δ x$ in the independent variable. In contradistinction to partial derivatives, $\frac{du}{dt}, \frac{du}{dx}$ are called total derivatives with respect to t and x respectively.^[4]

ILLUSTRATIVE EXAMPLE 1. Given $u = \sin \frac{x}{y}, x = e^t, y = t^2; \mbox{ find } \frac{du}{dt}.$

Solution. $\frac{\partial u}{\partial x} = \frac{1}{y} \cos \frac{x}{y}, \frac{\partial u}{\partial y} = -\frac{x}{y^2} \cos {\partial x}{\partial y}, \frac{dx}{dt} = e^t, \frac{dy}{dt} = 2t.$

Substituting in (51), $\frac{du}{dt} = (t - 2)\frac{e^t}{t^3} \cos \frac{e^t}{t^2}.$ Ans.

ILLUSTRATIVE EXAMPLE 2. Given $u = e^{ax} (y - z), y = a \sin x, z = \cos x; \mbox{ find } \frac{du}{dx}.$

Solution. \frac{\partial u}{\partial x} = ae^{ax} (y - z), \frac{\partial u}{\partial y} = e^{ax}, \pfrac{\partial u}{\partial z} = -e^{ax}; \frac{dy}{dx} = a \cos x, \frac{dz}{dx} = -\sin x.</math>

Substituting in (54),

$\frac{du}{dx} = ae^{ax}(y - z) + ae^{ax} \cos x + e^{ax} \sin x = e^{ax}(a^2 + l)\sin x.$ Ans.

NOTE. In examples like the above, u could, by substitution, be found explicitly in terms of the independent variable and then differentiated directly, but generally this process would be longer and in many cases could not be used at all.

Formulas (51) and (52) are very useful in all applications involving time rates of change of functions of two or more variables. The process is practically the same as that outlined in the rule given on p. 141 [§94], except that, instead of differentiating with respect to t (Third Step), we find the partial derivatives and substitute in (51) or (52). Let us illustrate by an example.

ILLUSTRATIVE EXAMPLE 3. The altitude of a circular cone is 100 inches, and decreasing at the rate of 10 inches per second; and the radius of the base is 50 inches, and increasing at the rate of 5 inches per second. At what rate is the volume changing?

Solution. Let x = radius of base, y = altitude; then $u = \frac{1}{3}\pi x^2 y$ = volume, $\frac{\partial u}{\partial x} = \frac{2}{3} \pi xy, \frac{\partial u}{\partial y} = \frac{1}{3}\pi x^2.$ Substitute in (51),

$\frac{du}{dt} = \frac{2}{3}\pi xy \frac{dx}{dt} + \frac{1}{3}\pi x^2 \frac{dy}{dt}.$

But x = 50, y = 100, $\frac{dx}{dt}$ = 5, $\frac{dy}{dt}$ = - 10.

∴ $\frac{du}{dt} = \frac{2}{3}\pi \cdot 5000 \cdot 5 - \frac{1}{3}\pi \cdot 2500 \cdot 10 = 15.15$ cu. ft. per sec., increase. Ans.

126. Total differentials. Multiplying (51) and (52) through by dt, we get

(55) $du = \frac{\partial du}{\partial dx} dx + \frac{\partial du}{\partial dy} dy,$

(56) $du = \frac{\partial du}{\partial dx} dx + \frac{\partial du}{\partial dy} dy + \frac{\partial u}{\partial z} dz,$

and so on.^[5] Equations (55) and (56) define the quantity du, which is called a total differential of u or a complete differential, and

$\frac{\partial u}{\partial x} dx, \frac{\partial u}{\partial y} dy, \frac{\partial u}{\partial z} dz$

are called partial differentials. These partial differentials are sometimes denoted by d_xu, d_yu, d_zu, so that (56) is also written

d u = d x u + d y u + d z u .

ILLUSTRATIVE EXAMPLE 1. Given $u = \arctan \frac{y}{x}$ , find dx.

Solution.	$\frac{\partial u}{\partial x} = -\frac{y}{x^2 + y^2}, \frac{\partial u}{\partial y} = \frac{x}{x^2 + y^2}.$
Substituting in (55),	$du = \frac{x dy - y dx}{x^2 - y^2}.$ Ans.

ILLUSTRATIVE EXAMPLE 2. The base and altitude of a rectangle are 5 and 4 inches respectively. At a certain instant they are increasing continuously at the rate of 2 inches and 1 inch per second respectively. At what rate is the area of the rectangle increasing at that instant?

Solution. Let x = base, y = altitude; then u = xy = area, $\frac{\partial u}{\partial x} = y, \frac{\partial u}{\partial y} = x.$

Substituting in (51),

(A) $\frac{du}{dt} = y \frac{dx}{dt} + x\frac{dy}{dt}.$

But x = 5 in., yy = 4 in., $\frac{dx}{dt}$ = 2 in. per sec., $\frac{dy}{dt}$ = 1 in. per sec.

∴ $\frac{du}{dt} = (8 + 5)$ sq. in. per sec. = 13 sq. in. per sec. Ans.

NOTE. Considering du as an infinitesimal increment of area due to the infinitesimal increments dx and dy, du is evidently the sum of two thin strips added on to the two sides. For, in du = ydx + xdy (multiplying (A) by dt),

[Image:Wag 126-1 expanding rectangle.png|right|171px]]

ydx = area of vertical strip, and

xdy = area of horizontal strip.

But the total increment Δu due to the increments dx and dy is evidently

Δ u = y d x + x d y + d x d y .

Hence the small rectangle in the upper right-hand corner (= dxdy) is evidently the difference between Δu and du. This figure illustrates the fact that the total increment and the total differential of a function of several variables are not in general equal.

127. Differentiation of implicit functions. The equation

(A) $f (x, y) = 0$

defines either x or y as an implicit function of the other.^[6] It represents any equation containing x and y when all its terms have been transposed to the first member. Let

(B)	$u = f (x, y);$
then	$\frac{du}{dx} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \frac{dy}{dx}.$	(53), p. 196 [§125]

But from (A), f(x, y) = 0. ∴ u = 0 and $\frac{du}{dx} = 0;$ that is,

(C) $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \frac{dy}{dx} = 0.$

Solving for $\frac{dy}{dx},$ ^[7] we get

(57) $\frac{dy}{dx} = -\frac{ \frac{\partial u}{\partial x} }{ \frac{\partial u}{\partial y} },$

a formula for differentiating implicit functions. This formula in the form (C) is equivalent to the process employed in § 62, p. 69, for differentiating implicit functions, and all the examples on p. 70 may be solved by using formula (57). Since

(D) f(x, y) = 0

for all admissible values of x and y, we may say that (57) gives the relative time rates of change of x and y which keep f (x, y) from changing at all. Geometrically this means that the point (x, y) must move on the curve whose equation is (D), and (57) determines the direction of its motion at any instant. Since

u = f(x, y),

we may write (57) in the form of

(57a) $\frac{dy}{dx} = -\frac{ \frac{\partial f}{\partial x} }{ \frac{\partial f}{\partial y} }, \frac{\partial f}{\partial y} \ne 0$

ILLUSTRATIVE EXAMPLE 1. Given $x 2 y 4 + sin y = 0$ , find $\frac{dy}{dx}$ .

Solution. Let $f (x, y) = x 2 y 4 + sin y .$

$\frac{\partial f}{\partial x} = 2xy^4, \frac{\partial f}{\partial y} = 4x^2 y^3 + \cos y.$ ∴ from (57a), $\frac{dy}{dx} = -\frac{2xy^4}{4x^2 y^3 + \cos y}.$ Ans.

ILLUSTRATIVE EXAMPLE 2. If x increases at the rate of 2 inches per second as it passes through the value x = 3 inches, at what rate must y change when y = 1 inch, in order that the function $2 x y 2 - 3 x 2 y$ shall remain constant?

Solution. Let $f (x, y) = 2 x y 2 - 3 x 2 y$ ; then

$\frac{\partial f}{\partial x} = 2y^2 - 6xy, \frac{\partial f}{\partial y} = 4xy - 3x^2.$

Substituting in (57a),

$\frac{dy}{dx} = -\frac{2Y^2 - 6xy}{4xy - 3x^2},$ or $\frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = -\frac{2y^2 - 6xy}{4xy - 3x^2}.$ By (33), p. 141 [§94]

But x = 3, y = 1, $\frac{dx}{dt} = 2.$ ∴ $\frac{dY}{dt} = -2\frac{2}{15},$ ft. per second. Ans.

Let P be the point (x, y, z) on the surface given by the equation

(E) $u = F (x, y, z) = 0,$

and let PC and AP be sections made by planes through P parallel to the YOZ- and XOZ-planes respectively. Along the curve AP, y is constant; therefore, from (E), z is an implicit function of x alone, and we have, from (57a),

(58) $\frac{\partial z}{\partial x} = -\frac{ \frac{\partial F}{\partial y} }{ \frac{\partial F}{\partial z} }.$

giving the slope at P of the curve AP, § 122, p. 190.

$\frac{\partial z}{\partial x}$ is used instead of $\frac{dz}{dx}$ in the first member, since z was originally, from (E), an implicit function of x and y; but (58) is deduced on the hypothesis that y remains constant.

Similarly, the slope at P of the curve PC is

(59) $\frac{\partial z}{\partial y} = -\frac{ \frac{\partial F}{\partial y} }{ \frac{\partial F}{\partial z} }$

↑ This will be better understood if the student again reads over §18, p. 14, on continuous functions of a single variable.
↑ The constant values are substituted in the function before differentiating
↑ This is really only a special case of a general theorem which may be stated as follows: If u is a function of the independent variables x, y, z,..., each of these in turn being a function of the independent variables r, s, t, ..., then (with certain assumptions as to continuity)

$\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r} + \frac{\partial u}{\partial z} \frac{\partial z}{\partial r} + \cdots,$

and similar expressions hold for $\frac{\partial u}{\partial s}, \frac{\partial u}{\partial t}$ etc.
↑ It should be observed that $\frac{\partial u}{\partial x}$ has a perfectly definite value for any point (x, y), while $\frac{du}{dx}$ depends not only on the point (x, y), but also on the particular direction chosen to reach that point. Hence

$\frac{\partial u}{\partial x}$ is called a point function; while

$\frac{du}{dx}$ is not called a point function unless it is agreed to approach the point from some particular direction.
↑ A geometric interpretation of this result will be given on p. 266 [§161].
↑ We assume that a small change in the value of x causes only a small change in the valne of y.
↑ It is assumed that $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ exist.

[0] This will be better understood if the student again reads over §18, p. 14, on continuous functions of a single variable.

[constant-1] The constant values are substituted in the function before differentiating

[2] This is really only a special case of a general theorem which may be stated as follows: If u is a function of the independent variables x, y, z,..., each of these in turn being a function of the independent variables r, s, t, ..., then (with certain assumptions as to continuity)

$\frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r} + \frac{\partial u}{\partial z} \frac{\partial z}{\partial r} + \cdots,$

and similar expressions hold for $\frac{\partial u}{\partial s}, \frac{\partial u}{\partial t}$ etc.

[3] It should be observed that $\frac{\partial u}{\partial x}$ has a perfectly definite value for any point (x, y), while $\frac{du}{dx}$ depends not only on the point (x, y), but also on the particular direction chosen to reach that point. Hence

$\frac{\partial u}{\partial x}$ is called a point function; while

$\frac{du}{dx}$ is not called a point function unless it is agreed to approach the point from some particular direction.

[4] A geometric interpretation of this result will be given on p. 266 [§161].

[5] We assume that a small change in the value of x causes only a small change in the valne of y.

[6] It is assumed that $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ exist.

[1]

[2]

[3]

[4]

[5]

[6]

[7]

Elements of the Differential and Integral Calculus/Chapter XV

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