CHAPTER 42
Partial Derivatives
42.1 If f(x, y) = 4x3 - 3x2y2 + 2x + 3y, find the partial derivatives £ and yf.
First, consider y constant. Then, differentiating with respect to x, we obtain ff = I2x2 — 6y2x + 2. If we
keep x fixed and differentiate with respect to y, we get fy = —6x2y + 3.
42.2 If f(x, y) = x5 In y, find £ and fy.
Differentiating with respect to x while keeping y fixed, we find that fx = 5x* In y. Differentiating with
respect to y while keeping x fixed, we get fy — xs/y.
42.3 For f(x, y) = 3x2-2x + 5, find fxandfy.
fx=6x-2 and fy = 0.
42.4 If f(x,y) = tan-l(x + 2y), find /.and/,.
42.5 For /(AC , y) = cos xy, find £ and /j,.
A = (-sin *y);y = -y sin *y /y = (-sin xy)x = -x sin xy
42.6 If /(r, 0) = r cos 0, find <?//<?r and df/dO.
and
42.7 If /(*,y) = find fx and /j,.
42.8 Find the first partial derivatives of f ( x , y, z) = xy2z3.
f, = y2*3 fy=2xyz3 /,=3xyV
42.9 Find the first partial derivatives of f ( u , v, t) — euv sin ut.
/„ = eu"(cos ut)t + ve"" sin ut = euv(t cos ut + v sin wf) /„ = ueu" sin ut f, = ue"v cos ut
42.10 Find the first partial derivatives of f ( x , y, z, u, u) = 2x + yz — ux + vy2.
A =2-« /, = z + 2yy A =y /. = -* /. = y2
42.11 Find the first partial derivatives of f ( x , y, u, v) = In (x/y) - ve"y.
Note that f ( x , y, u, v) = In x — In y — veuy. Then,
/„ = -yveuy /„ = -*""
42.12 Give an example of a function f(x, y) such that £(0,0) =/j,(0,0) = 0, but / is not continuous at (0,0).
Hence, the existence of the first partial derivatives does not ensure continuity.
376
, PARTIAL DERIVATIVES 377
Let
By Problem 41.62, f ( x , y) is discontinuous at the origin. Nevertheless,
42.13 If z is implicitly defined as a function of x and y by x2 + y2 — z2 = 3, find dzldx and dzldy.
By implicit differentiation with respect to x, 2x - 2z(dzldx) = 0, x = z(dzldx), dzldx = xlz. By im-
plicit differentiation with respect to y, 2y — 2z(dzldy) = 0, y = z(dzldy), dzldy = ylz.
42.14 If z is implicitly defined as a function of x and y by x sin z — z2y — I, find dzldx and dzldy.
By implicit differentiation with respect to x, x cos z (dzldx) + sin z - 2yz(dzldx~) - 0, (dzldx)(x cos z -
2yz) = -sin z, <?z/<?x = sin zl(2yz - x cos z). By implicit differentiation with respect to y, x cos z (dzldy) -
z2-2yz(dzIdy) = 0, (o>z/<?>')(;t cos z - 2}>z) = z2, <?z/<?y = z2/(xcos z -2>>z).
42.15 If z is defined as a function of x and y by xy - yz + xz = 0, find dzldx and dzldy.
By implicit differentiation with respect to x,
By implicit differentiation with respect to y,
42.16 If z is implicitly defined as a function of * and y by x2 + y 2 + z 2 = 1, show that
By implicit differentiation with respect to *, 2x + 2z(dzldx) = 0, dzldx=—xlz. By implicit differentia-
tion with respect to y, 2y + 2z(dzldy) = 0, dzldy = -ylz. Thus,
42.17 If z = In show that
Because we have and
Therefore,
42.18 If x = e2rcos6 and y = elr sin 6, find r,, r,,, 0 X , and Oy by implicit partial differentiation.
Differentiate both equations implicitly with respect to x. 1 = 2e2' (cosQ)rlt - e2r (sin 0)0,, 0 =
3
3e ' (sin 0)r, + e3r (cos 0)0,. From the latter, since e3r ¥= 0, 0 = 3 (sin 0)r, + (cos 0)0,. Now solve simulta-
neously for r, and 0,. r, =cos0/[e 2r (2 + sin2 e)], 0, = -3 sin 0/[e2r(2 + sin2 0)]. Now differentiate the
original equations for x and y implicitly with respect toy: 0 = 2e2r (cos0)ry - e2' (sin 0)0y, 1 =3e3r (sin 0)^ +
e 3r (cos0)0 v . From the first of these, since e 2l VO, we get 0 = 2(cos0)r,, -($^0)0^. Solving simulta-
neously for ry and 0y, we obtain ry = sin 0/[e3'(2 + sin2 0)] and Oy = 2cos 0/[e3r(2 + sin2 9)].
, 378 CHAPTER 42
42.19 If z = e"ysin(x/y) + ey"tcos(y/x),show that
It is easy to prove a more general result. Let z = f ( x / y ) , where/is an arbitrary differentiable function.
Then
and
by addition,
42.20 If z = xey'*, show that
In general, if z = xf(y/x), where / is differentiable,
42.21 If evaluate
and Problem 42.19 applies.
42.22 If find and
since, in general, Similarly,
Given a relationship F(x, y, z) = 0, where F has nonzero partial derivatives with respect to its arguments,
prove the cyclical formula (dxldy)(dyldz)(dz/dx) = -1.
Holding z constant, differentiate the functional equation on y: Fxxy + Fy=-0, or xy = -FyIFx. Similarly
(or by cyclical permutation of the variables), y: = ~Fl/Fy and zx=-F>t/F2. Then, by
multiplication, xyyzzx = —FyF.FJFxFyFI = — 1, which is the desired result.
42.24 If f(x, y) = Ix2y - 2x + 5y2, find £(1,2) and/v(l,2).
fr=6xy-6x2. Hence, £(1,2) = 12-6 = 6. /, = 3x2 + IQy. Hence, £(1,2) = 3 + 20 = 23.
42.25 If f ( x , y) = cos 3x sin 4y, find £(ir/12, 77/6) and f(ir/U, ir/6).
fr(x, y) = ~3 sin 3x sin 4y. Therefore,
/,.(•*» y) = 4 cos 3x cos 4y = 4
42.26 Find the slope of the tangent line to the curve that is the intersection of the surface z = x2 — y2 with the plane
x = 2, at the point (2,1,3).
In the plane x = 2, x is constant. Hence, the slope of the tangent line to the curve is the derivative
dzldy = -2y = -2(1) = -2.
42.27 Find the slope of the tangent line to the curve that is the intersection of the sphere x2 + y2 + z2 = 1 with the
plane y = f , at the point ( j , j, V2/2).
Since y is constant in the plane y = 5, the slope of the tangent line to the curve is dzldx. By implicit
differentiation of the equation x2 + y2 + z2 = l, we get 2x + 2z(dzldx) = 0. Hence, at the point
(\, \, V2/2), dzldx= -jc/z = -4/(V2/2)= -1/V2= -V2/2.
42.28 Find the slopes of the tangent lines to the curves cut from the surface z = 3x2 + 4y2 -6 by planes through the
point (1,1,1) and parallel to the xz- and yz-planes.
Partial Derivatives
42.1 If f(x, y) = 4x3 - 3x2y2 + 2x + 3y, find the partial derivatives £ and yf.
First, consider y constant. Then, differentiating with respect to x, we obtain ff = I2x2 — 6y2x + 2. If we
keep x fixed and differentiate with respect to y, we get fy = —6x2y + 3.
42.2 If f(x, y) = x5 In y, find £ and fy.
Differentiating with respect to x while keeping y fixed, we find that fx = 5x* In y. Differentiating with
respect to y while keeping x fixed, we get fy — xs/y.
42.3 For f(x, y) = 3x2-2x + 5, find fxandfy.
fx=6x-2 and fy = 0.
42.4 If f(x,y) = tan-l(x + 2y), find /.and/,.
42.5 For /(AC , y) = cos xy, find £ and /j,.
A = (-sin *y);y = -y sin *y /y = (-sin xy)x = -x sin xy
42.6 If /(r, 0) = r cos 0, find <?//<?r and df/dO.
and
42.7 If /(*,y) = find fx and /j,.
42.8 Find the first partial derivatives of f ( x , y, z) = xy2z3.
f, = y2*3 fy=2xyz3 /,=3xyV
42.9 Find the first partial derivatives of f ( u , v, t) — euv sin ut.
/„ = eu"(cos ut)t + ve"" sin ut = euv(t cos ut + v sin wf) /„ = ueu" sin ut f, = ue"v cos ut
42.10 Find the first partial derivatives of f ( x , y, z, u, u) = 2x + yz — ux + vy2.
A =2-« /, = z + 2yy A =y /. = -* /. = y2
42.11 Find the first partial derivatives of f ( x , y, u, v) = In (x/y) - ve"y.
Note that f ( x , y, u, v) = In x — In y — veuy. Then,
/„ = -yveuy /„ = -*""
42.12 Give an example of a function f(x, y) such that £(0,0) =/j,(0,0) = 0, but / is not continuous at (0,0).
Hence, the existence of the first partial derivatives does not ensure continuity.
376
, PARTIAL DERIVATIVES 377
Let
By Problem 41.62, f ( x , y) is discontinuous at the origin. Nevertheless,
42.13 If z is implicitly defined as a function of x and y by x2 + y2 — z2 = 3, find dzldx and dzldy.
By implicit differentiation with respect to x, 2x - 2z(dzldx) = 0, x = z(dzldx), dzldx = xlz. By im-
plicit differentiation with respect to y, 2y — 2z(dzldy) = 0, y = z(dzldy), dzldy = ylz.
42.14 If z is implicitly defined as a function of x and y by x sin z — z2y — I, find dzldx and dzldy.
By implicit differentiation with respect to x, x cos z (dzldx) + sin z - 2yz(dzldx~) - 0, (dzldx)(x cos z -
2yz) = -sin z, <?z/<?x = sin zl(2yz - x cos z). By implicit differentiation with respect to y, x cos z (dzldy) -
z2-2yz(dzIdy) = 0, (o>z/<?>')(;t cos z - 2}>z) = z2, <?z/<?y = z2/(xcos z -2>>z).
42.15 If z is defined as a function of x and y by xy - yz + xz = 0, find dzldx and dzldy.
By implicit differentiation with respect to x,
By implicit differentiation with respect to y,
42.16 If z is implicitly defined as a function of * and y by x2 + y 2 + z 2 = 1, show that
By implicit differentiation with respect to *, 2x + 2z(dzldx) = 0, dzldx=—xlz. By implicit differentia-
tion with respect to y, 2y + 2z(dzldy) = 0, dzldy = -ylz. Thus,
42.17 If z = In show that
Because we have and
Therefore,
42.18 If x = e2rcos6 and y = elr sin 6, find r,, r,,, 0 X , and Oy by implicit partial differentiation.
Differentiate both equations implicitly with respect to x. 1 = 2e2' (cosQ)rlt - e2r (sin 0)0,, 0 =
3
3e ' (sin 0)r, + e3r (cos 0)0,. From the latter, since e3r ¥= 0, 0 = 3 (sin 0)r, + (cos 0)0,. Now solve simulta-
neously for r, and 0,. r, =cos0/[e 2r (2 + sin2 e)], 0, = -3 sin 0/[e2r(2 + sin2 0)]. Now differentiate the
original equations for x and y implicitly with respect toy: 0 = 2e2r (cos0)ry - e2' (sin 0)0y, 1 =3e3r (sin 0)^ +
e 3r (cos0)0 v . From the first of these, since e 2l VO, we get 0 = 2(cos0)r,, -($^0)0^. Solving simulta-
neously for ry and 0y, we obtain ry = sin 0/[e3'(2 + sin2 0)] and Oy = 2cos 0/[e3r(2 + sin2 9)].
, 378 CHAPTER 42
42.19 If z = e"ysin(x/y) + ey"tcos(y/x),show that
It is easy to prove a more general result. Let z = f ( x / y ) , where/is an arbitrary differentiable function.
Then
and
by addition,
42.20 If z = xey'*, show that
In general, if z = xf(y/x), where / is differentiable,
42.21 If evaluate
and Problem 42.19 applies.
42.22 If find and
since, in general, Similarly,
Given a relationship F(x, y, z) = 0, where F has nonzero partial derivatives with respect to its arguments,
prove the cyclical formula (dxldy)(dyldz)(dz/dx) = -1.
Holding z constant, differentiate the functional equation on y: Fxxy + Fy=-0, or xy = -FyIFx. Similarly
(or by cyclical permutation of the variables), y: = ~Fl/Fy and zx=-F>t/F2. Then, by
multiplication, xyyzzx = —FyF.FJFxFyFI = — 1, which is the desired result.
42.24 If f(x, y) = Ix2y - 2x + 5y2, find £(1,2) and/v(l,2).
fr=6xy-6x2. Hence, £(1,2) = 12-6 = 6. /, = 3x2 + IQy. Hence, £(1,2) = 3 + 20 = 23.
42.25 If f ( x , y) = cos 3x sin 4y, find £(ir/12, 77/6) and f(ir/U, ir/6).
fr(x, y) = ~3 sin 3x sin 4y. Therefore,
/,.(•*» y) = 4 cos 3x cos 4y = 4
42.26 Find the slope of the tangent line to the curve that is the intersection of the surface z = x2 — y2 with the plane
x = 2, at the point (2,1,3).
In the plane x = 2, x is constant. Hence, the slope of the tangent line to the curve is the derivative
dzldy = -2y = -2(1) = -2.
42.27 Find the slope of the tangent line to the curve that is the intersection of the sphere x2 + y2 + z2 = 1 with the
plane y = f , at the point ( j , j, V2/2).
Since y is constant in the plane y = 5, the slope of the tangent line to the curve is dzldx. By implicit
differentiation of the equation x2 + y2 + z2 = l, we get 2x + 2z(dzldx) = 0. Hence, at the point
(\, \, V2/2), dzldx= -jc/z = -4/(V2/2)= -1/V2= -V2/2.
42.28 Find the slopes of the tangent lines to the curves cut from the surface z = 3x2 + 4y2 -6 by planes through the
point (1,1,1) and parallel to the xz- and yz-planes.
