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DEPARTMENT OF MATHEMATICS, IIT - GUWAHATI
Odd Semester of the Academic Year 2019- MA 101 Mathematics II Problem Sheet 2: Partial derivatives, tangent and normals, differentials, gradient, directional derivatives and chain rules etc. Instructor: Dr. J. C. Kalita and Dr. Sriparna Bandyopadhyay
- Let f (x, y) =
{ (^) x (^2) −xy x+y if^ x^ +^ y^6 = 0 0 if x + y = 0. Find
(a) fx(0, 0), fy (0, 0) (b) lim (x,y)→(0,0) fx(x, y), and check whether it is equal to fx(0, 0).
Solution: Note that fx(0, 0) if it exists, is given by: fx(0, 0) = lim h→ 0
f (h, 0) − f (0, 0) h
= lim h→ 0
h − 0 h
Similarly fy (0, 0) if it exists, is given by:
fy(0, 0) = lim k→ 0
f (0, k) − f (0, 0) k
= lim k→ 0
k
(b) Also fx(x, y) ( for x + y 6 = 0 ) if it exists, is given by: fx(x, y) = lim h→ 0
f (x + h, y) − f (x, y) h
Since (x, y) is such that x + y 6 = 0, if h sufficiently small such that x + y + h 6 = 0, then f (x + h, y) − f (x, y) h
(x+h)^2 −(x+h)y (x+h)+y −^
x^2 −xy x+y h
hx^2 + 2hxy − hy^2 + h^2 x + h^2 y h(x + y)(x + h + y)
x^2 + 2xy − y^2 + hx + hy (x + y)(x + h + y)
Since lim h→ 0 x^2 + 2xy − y^2 + hx + hy = x^2 + 2xy − y^2 and lim h→ 0 (x + y)(x + h + y) = (x + y)^2 6 = 0 for x + y 6 = 0,
fx(x, y) = lim h→ 0
f (x + h, y) − f (x, y) h
x^2 + 2xy − y^2 (x + y)^2
2 y^2 (x + y)^2
for x + y 6 = 0. Clearly lim (x,y)→(0,0)
fx(x, y) = lim (x,y)→(0,0)
2 y^2 (x + y)^2
does not exist (consider (x, y) → (0, 0) such that y = mx).
- Let f (x, y) =
√ x^2 + y^2.
(a) Find fx(x, y) and fy (x, y) for (x, y) 6 = (0, 0)
(b) Show that fx(0, 0) and fy(0, 0) do not exist.
Solution: (a) Note that fx(x, y) if it exists, is given by fx(x, y) = lim h→ 0
f (x + h, y) − f (x, y) h f (x + h, y) − f (x, y) h
√ (x + h)^2 + y^2 −
x^2 + y^2 h
=
√ (x + h)^2 + y^2 −
x^2 + y^2 )(
√ (x + h)^2 + y^2 +
x^2 + y^2 ) h(
√ (x + h)^2 + y^2 +
x^2 + y^2 )
h + 2x √ (x + h)^2 + y^2 +
x^2 + y^2
for (x, y) 6 = (0, 0). Hence fx(x, y) = lim h→ 0
f (x + h, y) − f (x, y) h
x √ x^2 + y^2
, for (x, y) 6 = (0, 0).
Similarly fy (x, y) = lim k→ 0
f (x, y + k) − f (x, y) k
y √ x^2 + y^2
, for (x, y) 6 = (0, 0).
Aliter: fx(x, y) for (x, y) 6 = (0, 0) may be obtained directly (without going through first principles) by taking the single variable derivative of f (x, y) with respect to x by treating y as constant. Similarly fy (x, y) for (x, y) 6 = (0, 0) may be obtained directly (without going through first principles) by taking the single variable derivative of f (x, y) with respect to y by treating x as constant.
(b) Also fx(0, 0) = lim h→ 0
f (h, 0) − f (0, 0) h
= lim h→ 0
|h| h
, does not exist. Similarly fy (0, 0) does not exist.
- Let
f (x, y) =
{ (^) xy 3 x^2 +y^6 if^ (x, y)^6 = (0,^ 0) 0 if (x, y) = (0, 0).
(a) Calculate fx(x, y) and fy (x, y) at all points where (x, y) 6 = (0, 0). (b) Compute all first and second order partial derivatives at (0, 0) if they exist. (c) Show that f is discontinuous at (0, 0).
Solution: (a) fx(x, y) =
y^3 (y^6 − x^2 ) (x^2 + y^6 )^2
and fy(x, y) =
3 xy^2 (x^2 − y^6 ) (x^2 + y^6 )^2
, for (x, y) 6 = (0, 0). (1)
fx(x, 0) = lim h→ 0
f (x + h, 0) − f (x, 0) h
= lim h→ 0
h
= 0 for x 6 = 0.
fy(0, y) = lim k→ 0
f (0, y + k) − f (0, y) k
= lim k→ 0
k
= 0 for y 6 = 0.
fy(x, 0) = lim k→ 0
f (x, k) − f (x, 0) k
= lim k→ 0
xk^3 x^2 +k^6 −^0 k
= lim k→ 0
xk^2 x^2 + k^6
= 0 for x 6 = 0.
fx(0, y) = lim h→ 0
f (h, y) − f (0, y) h
= lim h→ 0
hy^3 h^2 +y^6 −^0 h
y^3 y^6
y^3
for y 6 = 0. Aliter: The above expressions for fx(x, 0) = 0 for x 6 = 0,
Solution: Since the level surface is of the form f (x, y, z) = 0, where f (x, y, z) = z^2 − (x^2 + y^2 ), the tangent plane of the level surface at (x 0 , y 0 , z 0 ) is of the form: (z − z 0 )fz (x 0 , y 0 , z 0 ) + (x − x 0 )fx(x 0 , y 0 , z 0 ) + (y − y 0 )fy(x 0 , y 0 , z 0 ) = 0. ⇒ (z − z 0 )(2z 0 ) + (x − x 0 )(− 2 x 0 ) + (y − y 0 )(− 2 y 0 ) = 0, ⇒ zz 0 − xx 0 − yy 0 = 0 since z^20 − x^20 − y^20 = 0. Clearly (x, y, z) = (0, 0 , 0) satisfies the above equation for all points (x 0 , y 0 , z 0 ) of the cone.
- Find the equations of the tangent plane and normal line to the given surface at the specified point
(a) x^2 + y^2 − z^2 − 2 xy + 4xz = 4, (1, 0 , 1). (b) z + 1 = xey^ cos z, (1, 0 , 0).
Solution: (a) Since the level surface is of the form f (x, y, z) = 4, where f (x, y, z) = x^2 + y^2 − z^2 − 2 xy + 4xz, the tangent plane of the level surface at (x 0 , y 0 , z 0 ) = (1, 0 , 1) is of the form: (z − z 0 )fz (x 0 , y 0 , z 0 ) + (x − x 0 )fx(x 0 , y 0 , z 0 ) + (y − y 0 )fy(x 0 , y 0 , z 0 ) = 0. (1) where fx(x 0 , y 0 , z 0 ) = 2x 0 − 2 y 0 + 4z 0 , fy (x 0 , y 0 , z 0 ) = 2y 0 − 2 x 0 and fz(x 0 , y 0 , z 0 ) = − 2 z 0 + 4x 0. (1) implies ⇒ (z − 1)2 + (x − 1)6 + y(−2) = 0 or 2 z + 6x − 2 y = 8. The symmetric equations of the normal line is given by: (z − z 0 ) fz (x 0 , y 0 , z 0 )
(x − x 0 ) fx(x 0 , y 0 , z 0 )
(y − y 0 ) fy(x 0 , y 0 , z 0 )
, or (z − 1) 2
(x − 1) 6
(y − 0) − 2
(b) Since the level surface is of the form f (x, y, z) = −1, where f (x, y, z) = z − xey^ cos z, the tangent plane of the level surface at (x 0 , y 0 , z 0 ) = (1, 0 , 0) is of the form: (z − z 0 )fz (x 0 , y 0 , z 0 ) + (x − x 0 )fx(x 0 , y 0 , z 0 ) + (y − y 0 )fy(x 0 , y 0 , z 0 ) = 0, (1) where fx(x 0 , y 0 , z 0 ) = −ey^0 cos z 0 , fy (x 0 , y 0 , z 0 ) = −x 0 ey^0 cos z 0
and fz(x 0 , y 0 , z 0 ) = 1 + x 0 ey^0 sin z 0. (1) ⇒ (z − 0)1 + (x − 1)(−1) + (y − 0)(−1) = 0 or z − x − y = −1.
The symmetric equations of the normal line is given by: (z − z 0 ) fz (x 0 , y 0 , z 0 )
(x − x 0 ) fx(x 0 , y 0 , z 0 )
(y − y 0 ) fy(x 0 , y 0 , z 0 ) (z − 0) 1
(x − 1) − 1
(y − 0) − 1
- Suppose we are seeking the equation of the tangent plane to a surface S at the point P = (2, 1 , 3). You don’t have an equation for S, but you know that the curves
r 1 (t) =
〈 2 + 3t, 1 − t^2 , 3 − 4 t + t^2
〉
r 2 (t) =
〈 1 + t^2 , 2 t^3 − 1 , 2 t + 1
〉
both lie on S. Find an equation of the tangent plane at P. Solution: Clearly r 1 (0) corresponds to (2, 1 , 3), r 2 (1) corresponds to (2, 1 , 3).
Also the tangent plane to the surface S at (2, 1 , 3) contains the tangent lines to each of the curves r 1 (t) and r 2 (t) at (2, 1 , 3). The tangent vectors at (2, 1 , 3) for r 1 , r 2 at (2, 1 , 3) are given by:
r′ 1 (0) = 〈 3 , 0 , − 4 〉 , r′ 2 (1) = 〈 2 , 6 , 2 〉 , respectively. Hence a normal to the surface is one which is orthogonal to both these tangent vectors and is given by:
r′ 1 (0) × r′ 2 (1) =
∣∣ ∣∣ ∣∣ ∣
i j k 3 0 − 4 2 6 2
∣∣ ∣∣ ∣∣ ∣
0 5
5
10 z
15
y
0
20
x
-5 (^) -2 -1 0 1 2 3 4 5 -5 -4 -
z=x^2 -3xy-y^2
z=13x-
(2, 3, 13)
(a)
2
13
z
y
3 3.5 (^) 1.95 1. x
2.1^ 2.05^2 (^4) 2.2 2.
z=x^2 -3xy-y^2
(2, 3, 13)
z=13x-
(b)
- Find the directional derivative of the function at the given point in the direction of the vector v.
(a) f (x, y, z) =
√ x^2 + y^2 + z^2 , (1, 2 , −2), v = 〈− 6 , 6 , − 3 〉
(b) g(x, y, z) = x tan−^1
( (^) y
z
) , (1, 2 , −2), v = i + j − k.
Solution: (a) The directional derivative of f at (1, 2 , −2) along the unit vector u =
〈 −
〉 if it exists, is equal to the following limit:
Duf (1, 2 , −2) = lim t→ 0
f ((1, 2 , −2) + t(−^23 , 23 , −^13 )) − f (1, 2 , −2) t
f ((1, 2 , −2) + t(− 32 , 23 , −^13 )) − f (1, 2 , −2) t
√ (1 − 23 t)^2 + (2 + 23 t)^2 + (− 2 − 13 t)^2 − 3 t
√ 9 + t^2 + 83 t − 3 t
By applying L’Hospital’s rule we get the limit as
= lim t→ 0
2 t + (^83) 2
√ 9 + t^2 + 83 t
(b) The directional derivative of f at (1, 2 , −2) along the unit vector u =
〈 √^1 3 ,^ √^1 3 ,^ −^ √^1 3
〉
is equal to the following limit, if it exists.
Duf (1, 2 , −2) = lim t→ 0
f ((1, 2 , −2) + t( √^13 , √^13 , − √^13 )) − f (1, 2 , −2) t
f ((1, 2 , −2) + t( √^13 , √^13 , − √^13 )) − f (1, 2 , −2) t =
(1 + √t 3 )tan−^1 (−1) − tan−^1 (−1) t
, for t 6 = 0 and t sufficiently small.
√^ t 3 tan
t
tan−^1 (−1) = −
π 4
⇒ lim t→ 0
f ((1, 2 , −2) + t( √^13 , √^13 , (^) −^1 √ 3 )) − f (1, 2 , −2) t
π 4
Aliter: Note that if the existence of all directional derivatives of f at (x 0 , y 0 , z 0 ) is already guaranteed (say when f is differentiable at (x 0 , y 0 , z 0 ) ) then one can directly calculate the value of the directional derivative by the formula given below: Duf (x 0 , y 0 , z 0 ) = [∇f ](x 0 ,y 0 ,z 0 ) · u.
- Find the directional derivatives of the scalar field f (x, y) = x^3 − 3 xy along the parabola y = x^2 − x + 2 at the point (1,2).
Solution: A unit vector along the tangent to the parabola at (1, 2) can be obtained as follows: The slope of the tangent vector of the parabola at (1, 2) is given by dy dx
|x=1= (2x − 1)x=1 = 1. Hence the tangent line of the parabola at (1, 2) is given by y − 2 = 1(x − 1) or y − x = 1. Any two points on this tangent line can be taken as P 0 (x 0 , 1+x 0 ) and P 1 (2x 0 , 1+2x 0 ) (you can choose any arbitrary points in such a way that y = 1 + x). Therefore the vector
P 0 P 1 joining the points P 0 and P 1 is given by < 2 x 0 −x 0 , (1+2x 0 )−(1+x 0 ) >=< x 0 , x 0 > so that a unit vector along this tangent
is
〈 x 0 √ x^20 + x^20
x 0 √ x^20 + x^20
〉
〈 1 √ 2
〉
. (You can also obtain the unit tangent by
parametrizing the parabola by the vector r(t) =< t, t^2 − t + 1 >, so that a unit
vector at the point (1, 2) which corresponds to t = 1 is given by ˆu =
r′(t) |r′(t)|
∣∣ ∣∣ ∣t=1^ = 〈 1 √ 1 + (2t − 1)^2
2 t − 1 √ 1 + (2t − 1)^2
〉∣∣ ∣∣ ∣∣ t=
〈 1 √ 2
〉 .
Note that depending upon the orientation of the parabola, you may take this unit
If h 1 = h 2 6 = 0
then
f (h 1 , h 2 ) − fx(0, 0)h 1 − fy(0, 0)h 2 √ h^21 + h^22
( 1 −
) ( h 1 |h 1 |
) .
Hence lim (h 1 ,h 2 )→(0,0)
f (h 1 , h 2 ) − fx(0, 0)h 1 − fy(0, 0)h 2 √ h^21 + h^22
does not exist.
Hence f is not differentiable at (0, 0).
- Show that the following functions are differentiable at the respective points men- tioned below:
(a) Let
f (x, y) =
{ (^) x x+y if^ x^ +^ y^6 = 0 0 if x + y = 0. Show that f is differentiable at (2, 1) but not differentiable at (0, 0). (b) Show that f (x, y) =
x + ey^ is differentiable at (3, 0), where x, y is such that x + ey^ ≥ 0.
Solution: (a) f is differentiable at (2, 1) if
lim (h,k)→ 0
∣∣ ∣f (2 + h, 1 + k) − f (2, 1) − ∇f · 〈h, k〉
∣∣ ∣ √ h^2 + k^2
∣∣ ∣f (2 + h, 1 + k) − f (2, 1) − ∇f · 〈h, k〉
∣∣ ∣ √ h^2 + k^2
∣∣ ∣ (^) 3+2+hh+k − 23 − h 9 + 29 k
∣∣ ∣ √ h^2 + k^2 =
|hk + 2k^2 − h^2 | 9(3 + h + k)
h^2 + k^2
|hk + 2k^2 − h^2 | √ h^2 + k^2
for sufficiently small h and k.
Hence the above expression is ≤
|hk| √ h^2 + k^2
2 k^2 √ h^2 + k^2
h^2 √ h^2 + k^2
h^2 + k^2.
Also lim (h,k)→ 0
h^2 + k^2 = 0.
Since 0 ≤ lim (h,k)→ 0
∣∣ ∣f (2 + h, 1 + k) − f (2, 1) − ∇f · 〈h, k〉
∣∣ ∣ √ h^2 + k^2
≤ 4 lim (h,k)→ 0
h^2 + k^2 = 0,
lim (h,k)→ 0
∣∣ ∣f (2 + h, 1 + k) − f (2, 1) − ∇f · 〈h, k〉
∣∣ ∣ √ h^2 + k^2
Hence f is differentiable at (2, 1).
Aliter: Note that for all (x, y) such that x + y 6 = 0, fx(x, y) =
y (x + y)^2
and
fy(x, y) = −
x (x + y)^2
are continuous functions, hence fx, fy are continuous through- out some small neighborhood of (2, 1), hence f is differentiable at (2, 1).
If x = y 6 = 0, then f (x, x) =
x x + x
, hence lim x→ 0 f (x, x) =
, but f (0, 0) = 0, hence f is not continuous at (0, 0), which implies f is not differentiable at (0, 0).
(b) Note that for all (x, y) such that x + ey^ > 0, fx(x, y) =
x + ey^
and
fy(x, y) =
ey 2
x + ey^
are continuous, hence fx, fy are continuous throughout some small neighborhood of (3, 0), hence f is differentiable at (3, 0).
- Show that the following function is differentiable throughout R^2 and find the max- imum rate of change of f (x, y) = 6 − 3 x^2 − y^2 at the point (1,2) and the direction in which it occurs.
Solution: If f differentiable throughout R^2 , then the directional derivatives exist along any direction u at (1, 2). Then the directional derivative of f along any unit vector u = 〈u 1 , u 2 〉 at (1, 2), is given by Duf (1, 2) = ∇f · 〈u 1 , u 2 〉 =
∣∣ ∣∇f
∣∣ ∣ cos θ, where θ is the angle between ∇f and u. Hence the directional derivative is maximum when cos θ = 1, or θ = 0 or u is parallel to and in the same direction as ∇f .
Hence the required u =
∣∣ ∣∇f
∣∣ ∣ =
〈 − 6 √ 52
〉 .
To show that f is differentiable: Check that for (x, y) ∈ R^2 , fx(x, y) = − 6 x and fy(x, y) = − 2 y are continuous func- tions, hence f is differentiable throughout R^2. () (() For any (x 0 , y 0 ) ∈ R^2 |fx(x, y) − fx(x 0 , y 0 )| = 6 |x − x 0 | ≤ 6
√ (x − x 0 )^2 + (y − y 0 )^2. Hence given ǫ > 0 take δ = ǫ 6 , then |fx(x, y) − fx(x 0 , y 0 )| < ǫ if
√ (x − x 0 )^2 + (y − y 0 )^2 < δ which implies fx is continuous at (x 0 , y 0 ). Similarly one can show that fy is continuous at (x 0 , y 0 ).)
Aliter: f is differentiable at (x 0 , y 0 ) if
lim (h,k)→ 0
∣∣ ∣f (x 0 + h, y 0 + k) − f (x 0 , y 0 ) − [∇f ](x 0 ,y 0 ) · 〈h, k〉
∣∣ ∣ √ h^2 + k^2
∣∣ ∣f (x 0 + h, y 0 + k) − f (x 0 , y 0 ) − [∇f ](x 0 ,y 0 ) · 〈h, k〉
∣∣ ∣ √ h^2 + k^2
|− 3 h^2 − k^2 | √ h^2 + k^2
h^2 √ h^2 + k^2
k^2 √ h^2 + k^2 ≤ 3
h^2 + k^2 +
h^2 + k^2 (since h^2 , k^2 ≤ h^2 + k^2 ). = 4
h^2 + k^2.
Also lim (h,k)→ 0
h^2 + k^2 = 0.
Since 0 ≤ lim (h,k)→ 0
∣∣ ∣f (x 0 + h, y 0 + k) − f (x 0 , y 0 ) − [∇f ](x 0 ,y 0 ) · 〈h, k〉
∣∣ ∣ √ h^2 + k^2
≤ 4 lim (h,k)→ 0
h^2 + k^2 = 0,
∂g ∂θ
∂w ∂x
∂x ∂θ
∂w ∂y
∂y ∂θ
∂w ∂x
(−r sin θ) +
∂w ∂y
(r cos θ).
Again by applying the chain rule we get ∂^2 g ∂θ^2
∂w ∂x
(−r cos θ) + (−r sin θ)
( ∂^2 w ∂x^2
∂x ∂θ
∂^2 w ∂y∂x
∂y ∂θ
)
∂w ∂y
(−r sin θ)+
(r cos θ)
( ∂^2 w ∂y^2
∂y ∂θ
∂^2 w ∂x∂y
∂x ∂θ
) .
But we will write
∂^2 g ∂θ^2
as
∂^2 w ∂θ^2
∂^2 w ∂θ^2
= (−r cos θ)
∂w ∂x
( ∂^2 w ∂x^2
)
( ∂^2 w ∂y∂x
)
∂w ∂y
(r cos θ)^2
( ∂^2 w ∂y^2
)
( ∂^2 w ∂x∂y
)
. (3)
From (1) , (2) and (3) it follows that ∂^2 w ∂r^2
r
∂w ∂r
r^2
∂^2 w ∂θ^2
∂^2 w ∂x^2
∂^2 w ∂y^2
- Suppose w = f (u) where u =
x^2 − y^2 x^2 + y^2
. Assuming the existence of all the required first order partial derivatives of w and u show that xwx + ywy = 0.
Solution: By applying chain rule we get wx =
∂f ∂u
∂u ∂x
and wy =
∂f ∂u
∂u ∂y
∂u ∂x
|(x,y) 6 =(0,0) =
4 xy^2 (x^2 + y^2 )^2
and
∂u ∂y
|(x,y) 6 =(0,0) =
− 4 yx^2 (x^2 + y^2 )^2
Hence xwx + ywy =
4 x^2 y^2 (x^2 + y^2 )^2
∂f ∂u
− 4 x^2 y^2 (x^2 + y^2 )^2
∂f ∂u
- Implicit differentiation: If φ(x, y, z) = 0 defines z as an implicit function of x and y in a region R of the xy-plane, assuming the existence of all the required partial derivatives prove that
∂z ∂x
φx φz
and
∂z ∂y
φy φz
, where φz 6 = 0. Hence find
∂z ∂x and
∂z ∂y
when x
(^23)
(^23)
(^23) = 1.
Solution: Since φ(x, y, z) = 0, gives z implicitly as a function of x and y we write z = f (x, y). Hence φ(x, y, f (x, y)) = g(x, y) = 0. If we take u 1 (x, y) = x, u 2 (x, y) = y, and u 3 (x, y) = f (x, y) = z, then we get, φ(u 1 (x, y), u 2 (x, y), u 3 (x, y)) = g(x, y) = 0. By applying chain rule we get
∂φ ∂u 1
∂u 1 ∂x
∂φ ∂u 2
∂u 2 ∂x
∂φ ∂u 3
∂u 3 ∂x
∂g ∂x
= 0, or
∂φ ∂x
∂x ∂x
∂φ ∂y
∂y ∂x
∂φ ∂z
∂z ∂x
∂φ ∂x
∂φ ∂z
∂z ∂x
= 0 (since
∂y ∂x
Hence
∂z ∂x
φx φz
when φz 6 = 0,
and similarly
∂z ∂y
φy φz
, when φz 6 = 0.
If φ(x, y, z) = x
(^23)
(^23)
(^23) − 1 then φx =
x−^
1 (^3) , φy =^2 3
y−^
1 (^3) , φz =^2 3
z−^
1 (^3).
Hence
∂z ∂x
φx φz
z x
(^13) and
∂z ∂y
φy φz
z y
(^13) .
- Suppose that w =
r
f
( t −
r a
) and that r =
√ x^2 + y^2 + z^2. Assuming the existence of all the required second order partial derivatives, show that
∂^2 w ∂x^2
∂^2 w ∂y^2
∂^2 w ∂z^2
a^2
∂^2 w ∂t^2
Solution: w =
r
f
( t −
r a
) = g(x, y, z), say. By applying the chain rule we get
∂g ∂x
∂w ∂r
∂r ∂x ⇒
∂^2 g ∂x^2
∂x
∂w ∂r
∂r ∂x
( ∂r ∂x
) ( ∂ ∂x
∂w ∂r
)
( ∂w ∂r
) ( ∂ ∂x
∂r ∂x
) .
∂r ∂x
(( ∂ ∂r
∂w ∂r
) ∂r ∂x
)
∂w ∂r
( ∂^2 r ∂x^2
)
( ∂^2 w ∂r^2
) ( ∂r ∂x
) 2
( ∂w ∂r
) ( ∂^2 r ∂x^2
) .
Similarly
∂^2 g ∂y^2
( ∂^2 w ∂r^2
) ( ∂r ∂y
) 2
( ∂w ∂r
) ( ∂^2 r ∂y^2
) ,
and
∂^2 g ∂z^2
( ∂^2 w ∂r^2
) ( ∂r ∂z
) 2
( ∂w ∂r
) ( ∂^2 r ∂z^2
) .
∂^2 w ∂x^2
∂^2 w ∂y^2
∂^2 w ∂z^2
( ∂^2 w ∂r^2
) ( (
∂r ∂x
)^2 + (
∂r ∂y
)^2 + (
∂r ∂z
)^2
)
( ∂w ∂r
) ( ∂^2 r ∂x^2
∂^2 r ∂y^2
∂^2 r ∂z^2
) .
Note that
∂r ∂x
x √ x^2 + y^2 + z^2
∂^2 r ∂x^2
y^2 + z^2 (x^2 + y^2 + z^2 )
Similarly
∂r ∂y
y √ x^2 + y^2 + z^2
∂^2 r ∂y^2
x^2 + z^2 (x^2 + y^2 + z^2 )
3 2
and
∂r ∂z
z √ x^2 + y^2 + z^2
∂^2 r ∂z^2
x^2 + y^2 (x^2 + y^2 + z^2 )
get ∂^2 f (u, v) ∂t^2
= x
(( ∂ ∂u
∂f (u, v) ∂u
) ∂u ∂t
( ∂ ∂v
∂f (u, v) ∂u
) ∂v ∂t
)
y
(( ∂ ∂v
∂f (u, v) ∂v
) ∂v ∂t
( ∂ ∂u
∂f (u, v) ∂v
) ∂u ∂t
)
= n(n − 1)tn−^2 f (x, y) for all t ∈ R, where u = tx and v = ty. (2)
(2) ⇒ x^2
∂^2 f (u, v) ∂u^2
∂^2 f (u, v) ∂u∂v
∂^2 f (u, v) ∂v^2
= n(n − 1)tn−^2 f (x, y)
(since
∂^2 f (u, v) ∂u∂v
∂^2 f (u, v) ∂v∂u
Hence for t = 1 or u = x and v = y we get:
x^2
∂^2 f ∂x^2
∂^2 f ∂x∂y
∂^2 f ∂y^2
= n(n − 1)f (x, y).
(iii) fx(tx, ty) = lim h→ 0
f (tx + th, ty) − f (tx, ty) th
= lim h→ 0
tn(f (x + h, y) − f (x, y)) th
= tn−^1 lim h→ 0
f (x + h, y) − f (x, y) h
= tn−^1 fx(x, y).
