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Let f be defined on R 2 by f ( 0 , 0 ) = 0 and, for ( x , y ) ( 0 , 0 ) , f ( x , y ) = x 3 y / ( x 2 + y 2 ) .

  1. Prove that both partial derivatives f x and f y exist at each point in the plane.
  2. Show that f y x ( 0 , 0 ) = 1 and f x y ( 0 , 0 ) = 0 .
  3. Show that f x y exists at each point in the plane, but that it is not continuous at ( 0 , 0 ) .

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

Let f : S R be a real-valued function of two real variables, and let c = ( a , b ) S 0 be a point at which f attains a local maximum or a local minimum. Show that ifeither of the partial derivatives t i a l f / t i a l x or t i a l f / t i a l y exists at c , then it must be equal to 0.

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

Whenever we make a new definition about functions, the question arisesof 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.

(Chain Rule again) Suppose S is a subset of R 2 , that ( a , b ) is a point in the interior of S , and that f : S R is a real-valued function that is differentiable, as a function of two real variables, at the point ( a , b ) . Suppose that T is a subset of R , that c belongs to the interior of T , and that φ : T R 2 is differentiable at the point c and φ ( c ) = ( a , b ) . Write φ ( t ) = ( x ( t ) , y ( t ) ) . Then the composition f φ is differentiable at c 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 ) .

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 ) .

and from part (3) of [link] , write

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

or, in component form,

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

and

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

We also have that

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

and

lim h 0 θ y ( h ) h = 0 .

We will show that f φ is differentiable at c by showing that there exists a number L and a function θ satisfying the two conditions of part (3) of [link] .

Define

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

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 ) ) .

We define L = ( L 1 x ' ( c ) + L 2 y ' ( c ) ) and θ ( h ) = l 1 θ x ( h ) + L 2 θ y ( h ) + θ f ( k 1 ( h ) , k 2 ( h ) ) . By these definitions and the calculation above we have Equation (4.1)

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

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

  1. Finish the proof to the preceding theorem by showing that
    lim h 0 θ f ( k 1 ( h ) , k 2 ( h ) ) h = 0 .
    HINT: Review the corresponding part of the proof to [link] .
  2. Suppose f : S R is as in the preceding theorem and that φ is a real-valued function of a real variable.Suppose f is differentiable, as a function of two real variables, at the point ( a , b ) , and that φ is differentiable at the point c = f ( a , b ) . Let g = φ f . Find a formula for the partial derivatives of the real-valued function g of two real variables.
  3. (A generalized Mean Value Theorem) Suppose u is a real-valued function of two real variables, both of whose partial derivatives exist at each point in a disk B r ( a , b ) . Show that, for any two points ( x , y ) and ( x ' , y ' ) in B r ( a , b ) , there exists a point ( x ^ , y ^ ) on the line segment joining ( x , y ) to ( x ' , y ' ) such that
    u ( x , y ) - u ( x ' , y ' ) = t i a l u t i a l x ( x ^ , y ^ ) ( x - x ' ) + t i a l u t i a l y ( x ^ , y ^ ) ( y - y ' ) .
    HINT: Let φ : [ 0 , 1 ] R 2 be defined by φ ( t ) = ( 1 - t ) ( x ' , y ' ) + t ( x , y ) . Now use the preceding theorem.
  4. Verify that the assignment f t i a l f / t i a l x is linear; i.e., that
    t i a l ( f + g ) t i a l x = t i a l f t i a l x + t i a l g t i a l x .
    Check that the same is true for partial derivatives with respect to y .

Questions & Answers

what is math number
Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
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12, 17, 22.... 25th term
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12, 17, 22.... 25th term
Akash
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Carole
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AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
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Adu
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Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
kinnecy Reply
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
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Commplementary angles
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Nharnhar
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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