Explain why the function is differentiable at the given point. Then find the lin

postillan4

postillan4

Answered question

2021-09-08

Explain why the function is differentiable at the given point. Then find the linearization L(x, y) of the function at that point. f(x,y)=1+xIn(xy5),(2,3)

Answer & Explanation

Adnaan Franks

Adnaan Franks

Skilled2021-09-09Added 92 answers

Step 1
The Linearization L(x,y) for f(x,y) at (x0,y0) is
L(x,y)=f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)
Step 2
f(x,y)=1+xln(xy5)
f(2,3)=1+2ln(65)=1+2ln1=1
fx=ln(xy5)+xyxy5
fx(2,3)=ln(65)+665=6
fy=x2xy5
fy(2,3)=2265=4
Step 3
Linearization at (2,3) is
L(x,y)=f(2,3)+fx(2,3)(x2)+fy(2,3)(y3)Substitute
f(2,3)=1,fx(2,3)=6,fy(2,3)=4
L(x,y)=1+6(x2)+4(y3)
L(x,y)=1+6x12+4y12
L(x,y)=6x+4y23
Result
L(x,y)=6x+4y23

madeleinejames20

madeleinejames20

Skilled2023-05-28Added 165 answers

Step 1: Let's find the partial derivatives of f with respect to x and y:
fx=ln(xy5)
fy=xxy5
Step 2: Next, let's evaluate these partial derivatives at the given point (2,3):
fx|(2,3)=ln(2·35)=ln(1)=0
fy|(2,3)=22·35=21=2
Since both partial derivatives exist and are continuous at (2,3), the function f(x,y) is differentiable at that point.
Step 3: Now, let's find the linearization L(x,y) of the function f at the point (2,3). The linearization is given by:
L(x,y)=f(2,3)+fx|(2,3)(x2)+fy|(2,3)(y3)
Substituting the values we found earlier:
L(x,y)=1+0(x2)+2(y3)=1+2(y3)
Therefore, the linearization of the function f(x,y) at the point (2,3) is L(x,y)=1+2(y3).
Eliza Beth13

Eliza Beth13

Skilled2023-05-28Added 130 answers

Answer:
4x+4y19
Explanation:
To show why the function is differentiable at the given point (2,3) and find its linearization L(x,y), we need to evaluate the partial derivatives and substitute the point's coordinates into the linearization formula.
Let's start by calculating the partial derivatives of f(x,y) with respect to x and y using the product rule and chain rule:
fx=x(1+xln(xy5))
=ln(xy5)+x·xln(xy5)
=ln(xy5)+x·1xy5·x(xy5)
=ln(xy5)+x·1xy5·(y)
=ln(xy5)+xyxy5
Now, let's find the partial derivative with respect to y:
fy=y(1+xln(xy5))
=x·yln(xy5)
=x·1xy5·y(xy5)
=x·1xy5·(x)
=x2xy5
Now, let's substitute the coordinates (2,3) into the partial derivatives:
fx(2,3)=ln(2·35)+2·22·35=ln(65)+465=ln(1)+4=0+4=4
fy(2,3)=222·35=465=41=4
Now, we can write the linearization L(x,y) using the point-slope form of a linear equation:
L(x,y)=f(2,3)+fx(2,3)·(x2)+fy(2,3)·(y3)
L(x,y)=f(2,3)+4·(x2)+4·(y3)
L(x,y)=f(2,3)+4x8+4y12
L(x,y)=f(2,3)+4x+4y20
Since we're given the point (2,3), we substitute these values into the original function to find f(2,3):
f(2,3)=1+2ln(2·35)=1+2ln(65)=1+2ln(1)=1+2·0=1+0=1
Therefore, the linearization L(x,y) of the function at the point (2,3) is:
L(x,y)=1+4x+4y20
L(x,y)=4x+4y19.
Nick Camelot

Nick Camelot

Skilled2023-05-28Added 164 answers

To determine the differentiability of the function f(x,y)=1+xln(xy5) at the point (2,3), we need to check if the partial derivatives of f exist and are continuous in a neighborhood of that point.
First, let's find the partial derivatives of f with respect to x and y:
fx=x(1+xln(xy5))
Using the product rule, we have:
fx=0+ln(xy5)+x·x(ln(xy5))
Now, let's find x(ln(xy5)):
x(ln(xy5))=x(ln(xy5))·(xy5)x
Simplifying further:
x(ln(xy5))=x(ln(xy5))·y
Therefore, we have:
fx=ln(xy5)+x·x(ln(xy5))=ln(xy5)+x·x(ln(xy5))·y
Next, let's find fy:
fy=y(1+xln(xy5))
Applying the chain rule:
fy=x·y(ln(xy5))=x·y(ln(xy5))·(xy5)y
Simplifying further:
fy=x·y(ln(xy5))·x=x·y(ln(xy5))·x
Now, let's evaluate the partial derivatives at the point (2,3):
fx|(2,3)=ln(2·35)+2·x(ln(2·35))·3
fy|(2,3)=2·y(ln(2·35))·2
Now, let's find the values of the partial derivatives at the point (2,3):
fx|(2,3)=ln(65)+2·x(ln(65))·3=ln(1)+2·165·3=ln(1)+2·31=0+2·3=6
fy|(2,3)=2·y(ln(65))·2=2·165·2=2·2=4
Since both partial derivatives exist and are continuous at the point (2,3), the function f is differentiable at that point.
Next, let's find the linearization L(x,y) of the function f(x,y) at the point (2,3). The linearization can be expressed using the equation of a plane:
L(x,y)=f(2,3)+fx|(2,3)·(x2)+fy|(2,3)·(y3)
Substituting the values we found earlier:
L(x,y)=f(2,3)+6·(x2)+4·(y3)
Therefore, the linearization of the function f(x,y) at the point (2,3) is:
L(x,y)=f(2,3)+6·(x2)+4·(y3)

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