   Chapter 14.4, Problem 14E

Chapter
Section
Textbook Problem

Explain why the function is differentiable at the given point. Then find the linearization L(x, y) of the function at that point.14. f ( x , y ) = 1 + y 1 + x , (1, 3)

To determine

To find: The linearization L(x,y) of the function f(x,y)=1+y1+x at the point (1,3) and explain the reason why the given function is differentiable at the point (1,3).

Explanation

Theorem used:

“If the partial derivatives fx and fy exist near (a,b) and are continuous at (a,b) , then f is differentiable at (a,b) .”

Result used:

“Linearization of the function f(x,y) at (a,b) is L(x,y)=f(a,b)+fx(a,b)(xa)+fy(a,b)(yb) .”

Calculation:

Let the given function as z=1+y1+x . (1)

The point is P(a,b)=(1,3) .

Take partial derivative with respect to x in the equation (1),

fx(x,y)=(1+y)(1+x)1=(1+y)(1)(1+x)11=(1+y)(1+x)2=(1+y)(1+x)2

Take partial derivative with respect to y in the equation (1),

fy(x,y)=1(1+x)(0+1)=1(1+x)

Obtain the partial derivative of x at the given point (1,3) .

fx(a,b)=fx(1,3)=(1+3)(1+1)2=44=1

Obtain the partial derivative of y at the given point (1,3) .

fy(a,b)=fy(1,3)=11+1=12

Obtin the value of f(1,3)

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