   Chapter 14.5, Problem 57E

Chapter
Section
Textbook Problem

If f is homogeneous of degree n, show thatfx(tx, ty) = tn−1fx(x, y)

To determine

To show: The equation fx(tx,ty)=tn1fx(x,y) where the function f is homogeneous of degree n .

Explanation

Proof:

Given:

The function f is homogeneous of degree n .

Definition used: Homogeneous of degree n

“A function f is called homogeneous of degree n if it satisfies the equation

f(tx,ty)=tnf(x,y) for all t , where n is a positive integer and f has continuous second-order partial derivatives”.

Chain Rule:

“Suppose that z=f(x,y) is a differentiable function of x and y , where x=g(s,t) andy=h(s,t) are both differentiable functions of sandt . Then zs=zxxs+zyys and zt=zxxt+zyyt ”.

Calculation:

The given function is homogeneous of degree n . So, f(tx,ty)=tnf(x,y) .

Take a partial derivative with respect to x on both side of the equation f(tx,ty)=tnf(x,y) by using Chain Rule

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