Chapter 14, Problem 21RE

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Find all second partial derivatives of f.21. f(x, y, z) = xkylzm

To determine

To find: The second order partial derivative of the function f(x,y,z)=xkylzm .

Explanation

The given function is, f(x,y,z)=xkylzm .

Differentiate the given function with respect to x and obtain fx .

fx=x(xkylzm)=ylzmx(xk)=ylzm(kxk1)

fx=kxk1ylzm (1)

Differentiate the equation (1) with respect to x and obtain the second order derivative, fxx

2fx2=x(kxk1ylzm)=kylzmx(xk1)=kylzm[(k1)xk2]=k(k1)xk2ylzm

Thus, fxx(x,y,z)=k(k1)xk2ylzm .

Differentiate the given function with respect to y and obtain fy .

fy=y(xkylzm)=xkzmy(yl)=xkzm(lyl1)

fy=lxkyl1zm (2)

Differentiate the equation (2) with respect to y and obtain the second order derivative, fyy .

2fy2=y(lxkyl1zm)=lxkzmy(yl1)=lxkzm[(l1)yl2]=l(l1)xkyl2zm

Hence, fyy(x,y,z)=l(l1)xkyl2zm .

Differentiate the given function with respect to z and obtain fz .

fz=z(xkylzm)=xkylz(zm)=xkyl(mzm1)

fz=mxkylzm1 (3)

Differentiate the equation (3) with respect to z and obtain the second order derivative, fzz

2fz2=z(mxkylzm1)=mxkyly(zm1)=mxkyl[(m1)zm2]=m(m1)xkylzm2

Hence, fzz(x,y,z)=m(m1)xkylzm2 .

Differentiate equation (1) with respect to y and obtain the partial derivative, fxy .

2fxy=y(kxk1ylzm)=kxk1zmy(yl)=kxk1zm[lyl1]=klxk1yl1zm

Therefore, fxy(x,y,z)=klxk1yl1zm .

Differentiate equation (2) with respect to x and obtain the partial derivative, fyx .

2fyx=x(lxkyl1zm)=lyl1zmx(xk)=lyl1zm[kxk1]=klxk1yl1zm

Thus, fyx(x,y,z)=klxk1yl1zm

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