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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Finding Second Partial Derivatives In Exercises 55–60, find the second partial derivatives.

f ( x , y , z ) = 3 y z x + z

To determine

To calculate: The second partial derivative of function f(x,y,z)=3yzx+z.

Explanation

Given Information:

The provided function is f(x,y,z)=3yzx+z.

Formula used:

The power rule of differentiation with respect to x by holding y constant,

x[f(x,y)]n=n[f(x,y)]n1xf(x,y)

The power rule of differentiation with respect to y by holding x constant,

y[f(x,y)]n=n[f(x,y)]n1yf(x,y)

The quotient rule of differentiation with respect to x by holding y and z constant,

x[f(x,y,z)g(x,y,z)]=[g(x,y,z)xf(x,y,z)f(x,y,z)xg(x,y,z)][g(x,y,z)]2

The quotient rule of differentiation with respect to y by holding x and z constant,

y[f(x,y,z)g(x,y,z)]=[g(x,y,z)yf(x,y,z)f(x,y,z)yg(x,y,z)][g(x,y,z)]2

The quotient rule of differentiation with respect to z by holding x and y constant,

z[f(x,y,z)g(x,y,z)]=[g(x,y,z)zf(x,y,z)f(x,y,z)zg(x,y,z)][g(x,y,z)]2

Calculation:

Consider the function,

f(x,y,z)=3yzx+z

Now apply, quotient rule of differentiation with respect to x by holding y constant,

xf(x,y,z)=x(3yzx+z)=(x+z)x(3yz)(3yz)x(x+z)(x+z)2=(x+z)(0)(3yz)(1)(x+z)2=3yz(x+z)2

The first partial derivative of function f(x,y,z)=3yzx+z with respect to x,

xf(x,y,z)=3yz(x+z)2

Now again differentiate with respect to x by holding y and z constant,

22xf(x,y,z)=x[3yz(x+z)2]=(x+z)2x(3yz)(3yz)x(x+z)2(x+z)4=(x+z)2(0)(6yz)(x+z)(x+z)4=6yz(x+z)3

Consider the function,

f(x,y,z)=3yzx+z

The first partial derivative of function f(x,y) with respect to y by holding x and z constant,

yf(x,y,z)=y(3yzx+z)=(3zx+z)y(y)=3zx+z

The first partial derivative of function f(x,y,z)=3yzx+z with respect to y,

f(x,y,z)=3zx+z

Now again differentiate with respect to y by holding x and z constant,

22yf(x,y,z)=y[3z(x+z)]=(x+z)y(3z)(3z)y(x+z)(x+z)=(x+z)(0)+(3z)(1)(x+z)=0

Consider the function,

f(x,y,z)=3yzx+z

Differentiate f(x,y,z) with respect to z by holding x and y constant,

zf(x,y,z)=z(3yzx+z)=(x+z)z(3yz)(3yz)z(x+z)(x+z)2=(x+z)(3y)(3yz)(1)(x+z)2=3xy(x+z)2

The first partial derivative of function f(x,y,z)=3yzx+z with respect to z,

zf(x,y,z)=3xy(x+z)2

Now again differentiate with respect to z by holding x and y constant,

22zf(x,y,z)=z[3xy(x+z)2]=(x+z)2z(3zy)(3zy)z(x+z)2(x+z)4=(x+z)2(3y)(6zy)(x+z)(x+z)4=6xy(x+z)3

Consider the partial derivative f(x,y,z) with respect to z,

zf(x,y,z)=3xy(x+z)2

Now differentiate with respect to x by holding y and z constant,

2xzf(x,y,z)=x3xy(x+z)2=(x+z)2x(3xy)(3xy)x(x+z)2(x+z)4=(x+z)2(3y)(6xy)(x+z)(x+z)4=3y(xz)(x+z)3

The first partial derivative of function f(x,y,z)=3yzx+z with respect to x,

xf(x,y,z)=3yz(x+z)2

Now differentiate the above equation with respect to z by holding x and z constant

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