Finding Second Partial Derivatives In Exercises 55–60, find the second partial derivatives.

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

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

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)]n−1⋅∂∂xf(x,y)

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

∂∂y[f(x,y)]n=n[f(x,y)]n−1⋅∂∂yf(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,

∂2∂2xf(x,y,z)=∂∂x[−3yz(x+z)2]=−(x+z)2∂∂x(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,

∂2∂2yf(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,

∂2∂2zf(x,y,z)=∂∂z[3xy(x+z)2]=−(x+z)2∂∂z(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,

∂2∂x∂zf(x,y,z)=∂∂x3xy(x+z)2=−(x+z)2∂∂x(3xy)−(3xy)∂∂x(x+z)2(x+z)4=−(x+z)2(3y)−(6xy)(x+z)(x+z)4=−3y(x−z)(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