Finding Second Partial Derivatives In Exercises 57-60, find the nine second partial derivatives. See Example 7.

w = x y x + y + z

To calculate: The nine second partial derivatives of the function w=xy(x+y+z).

Given Information:

The function with these variables is w=xy(x+y+z).

Formula used:

For the given function of these variables

w=f(x,y,z)

(i)fxx=∂2f∂x2 [double partial derivative with respect to x, keeping y & z are constant].

(ii)fxy=∂2f∂x∂y [fxy is double partial derivative, getting by first differentiating with y and keeping x & z constant then double differentiating with respect to x, keeping y & z constant].

(iii)fxz=∂2f∂x∂z [fxz is double partial derivative getting by first differentiating with respect to z keeping y & x constant then double differentiating with respect to x keeping y & z constant].

(iv)fyx=∂2f∂y∂x [fyx is double partial derivative getting by first differentiating with respect to x keeping y & z constant then double differentiating with respect to y keeping x & z constant].

(v)fyy=∂2f∂y2 [fyy is double partial derivative with respect to y keeping x & z are constant].

(vi)fyz=∂2f∂y∂z [fyz is double partial derivative getting by first differentiating with respect to z keeping y & x constant then double differentiating with respect to y keeping x & z constant].

(vii)fzx=∂2f∂z∂x [fxz is double partial derivative getting by first differentiating with respect to x keeping y & z constant then double differentiating with respect to z keeping x & y constant].

(viii)fzy=∂2∂z∂y [First differentiate with respect to y keeping x & z constant then double differentiating with respect to z keeping x & y constant].

(ix)fzz=∂2f∂z2 [Double differentiation with respect to z keeping x & y constant].

Calculation:

Consider the primary equation,

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

Partially differentiating f with respect to x keeping y and z constant for fx,

fx=∂∂x[xyx+y+z]=(x+y+z)∂(xy)∂x−(xy)∂(x+y+z)∂x(x+y+z)2=(x+y+z)y−xy(x+y+z)2=(y+z)y(x+y+z)2

And,

fy=∂∂y[xyx+y+z]=(x+y+z)∂(xy)∂y−(xy)∂(x+y+z)∂y(x+y+z)2=(x+y+z)x−xy(x+y+z)2=(x+z)x(x+y+z)2

And,

fz=∂∂z[xyx+y+z]=(x+y+z)∂(xy)∂z−(xy)∂(x+y+z)∂z(x+y+z)2=0−xy(x+y+z)2=−xy(x+y+z)2

Now for double partial differentiation,

fxx=∂∂x(y(y+z)(x+y+z)2)=(x+y+z)2∂(y2+yz)∂x−y(y+z)∂(x+y+z)2∂x[(x+y+z)2]2=(x+y+z)2×0−y(y+z)×2(x+y+z)×1(x+y+z)4=−2y(y+z)(x+y+z)3

And,

fxy=∂∂x(x(x+z)(x+y+z)2)=(x+y+z)2∂x(x+z)∂x−x(x+z)∂(x+y+z)2∂x(x+y+z)4=(x+y+z)2(2x+z)−x(x+z)2(x+y+z)(x+y+z)4=2x2+2xy+2xz+zx+zy+z2−2x2−2xz(x+y+z)3

Solve furthermore,

fxy=2xy+zx+zy+z2(x+y+z)3

And,

fxz=∂∂x(−xy(x+y+z)2)=(x+y+z)2∂(−xy)∂x−(−xy)∂(x+y+z)2∂x(x+y+z)4=−y(x+y+z)2+2xy(x+y+z)(x+y+z)4=2xy−y(x+y+z)(x+y+z)3

Solve furthermore,

fxz=2xy−yx−y2−yz(x+y+z)3=xy−y2−yz(x+y+z)3

Now,

fyx=∂∂y((y+z)y(x+y+z)2)