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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 57-60, find the nine second partial derivatives. See Example 7.

w = x y x + y + z

To determine

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

Explanation

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=2fx2 [double partial derivative with respect to x, keeping y & z are constant].

(ii)fxy=2fxy [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=2fxz [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=2fyx [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=2fy2 [fyy is double partial derivative with respect to y keeping x & z are constant].

(vi)fyz=2fyz [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=2fzx [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=2zy [First differentiate with respect to y keeping x & z constant then double differentiating with respect to z keeping x & y constant].

(ix)fzz=2fz2 [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)yxy(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)xxy(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=0xy(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)xy(y+z)(x+y+z)2x[(x+y+z)2]2=(x+y+z)2×0y(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)2x(x+z)xx(x+z)(x+y+z)2x(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+z22x22xz(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)2x(x+y+z)4=y(x+y+z)2+2xy(x+y+z)(x+y+z)4=2xyy(x+y+z)(x+y+z)3

Solve furthermore,

fxz=2xyyxy2yz(x+y+z)3=xyy2yz(x+y+z)3

Now,

fyx=y((y+z)y(x+y+z)2)

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