   Chapter 7.4, Problem 58E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding Second Partial Derivatives InExercises 57-60, find the nine second partial derivatives. See Example 7. w = x 2 y 3 + 2 x y z − 3 y z

To determine

To calculate: The nine second partial derivatives of the function w=x2y3+2xyz3yz.

Explanation

Given Information:

The function with the variables is w=x2y3+2xyz3yz

Formula used:

For the given function of 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 given equation

w=x2y3+2xyz3yz

Consider, w=f(x,y,z)=x2y3+2xyz3yz

Now, partially differentiating f with respect to x keeping y & z constant for fx,

fx(x,y,z)=x[xy23+2xyz3yz]fx=2xy3+2yz0=2xy3+2yz

Differentiating primary equation with respect to y keeping x & z constant to get fy

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