   Chapter 14, Problem 2T Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Solutions

Chapter
Section Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

Find all first and second partial derivatives of z   = f ( x , y ) =   5 x   −   9 y 2 + 2 ( x y   + l ) 5

To determine

To calculate: All the first and second partial derivatives of z=f(x,y)=5x9y2+2(xy+1)5.

Explanation

Given Information:

The provided function is z=f(x,y)=5x9y2+2(xy+1)5.

Formula used:

For a function f(x,y), the partial derivative of f with respect to x is calculated by taking the derivative of f(x,y) with respect to x and keeping the other variable y constant. The partial derivative of f with respect to x is denoted by fx and the partial derivative of f with respect to y is denoted by fy.

For a function z(x,y), the second partial derivative,

(1) When both derivatives are taken with respect to x is zxx=2zx2=x(zx).

(2) When both derivatives are taken with respect to y is zyy=2zy2=y(zy).

(3) When first derivative is taken with respect to x and second derivative is taken with respect to y is zxy=2zyx=y(zx).

(4) When first derivative is taken with respect to y and second derivative is taken with respect to x is zyx=2zxy=x(zy).

Power of x rule for a real number n is such that, if f(x)=xn then f(x)=nxn1.

Constant function rule for a constant c is such that, if f(x)=c then f(x)=0.

Chain rule for function f(x)=u(v(x)) is f(x)=u(v(x))v(x).

Coefficient rule for a constant c is such that, if f(x)=cu(x), where u(x) is a differentiable function of x, then f(x)=cu(x).

Calculation:

Consider the function, z=f(x,y)=5x9y2+2(xy+1)5.

Recall that, for a function f(x,y), the partial derivative of f with respect to x is calculated by taking the derivative of f(x,y) with respect to x and keeping the other variable y constant.

Use the power of x rule for derivatives, the constant function rule, the chain rule and the coefficient rule,

zx=zx=5+2(5(xy+1)4)(y)=5+10y(xy+1)4

And,

zy=zy=9(2y)+2(5(xy+1)4)(x)=18y+10x(xy+1)4

Hence, the first partial derivatives of the function z=f(x,y)=5x9y2+2(xy+1)5 are zx=5+10y(xy+1)4 and zy=18y+10x(xy+1)4.

Recall that, for a function z(x,y), the second partial derivative, when both derivatives are taken with respect to x is zxx=2zx2=x(zx).

Use the power of x rule for derivatives, the constant function rule and the coefficient rule.

zxx=2zx2=x(zx)=x(5+2(5(xy+1)4)(y))=x(5+10y(xy+1)4)

Simplify it further,

zxx=10y(4(xy+1)3(y))=40y2(xy+1)3

Recall that, for a function z(x,y), the second partial derivative, when first derivative is taken with respect to x and second derivative is taken with respect to y is zxy=2zyx=y(zx)

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