In Exercises 9-14, find all first and second partial derivatives.

z = x e x y

To calculate: The first and second partial derivatives of function z=xexy

Given Information:

Given function with two variable x, y is z=xexy

Formula used:

The first partial derivative of z=z(x,y) are denoted by

Partial derivative with respect to x, keeping y constant.

∂z∂x=zx(x,y)∂z∂x=∂∂x[z(x,y)]

Partial derivative with respect to y, keeping x constant

∂z∂y=zy(x,y)∂z∂y=∂∂y[z(x,y)]

The value of partial derivative at the point (a,b) is denoted by

∂z∂x|(a,b)= zx(a,b) and ∂z∂y|(a,b)=zy(a,b)

Finding second partial derivative of function z(x,y)

∂2z∂x2= zxx∂2z∂x2 ​= ∂∂x[∂z(x,y)∂x]

Second partial derivative of ∂z(x,y)∂x with respect to x, keeping y constant.

∂2z∂y2= zyy(x,y) ∂2z∂y2​= ∂∂y[∂z(x,y)∂y]

second partial derivatives of ∂z(x,y)∂y with respect to y, keeping x constant.

∂2z∂y∂x=zyx(x,y) ,∂2z∂y∂x​ = ∂∂y[∂z(x,y)∂x]

second partial derivatives of ∂z(x,y)∂y with respect to x, keeping y constant.

Product rule,

If f(x,y)=f1(x,y)×f2(x,y) are two function of x, y then,

∂f(x,y)∂x=f1(x,y)×∂f2(x,y)∂x+f2(x,y)×∂f1(x,y)∂x

Calculation:

Consider the given equation:

z(x,y)=xexy

Differentiating primary equation z(x,y) with respect to x, keeping y as constant

zx(x,y)=∂∂x(x⋅exy)

Applying product rule to differentiate above function

Product rule,

If f(x,y)=f1(x,y)×f2(x,y) are two function of x, y then,

∂f(x,y)∂x=f1(x,y)×∂f2(x,y)∂x+f2(x,y)×∂f1(x,y)∂x

Then,

zx(x,y)=∂∂xz(x,y)zx(x,y)=x⋅∂exy∂x+exy∂x∂xzx(x,y)=x⋅(exy)⋅(y)+exy∂∂xz(x,y)=exy[xy+1]

Differentiating given equation with respect to y, keeping x as constant

zy(x,y)=∂∂yz(x,y)zy(x,y)=∂∂y[x⋅exy]zy(x,y)=x⋅∂∂y(exy)zy(x,y)=x⋅(exy)(x)

Further simplify,

zy(x,y)=x2exy

Hence, first partial derivatives are:

zx(x,y)=exy[xy+1]zy(x,y)=x2exy

Differentiating zx(x,y)=exy⋅[xy+1] with respect to x to find double partial derivative

keeping y as constant

zxx(x,y)=∂∂x[exy(xy+1)]

Applying product rule to differentiate above function

Product rule,

If f(x,y)=f1(x,y)×f2(x,y) are two function of x, y then,

∂f(x,y)∂x=f