Chapter 14, Problem 5P

Find the gradient vector and Hessian matrix for each of the following functions:

(a) $f\left(x,y\right)=2x{y}^2+3e^{xy}$

(b) $f\left(x,y,z\right)=x^2+y^2+2z^2$

(c) $f\left(x,y\right)=\text{In}\left(x^2+2xy+3y^2\right)$

To determine

To calculate: The gradient vector and Hessian matrix for the function $f\left(x,y\right)=2x{y}^2+3e^{xy}$.

Answer to Problem 5P

Solution:

The gradient vector for the function $f\left(x,y\right)=2x{y}^2+3e^{xy}$ is $\left\{\begin{array}{l}2y^2+3y{e}^{xy}\\ 4xy+3x{e}^{xy}\end{array}\right\}$.

The Hessian matrix for the function $f\left(x,y\right)=2x{y}^2+3e^{xy}$ is,

$\left[\begin{array}{cc}3y^2{e}^{xy}& 4y+3yx{e}^{xy}+3e^{xy}\\ 4y+3yx{e}^{xy}+3e^{xy}& 4x+3x^2{e}^{xy}\end{array}\right]$

Explanation of Solution

Given:

The function $f\left(x,y\right)=2x{y}^2+3e^{xy}$.

Formula used:

The gradient vector for the function $f\left(x,y\right)$ is,

$\nabla f\left(x,y\right)=\left\{\begin{array}{l}\frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\end{array}\right\}$

The Hessian matrix for the function $f\left(x,y\right)$ is given by,

$H=\left[\begin{array}{cc}\frac{{\partial }^{2}f}{\partial x^2}& \frac{{\partial }^{2}f}{\partial x\partial y}\\ \frac{{\partial }^{2}f}{\partial y\partial x}& \frac{{\partial }^{2}f}{\partial y^2}\end{array}\right]$

Calculation:

Consider the function,

$f\left(x,y\right)=2x{y}^2+3e^{xy}$

Partial differentiate the above function with respect to x,

$\begin{array}{c}\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left(2x{y}^2+3e^{xy}\right)\\ =\frac{\partial }{\partial x}\left(2x{y}^2\right)+\frac{\partial }{\partial x}\left(3e^{xy}\right)\\ =2y^2+3y{e}^{xy}\end{array}$

Again, partial differentiate the above equation with x,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial x^2}=\frac{\partial }{\partial x}\left(2y^2+3e^{xy}\right)\\ =\frac{\partial }{\partial x}\left(2y^2\right)+\frac{\partial }{\partial x}\left(3y{e}^{xy}\right)\\ =0+3y^2{e}^{xy}\\ =3y^2{e}^{xy}\end{array}$

Partial differentiate the $\frac{\partial f}{\partial x}=2y^2+3y{e}^{xy}$ with respect to y,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{\partial }{\partial y}\left(2y^2+3y{e}^{xy}\right)\\ =\frac{\partial }{\partial y}\left(2y^2\right)+\frac{\partial }{\partial y}\left(3y{e}^{xy}\right)\\ =4y+3yx{e}^{xy}+3e^{xy}\end{array}$

Now, partial differentiate the function $f\left(x,y\right)=2x{y}^2+3e^{xy}$ with respect to y,

$\begin{array}{c}\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left(2x{y}^2+3e^{xy}\right)\\ =\frac{\partial }{\partial y}\left(2x{y}^2\right)+\frac{\partial }{\partial y}\left(3e^{xy}\right)\\ =2x\times 2y+3x{e}^{xy}\\ =4xy+3x{e}^{xy}\end{array}$

Again, partial differentiate the above equation with y,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial y^2}=\frac{\partial }{\partial y}\left(4xy+3x{e}^{xy}\right)\\ =\frac{\partial }{\partial y}\left(4xy\right)+\frac{\partial }{\partial y}\left(3x{e}^{xy}\right)\\ =4x+3x^2{e}^{xy}\end{array}$

Partial differentiate the $\frac{\partial f}{\partial y}=4xy+3x{e}^{xy}$ with respect to x,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}\left(4xy+3x{e}^{xy}\right)\\ =\frac{\partial }{\partial x}\left(4xy\right)+\frac{\partial }{\partial x}\left(3x{e}^{xy}\right)\\ =4y+3yx{e}^{xy}+3e^{xy}\end{array}$

Therefore, the gradient vector for the function is,

$\nabla f\left(x,y\right)=\left\{\begin{array}{l}2y^2+3y{e}^{xy}\\ 4xy+3x{e}^{xy}\end{array}\right\}$

And, the Hessian matrix for the function is,

$H=\left[\begin{array}{cc}3y^2{e}^{xy}& 4y+3yx{e}^{xy}+3e^{xy}\\ 4y+3yx{e}^{xy}+3e^{xy}& 4x+3x^2{e}^{xy}\end{array}\right]$

To determine

To calculate: The gradient vector and Hessian matrix for the function $f\left(x,y,z\right)=x^2+y^2+2z^2$.

Answer to Problem 5P

Solution:

The gradient vector for the function $f\left(x,y,z\right)=x^2+y^2+2z^2$ is $\left\{\begin{array}{l}2x\\ 2y\\ 4z\end{array}\right\}$.

The Hessian matrix for the function $f\left(x,y,z\right)=x^2+y^2+2z^2$ is $\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 4\end{array}\right]$.

Explanation of Solution

Given:

The function $f\left(x,y,z\right)=x^2+y^2+2z^2$.

Formula used:

The gradient vector for the function $f\left(x,y,z\right)$ is,

$\nabla f\left(x,y,z\right)=\left\{\begin{array}{l}\frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\\ \frac{\partial f}{\partial z}\end{array}\right\}$

The Hessian matrix for the function $f\left(x,y,z\right)$ is given by,

$H=\left[\begin{array}{ccc}\frac{{\partial }^{2}f}{\partial x^2}& \frac{{\partial }^{2}f}{\partial x\partial y}& \frac{{\partial }^{2}f}{\partial x\partial z}\\ \frac{{\partial }^{2}f}{\partial y\partial x}& \frac{{\partial }^{2}f}{\partial y^2}& \frac{{\partial }^{2}f}{\partial y\partial z}\\ \frac{{\partial }^{2}f}{\partial z\partial x}& \frac{{\partial }^{2}f}{\partial z\partial y}& \frac{{\partial }^{2}f}{\partial z^2}\end{array}\right]$

Calculation:

Consider the function,

$f\left(x,y,z\right)=x^2+y^2+2z^2$

Partial differentiate the above function with respect to x,

$\begin{array}{c}\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left(x^2+y^2+2z^2\right)\\ =\frac{\partial }{\partial x}\left(x^2\right)+\frac{\partial }{\partial x}\left(y^2\right)+\frac{\partial }{\partial x}\left(2z^2\right)\\ =2x+0+0\\ =2x\end{array}$

Again, partial differentiate the above equation with x,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial x^2}=\frac{\partial }{\partial x}\left(2x\right)\\ =2\end{array}$

Partial differentiate the $\frac{\partial f}{\partial x}=2x$ with respect to y,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{\partial }{\partial y}\left(2x\right)\\ =0\end{array}$

Partial differentiate the $\frac{\partial f}{\partial x}=2x$ with respect to z,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial z\partial x}=\frac{\partial }{\partial z}\left(2x\right)\\ =0\end{array}$

Now, partial differentiate the function $f\left(x,y,z\right)=x^2+y^2+2z^2$ with respect to y,

$\begin{array}{c}\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left(x^2+y^2+2z^2\right)\\ =\frac{\partial }{\partial y}\left(x^2\right)+\frac{\partial }{\partial y}\left(y^2\right)+\frac{\partial }{\partial y}\left(2z^2\right)\\ =0+2y+0\\ =2y\end{array}$

Again, partial differentiate the above equation with y,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial y^2}=\frac{\partial }{\partial y}\left(2y\right)\\ =2\end{array}$

Partial differentiate the $\frac{\partial f}{\partial y}=2y$ with respect to x,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}\left(2y\right)\\ =0\end{array}$

Partial differentiate the $\frac{\partial f}{\partial y}=2y$ with respect to z,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial z\partial y}=\frac{\partial }{\partial z}\left(2y\right)\\ =0\end{array}$

Now, partial differentiate the function $f\left(x,y,z\right)=x^2+y^2+2z^2$ with respect to z,

$\begin{array}{c}\frac{\partial f}{\partial z}=\frac{\partial }{\partial z}\left(x^2+y^2+2z^2\right)\\ =\frac{\partial }{\partial z}\left(x^2\right)+\frac{\partial }{\partial z}\left(y^2\right)+\frac{\partial }{\partial z}\left(2z^2\right)\\ =0+0+2\times 2z\\ =4z\end{array}$

Again, partial differentiate the above equation with z,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial z^2}=\frac{\partial }{\partial z}\left(4z\right)\\ =4\end{array}$

Partial differentiate the $\frac{\partial f}{\partial y}=4z$ with respect to x,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial x\partial z}=\frac{\partial }{\partial x}\left(4z\right)\\ =0\end{array}$

Partial differentiate the $\frac{\partial f}{\partial y}=4z$ with respect to y,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial y\partial z}=\frac{\partial }{\partial y}\left(4z\right)\\ =0\end{array}$

Therefore, the gradient vector for the function is,

$\nabla f\left(x,y\right)=\left\{\begin{array}{l}2x\\ 2y\\ 4z\end{array}\right\}$

And, the Hessian matrix for the function is,

$H=\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 4\end{array}\right]$

To determine

To calculate: The gradient vector and Hessian matrix for the function $f\left(x,y\right)=\ln \left(x^2+2xy+3y^2\right)$.

Answer to Problem 5P

Solution:

The gradient vector for the function $f\left(x,y\right)=\ln \left(x^2+2xy+3y^2\right)$ is $\left\{\begin{array}{l}2y^2+3y{e}^{xy}\\ 4xy+3x{e}^{xy}\end{array}\right\}$.

The Hessian matrix for the function $f\left(x,y\right)=\ln \left(x^2+2xy+3y^2\right)$ is,

$\frac{\left[\begin{array}{cc}-2x^2-4xy+2y^2& -2x^2-12xy-6y^2\\ -2x^2-12xy-6y^2& 2x^2-12xy-18y^2\end{array}\right]}{{\left(x^2+2xy+3y^2\right)}^2}$

Explanation of Solution

Given:

The function $f\left(x,y\right)=\ln \left(x^2+2xy+3y^2\right)$.

Formula used:

The gradient vector for the function $f\left(x,y\right)$ is,

$\nabla f\left(x,y\right)=\left\{\begin{array}{l}\frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\end{array}\right\}$

The Hessian matrix for the function $f\left(x,y\right)$ is given by,

$H=\left[\begin{array}{cc}\frac{{\partial }^{2}f}{\partial x^2}& \frac{{\partial }^{2}f}{\partial x\partial y}\\ \frac{{\partial }^{2}f}{\partial y\partial x}& \frac{{\partial }^{2}f}{\partial y^2}\end{array}\right]$

Calculation:

Consider the function,

$f\left(x,y\right)=\ln \left(x^2+2xy+3y^2\right)$

Partial differentiate the above function with respect to x,

$\begin{array}{c}\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left\{\ln \left(x^2+2xy+3y^2\right)\right\}\\ =\frac{1}{x^2+2xy+3y^2}\cdot \frac{\partial }{\partial x}\left(x^2+2xy+3y^2\right)\\ =\frac{1}{x^2+2xy+3y^2}\left(2x+2y\right)\\ =\frac{2x+2y}{x^2+2xy+3y^2}\end{array}$

Again, partial differentiate the above equation with x,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{2x+2y}{x^2+2xy+3y^2}\right)\\ =\frac{\left(x^2+2xy+3y^2\right)\frac{\partial }{\partial x}\left(2x+2y\right)-\left(2x+2y\right)\frac{\partial }{\partial x}\left(x^2+2xy+3y^2\right)}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{\left(x^2+2xy+3y^2\right)\times 2-\left(2x+2y\right)\left(2x+2y\right)}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}$

Simplify furthermore,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial x^2}=\frac{\left(2x^2+4xy+6y^2\right)-\left(4x^2+8xy+4y^2\right)}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{2x^2+4xy+6y^2-4x^2-8xy-4y^2}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{-2x^2-4xy+2y^2}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}$

Partial differentiate the $\frac{\partial f}{\partial x}=\frac{2x+2y}{x^2+2xy+3y^2}$ with respect to y,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{\partial }{\partial y}\left(\frac{2x+2y}{x^2+2xy+3y^2}\right)\\ =\frac{\left(x^2+2xy+3y^2\right)\frac{\partial }{\partial y}\left(2x+2y\right)-\left(2x+2y\right)\frac{\partial }{\partial y}\left(x^2+2xy+3y^2\right)}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{\left(x^2+2xy+3y^2\right)\times 2-\left(2x+2y\right)\left(2x+6y\right)}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}$

Simplify furthermore,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{\left(2x^2+4xy+6y^2\right)-\left(4x^2+16xy+12y^2\right)}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{2x^2+4xy+6y^2-4x^2-16xy-12y^2}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{-2x^2-12xy-6y^2}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}$

Now, partial differentiate the function $f\left(x,y\right)=\ln \left(x^2+2xy+3y^2\right)$ with respect to y,

$\begin{array}{c}\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left\{\ln \left(x^2+2xy+3y^2\right)\right\}\\ =\frac{1}{x^2+2xy+3y^2}\cdot \frac{\partial }{\partial y}\left(x^2+2xy+3y^2\right)\\ =\frac{1}{x^2+2xy+3y^2}\times \left(2x+6y\right)\\ =\frac{2x+6y}{x^2+2xy+3y^2}\end{array}$

Again, partial differentiate the above equation with y,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial y^2}=\frac{\partial }{\partial y}\left(\frac{2x+6y}{x^2+2xy+3y^2}\right)\\ =\frac{\left(x^2+2xy+3y^2\right)\frac{\partial }{\partial y}\left(2x+6y\right)-\left(2x+6y\right)\frac{\partial }{\partial y}\left(x^2+2xy+3y^2\right)}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{\left(x^2+2xy+3y^2\right)\times 6-\left(2x+6y\right)\left(2x+6y\right)}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{\left(6x^2+12xy+18y^2\right)-\left(4x^2+24xy+36y^2\right)}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}$

Simplify furthermore,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial y^2}=\frac{6x^2+12xy+18y^2-4x^2-24xy-36y^2}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{2x^2-12xy-18y^2}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}$

Partial differentiate the $\frac{\partial f}{\partial y}=\frac{2x+6y}{x^2+2xy+3y^2}$ with respect to x,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{2x+6y}{x^2+2xy+3y^2}\right)\\ =\frac{\left(x^2+2xy+3y^2\right)\frac{\partial }{\partial x}\left(2x+6y\right)-\left(2x+6y\right)\frac{\partial }{\partial x}\left(x^2+2xy+3y^2\right)}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{\left(x^2+2xy+3y^2\right)\times 2-\left(2x+6y\right)\left(2x+2y\right)}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{\left(2x^2+4xy+6y^2\right)-\left(4x^2+16xy+12y^2\right)}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}$

Simplify furthermore,

$\begin{array}{c}\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{2x^2+4xy+6y^2-4x^2-16xy-12y^2}{{\left(x^2+2xy+3y^2\right)}^2}\\ =\frac{-2x^2-12xy-6y^2}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}$

Therefore, the gradient vector for the function is,

$\nabla f\left(x,y\right)=\left\{\begin{array}{l}\frac{2x+2y}{x^2+2xy+3y^2}\\ \frac{2x+6y}{x^2+2xy+3y^2}\end{array}\right\}$

And, the Hessian matrix for the function is,

$\begin{array}{c}H=\left[\begin{array}{cc}\frac{-2x^2-4xy+2y^2}{{\left(x^2+2xy+3y^2\right)}^2}& \frac{-2x^2-12xy-6y^2}{{\left(x^2+2xy+3y^2\right)}^2}\\ \frac{-2x^2-12xy-6y^2}{{\left(x^2+2xy+3y^2\right)}^2}& \frac{2x^2-12xy-18y^2}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}\right]\\ =\frac{\left[\begin{array}{cc}-2x^2-4xy+2y^2& -2x^2-12xy-6y^2\\ -2x^2-12xy-6y^2& 2x^2-12xy-18y^2\end{array}\right]}{{\left(x^2+2xy+3y^2\right)}^2}\end{array}$