In order to find the divergence, we need to remember the formula.

Divergence Formula:

\(\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\), where \(\displaystyle P\), \(\displaystyle Q\), and \(\displaystyle R\) correspond to the components of a given vector field \(\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}\).

Now lets apply this to our situation.

\(\displaystyle div\vec{F}=\frac{\partial}{\partial x}\Big(yz\ln(x)\Big)+\frac{\partial}{\partial y}\Big(2xyz\Big)+\frac{\partial}{\partial z}\Big(z\Big)\)

\(\displaystyle div\vec{F}=\frac{yz}{x}+2xz+1\)

In order to find the divergence, we need to remember the formula.

Divergence Formula:

\(\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\), where \(\displaystyle P\), \(\displaystyle Q\), and \(\displaystyle R\) correspond to the components of a given vector field \(\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}\).

Now lets apply this to our situation.

\(\displaystyle div\vec{F}=\frac{\partial}{\partial x}\Big(x^2z\ln(y)\Big)+\frac{\partial}{\partial y}\Big(x2^{y^2}z\Big)+\frac{\partial}{\partial z}\Big(x^x\Big)\)

\(\displaystyle div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}+0\)

\(\displaystyle div\vec{F}=2xz\ln(y)+2\log(2)xyz2^{y^2}\)

In order to find the divergence, we need to remember the formula.

Divergence Formula:

\(\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\), where \(\displaystyle P\), \(\displaystyle Q\), and \(\displaystyle R\) correspond to the components of a given vector field \(\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}\).

Now lets apply this to our situation.

\(\displaystyle div\vec{F}=\frac{\partial}{\partial x}\Big(xy\Big)+\frac{\partial}{\partial y}\Big(xz\Big)+\frac{\partial}{\partial z}\Big(xyz\Big)\)

\(\displaystyle div \vec{F}=y+0+xy\)

\(\displaystyle div \vec{F}=y+xy=y(1+x)\)

In order to find the divergence, we need to remember the formula.

Divergence Formula:

\(\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\), where \(\displaystyle P\), \(\displaystyle Q\), and \(\displaystyle R\) correspond to the components of a given vector field \(\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}\).

Now lets apply this to our situation.

\(\displaystyle div\vec{F}=\frac{\partial}{\partial x}\Big(xyz\Big)+\frac{\partial}{\partial y}\Big(x^2z^{-4}\Big)+\frac{\partial}{\partial z}\Big(y^{-10}z^3\Big)\)

\(\displaystyle div\vec{F}={yz}+0+3y^{-10}z^2=yz+\frac{3z^2}{y^{10}}\)

All we need to do is calculate the partial derivatives and add them together.

\(\displaystyle div\vec{F}=\frac{\partial}{\partial x}(10x^2\ln(yz))+\frac{\partial}{\partial y}(xyz)+\frac{\partial}{\partial z}(xz\ln(y))\)

\(\displaystyle div\vec{F}=20x\ln(yz)+xz+x\ln(y)\)

Given a vector field 

\(\displaystyle \vec{F}=P\vec{i}+Q\vec{j}+R\vec{k}\)

we find its divergence by taking the dot product with the gradient operator:

\(\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\)

We know that \(\displaystyle P=yz, Q=xz, R=xy\), so we have 

\(\displaystyle div\vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=0+0+0=0\)

We know,

\(\displaystyle div F(x,y,z)=\frac{\partial{F}}{\partial{x}}+\frac{\partial{F}}{\partial{y}}+\frac{\partial{F}}{\partial{z}}\)

Use this to obtain the correct answer

\(\displaystyle div(\vec{F})=\frac{\partial{F}}{\partial{x}}+\frac{\partial{F}}{\partial{y}}+\frac{\partial{F}}{\partial{z}}\)

using this formula we have

\(\displaystyle div(\vec{F})=2+1+2=5\)

The divergence of a vector is given by

\(\displaystyle div \vec{F}=\bigtriangledown \cdot \vec{F}\)

where \(\displaystyle \bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle\)

So, we take the partial derivative of each component of our vector with respect to x, y, and z respectively and add them together:

\(\displaystyle f_x(2x)=2\)

\(\displaystyle f_y(3xy)=3x\)

\(\displaystyle f_z(z^2\cos(z))=2z\cos(z)-z^2\sin(z)\)

The derivatives were found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)

The divergence of a curve is given by

\(\displaystyle div \vec{F}=\bigtriangledown \cdot \vec{F}\)

where \(\displaystyle \bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle\)

Taking the dot product of the gradient and the curve, we end up summing the respective partial derivatives (for example, the x coordinate's partial derivative with respect to x is found).

The partial derivatives are:

\(\displaystyle f_x=0\)

\(\displaystyle f_y=2ye^x\)

\(\displaystyle f_z=2z\)

The following rules were used to find the derivatives:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)