Step 1

Consider the provided function,

\(\displaystyle{S}{\left({y},{z}\right)}={z}^{{2}}{\tan{{y}}}{z}\)

Find the first partial derivatives.

So, we find the expression for \(\displaystyle\frac{{\partial{S}}}{{\partial{y}}}{\quad\text{and}\quad}\frac{{\partial{S}}}{{\partial{z}}}\).

First we find the expression for \(\displaystyle\frac{{\partial{S}}}{{\partial{y}}}\).

\(\displaystyle\frac{{\partial{S}}}{{\partial{y}}}=\frac{{\partial}}{{\partial{y}}}{\left({z}^{{2}}{\tan{{y}}}{z}\right)}\)

Step 2

Simplifying further,

\(\displaystyle\frac{{\partial{S}}}{{\partial{y}}}=\frac{{\partial}}{{\partial{y}}}{\left({z}^{{2}}{\tan{{y}}}{z}\right)}\)

\(\displaystyle={z}^{{2}}\frac{\partial}{{\partial{y}}}{\left({\tan{{y}}}{z}\right)}\)

\(\displaystyle={z}^{{2}}{{\sec}^{{2}}{y}}{z}\frac{\partial}{{\partial{y}}}{\left({y}{z}\right)}\)

\(\displaystyle={z}^{{2}}{{\sec}^{{2}}{y}}{z}\cdot{z}\)

\(\displaystyle={z}^{{3}}{{\sec}^{{2}}{y}}{z}\)

Now, find the expression for \(\displaystyle\frac{{\partial{S}}}{{\partial{z}}}\).

\(\displaystyle\frac{{\partial{S}}}{{\partial{z}}}=\frac{{\partial}}{{\partial{z}}}{\left({z}^{{2}}{\tan{{y}}}{z}\right)}\)

Step 3

Simplifying further,

\(\displaystyle\frac{{\partial{S}}}{{\partial{z}}}=\frac{{\partial}}{{\partial{z}}}{\left({z}^{{2}}\right)}{\tan{{\left({y}{z}\right)}}}+\frac{\partial}{{\partial{z}}}{\left({\tan{{\left({y}{z}\right)}}}\right)}{z}^{{2}}\)

\(\displaystyle={2}{z}\cdot{\tan{{\left({y}{z}\right)}}}+{z}^{{2}}\cdot{{\sec}^{{2}}{\left({y}{z}\right)}}{y}\)

\(\displaystyle={2}{z}{\tan{{\left({y}{z}\right)}}}+{{\sec}^{{2}}{\left({y}{z}\right)}}{y}{z}^{{2}}\)

Hence.

Consider the provided function,

\(\displaystyle{S}{\left({y},{z}\right)}={z}^{{2}}{\tan{{y}}}{z}\)

Find the first partial derivatives.

So, we find the expression for \(\displaystyle\frac{{\partial{S}}}{{\partial{y}}}{\quad\text{and}\quad}\frac{{\partial{S}}}{{\partial{z}}}\).

First we find the expression for \(\displaystyle\frac{{\partial{S}}}{{\partial{y}}}\).

\(\displaystyle\frac{{\partial{S}}}{{\partial{y}}}=\frac{{\partial}}{{\partial{y}}}{\left({z}^{{2}}{\tan{{y}}}{z}\right)}\)

Step 2

Simplifying further,

\(\displaystyle\frac{{\partial{S}}}{{\partial{y}}}=\frac{{\partial}}{{\partial{y}}}{\left({z}^{{2}}{\tan{{y}}}{z}\right)}\)

\(\displaystyle={z}^{{2}}\frac{\partial}{{\partial{y}}}{\left({\tan{{y}}}{z}\right)}\)

\(\displaystyle={z}^{{2}}{{\sec}^{{2}}{y}}{z}\frac{\partial}{{\partial{y}}}{\left({y}{z}\right)}\)

\(\displaystyle={z}^{{2}}{{\sec}^{{2}}{y}}{z}\cdot{z}\)

\(\displaystyle={z}^{{3}}{{\sec}^{{2}}{y}}{z}\)

Now, find the expression for \(\displaystyle\frac{{\partial{S}}}{{\partial{z}}}\).

\(\displaystyle\frac{{\partial{S}}}{{\partial{z}}}=\frac{{\partial}}{{\partial{z}}}{\left({z}^{{2}}{\tan{{y}}}{z}\right)}\)

Step 3

Simplifying further,

\(\displaystyle\frac{{\partial{S}}}{{\partial{z}}}=\frac{{\partial}}{{\partial{z}}}{\left({z}^{{2}}\right)}{\tan{{\left({y}{z}\right)}}}+\frac{\partial}{{\partial{z}}}{\left({\tan{{\left({y}{z}\right)}}}\right)}{z}^{{2}}\)

\(\displaystyle={2}{z}\cdot{\tan{{\left({y}{z}\right)}}}+{z}^{{2}}\cdot{{\sec}^{{2}}{\left({y}{z}\right)}}{y}\)

\(\displaystyle={2}{z}{\tan{{\left({y}{z}\right)}}}+{{\sec}^{{2}}{\left({y}{z}\right)}}{y}{z}^{{2}}\)

Hence.