# Find the first partial derivatives of the following functions. s(y, z) = z^2 tan yz

Question
Derivatives
Find the first partial derivatives of the following functions.
$$\displaystyle{s}{\left({y},{z}\right)}={z}^{{2}}{\tan{{y}}}{z}$$

2021-03-10
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.

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