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

Question
Derivatives
asked 2021-03-09
Find the first partial derivatives of the following functions.
\(\displaystyle{s}{\left({y},{z}\right)}={z}^{{2}}{\tan{{y}}}{z}\)

Answers (1)

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.
0

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