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

Find the first partial derivatives of the following functions.
$s\left(y,z\right)={z}^{2}\mathrm{tan}yz$
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Step 1
Consider the provided function,
$S\left(y,z\right)={z}^{2}\mathrm{tan}yz$
Find the first partial derivatives.
So, we find the expression for $\frac{\partial S}{\partial y}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial S}{\partial z}$.
First we find the expression for $\frac{\partial S}{\partial y}$.
$\frac{\partial S}{\partial y}=\frac{\partial }{\partial y}\left({z}^{2}\mathrm{tan}yz\right)$
Step 2
Simplifying further,
$\frac{\partial S}{\partial y}=\frac{\partial }{\partial y}\left({z}^{2}\mathrm{tan}yz\right)$
$={z}^{2}\frac{\partial }{\partial y}\left(\mathrm{tan}yz\right)$
$={z}^{2}{\mathrm{sec}}^{2}yz\frac{\partial }{\partial y}\left(yz\right)$
$={z}^{2}{\mathrm{sec}}^{2}yz\cdot z$
$={z}^{3}{\mathrm{sec}}^{2}yz$
Now, find the expression for $\frac{\partial S}{\partial z}$.
$\frac{\partial S}{\partial z}=\frac{\partial }{\partial z}\left({z}^{2}\mathrm{tan}yz\right)$
Step 3
Simplifying further,
$\frac{\partial S}{\partial z}=\frac{\partial }{\partial z}\left({z}^{2}\right)\mathrm{tan}\left(yz\right)+\frac{\partial }{\partial z}\left(\mathrm{tan}\left(yz\right)\right){z}^{2}$
$=2z\cdot \mathrm{tan}\left(yz\right)+{z}^{2}\cdot {\mathrm{sec}}^{2}\left(yz\right)y$
$=2z\mathrm{tan}\left(yz\right)+{\mathrm{sec}}^{2}\left(yz\right)y{z}^{2}$
Hence.