Find the partial derivative $$ \frac{\partial z}{\partial y} $$ for the function $$ z=\cos \left(x^{15}+y^{15}\right) $$

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Let $$ u = x^{15} + y^{15} $$. Then we have
$$
z = \cos(u).
$$
Using the chain rule, the partial derivative with respect to $$ y $$ is
$$
\frac{\partial z}{\partial y} = \frac{d}{du} \cos(u) \cdot \frac{\partial u}{\partial y}.
$$
The derivative of $$ \cos(u) $$ with respect to $$ u $$ is
$$
\frac{d}{du} \cos(u) = -\sin(u).
$$
Next, differentiate $$ u = x^{15} + y^{15} $$ with respect to $$ y $$:
$$
\frac{\partial u}{\partial y} = 15y^{14}.
$$
Therefore, combining the results gives
$$
\frac{\partial z}{\partial y} = -\sin(u) \cdot 15y^{14} = -15y^{14}\sin(x^{15}+y^{15}).
$$
The partial derivative $$ \frac{\partial z}{\partial y} $$ is $$ -15y^{14}\sin(x^{15}+y^{15}) $$.

Find the partial derivative for the function

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