Answer
$f_{x}=32xy(5y^{3}+2x^{2}y)^{7}$
and
$f_{y}=8(5y^{3}+2x^{2}y)^{7} (15y^{2}+2x^{2})$
Work Step by Step
Given: $f(x,y)=(5y^{3}+2x^{2}y)^{8}$
We need to find the first partial derivatives $f_{x}$ and $f_{y}$.
Differentiate the function with respect to $x$ keeping $y$ constant.
$f_{x}=8(5y^{3}+2x^{2}y)^{8-1}\times4xy$
$f_{x}=32xy8(5y^{3}+2x^{2}y)^{7}$
Differentiate the function with respect to $y$ keeping $x$ constant.
$f_{y}=8(5y^{3}+2x^{2}y)^{8-1}\times (15y^{2}+2x^{2})$
Hence, $f_{x}=32xy(5y^{3}+2x^{2}y)^{7}$
and
$f_{y}=8(5y^{3}+2x^{2}y)^{7} (15y^{2}+2x^{2})$
