Answer
$f_{yxy}=-50cos(2x+5y)$
Work Step by Step
Consider the function $f(x,y)=sin(2x+5y)$
Let us start by finding $f_{y}(x,y)$ by differentiating $f(x,y) $with respect to $y$ keeping $x$ constant.
As we know
$f_{y}=\frac{∂}{∂y}f(x,y) $
$=\frac{∂}{∂y}[sin(2x+5y)]$
$=5cos(2x+5y)$
Now, let us start by finding $f_{y}(x,y)$ by differentiating $f(x,y) $with respect to $x$ keeping $y$ constant.
$f_{yx}=\frac{∂}{∂x}[5cos(2x+5y)]=-10sin(2x+5y)$
$f_{yxy}=\frac{∂}{∂y}[f_{yx}]$
$=\frac{∂}{∂y}[-10sin(2x+5y)]$
Hence, $f_{yxy}=-50cos(2x+5y)$
