Step 1: Find the partial derivative of $f(x,y)$ with respect to $y$. The given function is $f(x,y) = e^{2y} \cos(xy)$. We use the product rule $\frac{\partial}{\partial y}(uv) = \frac{\partial u}{\partial y}v + u\frac{\partial v}{\partial y}$. Let $u = e^{2y}$ and $v = \cos(xy)$. First, find $\frac{\partial u}{\partial y}$: $$ \frac{\partial}{\partial y}(e^{2y}) = e^{2y} \cdot \frac{\partial}{\partial y}(2y) = 2e^{2y} $$ Next, find $\frac{\partial v}{\partial y}$: $$ \frac{\partial}{\partial y}(\cos(xy)) = -\sin(xy) \cdot \frac{\partial}{\partial y}(xy) = -\sin(xy) \cdot x = -x\sin(xy) $$ Now, apply the product rule: $$ f_y(x,y) = (2e^{2y})\cos(xy) + e^{2y}(-x\sin(xy)) $$ $$ f_y(x,y) = 2e^{2y}\cos(xy) - xe^{2y}\sin(xy) $$ Factor out $e^{2y}$: $$ f_y(x,y) = e^{2y}(2\cos(xy) - x\sin(xy)) $$ Step 2: Evaluate $f_y(x,y)$ at the point $(0, \pi)$. Substitute $x=0$ and $y=\pi$ into the expression for $f_y(x,y)$: $$ f_y(0, \pi) = e^{2(\pi)}(2\cos(0 \cdot \pi) - 0 \cdot \sin(0 \cdot \pi)) $$ $$ f_y(0, \pi) = e^{2\pi}(2\cos(0) - 0 \cdot \sin(0)) $$ Since $\cos(0) = 1$ and $\sin(0) = 0$: $$ f_y(0, \pi) = e^{2\pi}(2(1) - 0(0)) $$ $$ f_y(0, \pi) = e^{2\pi}(2 - 0) $$ $$ f_y(0, \pi) = 2e^{2\pi} $$ The final answer is $\boxed{2e^{2\pi}}$. Send me the next one 📸