I need to find the stationary point, but I'm unsure the partial derivatives.
$\displaystyle f(x)=e^{-x^2-y^2}$
Any help will be appreciated
$\displaystyle f(x)=e^{-x^2-y^2}$
With respect to x hold y constant and vice versa. Consider when $\displaystyle y= e^{f(x)} \Rightarrow \frac{dy}{dx}= f'(x)e^{f(x)}$
so
$\displaystyle \frac{\partial f}{\partial x} = -2xe^{-x^2-y^2}
$
can you find
$\displaystyle \frac{\partial f}{\partial y} $ ?