What will be the Hessian matrix of $(\ln(x+y))^{xy}$ at $(x_0,y_0)=(1,1)$?

Question

I should compute the Hessian matrix at this point, but in the very first step, for $f'_x$ I get a pretty long expression, can I maybe somehow find a "tricky" way to solve this example?

Answer 1

Since $f(x,y) = f(y,x)$, you have $f_x(x,y) = f_y(y,x)$, so we get $f_{xx}(x,y) = f_{yy}(y,x)$ and $f_{xy}(x,y)= f_{yx}(y,x)$. If you want $x_0=y_0= 1$, you only have to compute $f_{xx}(1,1)$ and $f_{xy}(1,1)$. Write $$\ln f(x,y) = xy \ln(\ln(x+y))$$and go for it. You'll have $$f_x(x,y) = f(x,y)\left(y \ln(\ln(x+y)) + \frac{1}{\ln(x+y)} \frac{xy}{x+y}\right).$$Differentiate both sides of that with respect to $x$ and $y$ and then make $x=y=1$. Do not substitute the explicit expression for $f(x,y)$ in the expression above, and do not try to simplify the expression for $f_x(x,y)$! Write $$f_{xx}(x,y) = f_x(x,y)\left(y \ln(\ln(x+y)) + \frac{1}{\ln(x+y)} \frac{xy}{x+y}\right) + f(x,y) \frac{\partial}{\partial x} \text{(previous crap)},$$and plug directly values for $f_x(1,1)$ and $f(1,1)$ which are easy to compute. Same strategy for $f_{xy}(1,1)$.