How to find the partial derivative value of this function?

Question

Question: Let $ f(x,y) = 2xy + e^{x^2-y^2}$, Compute $ f_{xy} (1,1) $

My work: $f_x = 2xy + 2xe^{x^2-y^2} $ Then $f_{xy} = dy (f_x)$ I get $f_{xy} = 2x + 4xye^{x^2-y^2}$

But when I evaluate that result at (1,1) I get 6 when the right answer is -2. Can someone tell me where I went wrong?

Answer 1

Actually it is $$f_x=2(y+xe^{x^2-y^2})$$ and $$f_{xy}=2(1-2xye^{x^2-y^2})$$

Answer 2

$f_x = \pmb{2y} + 2xe^{x^2-y^2}$ since $\frac{\partial (2xy)}{\partial x}=2y$ so $f_{xy} = \frac{\partial f_x}{\partial y}= {2 \pmb{- 4xye^{x^2-y^2}}}$ since $\frac{\partial (e^{x^2-y^2})}{\partial y}=-2ye^{x^2-y^2}$.

Answer 3

My work: $f_x = 2xy + 2xe^{x^2-y^2} $ Then $f_{xy} = dy (f_x)$ I get $f_{xy} = 2x + 4xye^{x^2-y^2}$

While computing the partial derivative with respect to $x$, you applied the chain rule correctly to form

$$\frac{\partial}{\partial x}(e^{x^2-y^2})=2xe^{x^2-y^2}$$ but forgot that $2xy$ is also function that depends on $x$ $$\frac{\partial}{\partial x}(2xy)=\color{blue}{2y}$$ therefore the first partial derivative is

$$f_x = \color{blue}{2y} + 2xe^{x^2-y^2} $$

Next, you made a small mistake in computing the partial derivative of $2xe^{x^2-y^2}$ with respect to $y$. Namely, when you apply the chain rule, the $-y^2$ in the exponent becomes $-2y$ in the multiplier

$$\frac{\partial}{\partial y}(2xe^{x^2-y^2})=(2x)(\color{blue}{-2y})e^{x^2-y^2}=\color{blue}{-4xy}e^{x^2-y^2}$$

so that

$$f_{xy}=2{-4xy}e^{x^2-y^2}$$

hence

$$f_{xy}(1,1)=2-4e^{0}=-2$$