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Using the definition, prove that the function $f : R^2 → R$ defined by $f(x, y) = xy$ is differentiable on $R^2$. Compute the total derivative at each point of $R^2$.

The partial derivatives exist and are continuous. I know how to compute at a certain point,but I could not find the total derivative at each point.

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  • $\begingroup$ the total derivative is just dual to the gradient; $$df_{(x,y)}(\delta x,\delta y)=\delta x\cdot\left(\frac{\partial f}{\partial x}\right)(x,y)\,+\delta y\cdot\left(\frac{\partial f}{\partial y}\right)(x,y)\,\delta y$$ $\endgroup$ – oldrinb Jul 18 '15 at 20:30
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$d f(x,y) = {\partial f \over \partial x} dx + {\partial f \over \partial y} dy = y dx + x dy$.

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The total derivative of $f$ is \begin{align} f' &= J f \\ &= (\partial_x f, \partial_y f) \\ &= (\mbox{grad } f )^t \\ &= (y, x) \end{align} where $Jf$ is the Jacobian of $f$ which reduces to the transposed gradient for scalar valued functions.

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  • $\begingroup$ How do I show the function is differentiable by using the definition of differentiability @mvw $\endgroup$ – abcdef Jul 19 '15 at 21:16
  • $\begingroup$ Which definition do you use? $\endgroup$ – mvw Jul 19 '15 at 23:02

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