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I was working on problems that made me find partial derivative of the functions. For example, $\mathbf{f}(\mathbf{x})=x_1 \cos x_3$ becomes $0$ when I take partial derivative with respect to $x_2$. Does this mean that the function does not have partial derivative with respect to $x_2$? How would I try to show a given function is not partially differentiable?

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    $\begingroup$ The partial derivative is 0 because the function does not change when $x_{2}$ changes. It doesn't rely on $x_{2}$. Think about $f(x) = 7$. The derivative with respect to $x$ is 0 because the function does not change with $x$. $\endgroup$ – user367387 May 2 '17 at 17:06
  • $\begingroup$ In one variable, functions whose derivative is $\equiv 0$ are constants. Of course constants are differentiable. $\endgroup$ – zhw. May 2 '17 at 17:10
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A derivative being zero and not existing are two different things. If it is zero, it is no different than being constant in one direction, e.g., a saddle point. Not existing is less common, and occurs when the limit from different directions does not exist, for instance, such as in $y=|x|$. In that case you would say the (partial) derivative is "not well defined", but you can still talk about derivatives while specifying how you take it, such as "from the positive direction".

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