I need to prove that the following function is differentiable and find $df$.
$$
f(x,y,z)=\frac{\sqrt{x^2-y^2}}{z^2+x+y}
$$
I found all partial derivatives:
$$
\frac{\partial f}{\partial x}=\frac{x(y+z^2)+y^2}{(z^2+x+y)^2\sqrt{x^2-y^2}} \\
\frac{\partial f}{\partial y}=-\frac{y(x+z^2)+x^2}{(z^2+x+y)^2\sqrt{x^2-y^2}} \\
\frac{\partial f}{\partial z}=-\frac{2z\sqrt{x^2-y^2}}{(z^2+x+y)^2}
$$
And then wrote something like this:
$$
\text{If}\
\begin{cases}
z^2+x+y\ne 0 \\
x^2-y^2 \ne 0
\end{cases}
\ \ \text{then}\ \
\exists df=\frac{\partial f}{\partial x}dx+
\frac{\partial f}{\partial y}dy+
\frac{\partial f}{\partial z}dz
$$
The problem is I am not sure if my last line of solution is correct. Maybe the initial function has differentials even if the conditions above are not met. Can anyone explain if I am right?
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Do you mean "find $df$" rather than "find $dx$"? – user10354138 May 19 '19 at 13:46
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Oh, yes. My mistake. Thank you! – Bonrey May 19 '19 at 13:47
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At "calculus" level, you probably need impose $x^2>y^2$ instead of letting $x^2-y^2\neq 0$ (and play with $\mathbb{C}$) – user10354138 May 19 '19 at 13:51
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Let's suppose that I am working on $\mathbb{R}$ – Bonrey May 19 '19 at 13:55
1 Answers
I think you've completely answered the question.
Firstly, logically speaking, if you want to know whether your last line is correct, you shouldn't care about the points that don't meet those two conditions, as you make no claim about them. And since, for every point that meets those conditions, you have continuous partial derivatives in a neighborhood, your sentence is indeed correct.
And secondly, I don't think you're supposed to check anything on the remaining points: I think "differentiable" here means differentiable at points that respect both your conditions (these are the points in the interior of your function domain). I claim this because of two reasons:
1) It doesn't make sense to talk about differentiability of a function outside the function's domain, so checking differentiability at points that don't respect your first condition (and therefore are outside the function's domain) doesn't make sense.
2) The points for which the second condition doesn't hold are in the boundary of your domain and you usually don't define derivative at these points either, at least for multivariable functions. It's still an interesting case though, so I suggest you read
Are derivatives defined at boundaries?
I hope this helps.
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