Does this function $f:\mathbb{R^2}\to\mathbb{R}$ defined by: $$\frac{\sin(x^2)+\cos(y^2)}{\sqrt{x^2+y^2}}$$ differentiable at the origin? Thank you
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1The function doesn't seem to be defined at the origin, for instance, if you set $ y = 0 $ then it blows up as $ x \rightarrow 0 $. – Kevin May 01 '17 at 15:34
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Then we can't calculate partial derivative at the origin? – Theory Nombre May 01 '17 at 15:39
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Right, if a function isn't defined at $ x $, then its derivative isn't defined at $ x $ either. – Kevin May 01 '17 at 15:43
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The function $$ f(x) = \frac{\sin \left(x^2\right)+\cos \left(y^2\right)}{\sqrt{x^2+y^2}} $$ is not differentiable at the origin.
The iterated limit demonstrates the problem.
$$ \begin{align} \lim_{\color{red}{x}\to 0} f(x,y) &= \frac{\cos \left(y^2\right)}{\sqrt{y^2}} \\[3pt] \lim_{\color{blue}{y}\to 0} \left( \lim_{\color{red}{x}\to 0} f(x,y) \right) &= \infty \\ \end{align} $$
$$ \begin{align} \lim_{\color{blue}{y}\to 0} f(x,y) &= \frac{\sin \left(x^2\right)+1}{\sqrt{x^2}} \\[3pt] \lim_{\color{red}{x}\to 0} \left( \lim_{\color{blue}{y}\to 0} f(x,y) \right) &= \infty \\ \end{align} $$
The function is growing rapidly near the origin.
Hint: Maclaurin series
$$ \sin x^{2} = x^{2} - \frac{x^{2}}{6} + \mathcal{O}\left( x^{8} \right) $$
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