# Differentiable points of function

• March 1st 2011, 12:55 PM
Combinatus
Differentiable points of function
The problem is stated as follows:

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The function $f(x,y)$ is defined in $R^2$ as $f(0,0) = 0$ and $f(x,y) = \frac{(\sin^2{x})(\sin^2{y})}{x^2 + y^2}$ if $(x,y) \ne (0,0)$.

In what points is $f(x,y)$ differentiable?

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It's easily shown that the two first partial derivatives are continuous except at the origin, so f is differentiable at every point that isn't the origin.

Using the definition of differentiability for the origin,

$\lim_{(h,k) \to (0,0)}{\frac{f(h,k) - f(0,0) - h f_1(0,0) - k f_2(0,0)}{\sqrt{{h}^2 + {k}^2}}} = \lim_{(h,k) \to (0,0)}{\frac{f(h,k)}{\sqrt{{h}^2 + {k}^2}}} = \lim_{(h,k) \to (0,0)}{\frac{\sin^2{h}\sin^2{k}}{(h^2 + k^2)^{3/2}}}$

After trying Maclaurin expansions of the numerator and denominator without getting anywhere (perhaps due to arithmetical mistakes), and fooling around a bit without any more luck, I switched to polar coordinates, i.e. $h = r\cos{\theta}, k = r\sin{\theta}$. Since r approaches 0 regardless of the angle $\theta$, the limit is equivalent to showing that $\lim_{r \to 0}\frac{\sin^2{(r\cos{\theta})}\sin^2{(r\sin{\thet a})}}{r^3} = 0$.

By using L'Hôpital's rule three times, the latter limit can be shown, and while the operations are simple, a considerable amount of arithmetic is required to show differentiability in the origin (and hence, in every point in $R^2$) like this. I strongly suspect that I have missed at least one much easier solution to this. Hints?
• March 1st 2011, 01:04 PM
girdav
Use the inequality $|\sin t|\leq |t|$ for $t\in\mathbb R$.
• March 1st 2011, 01:37 PM
Combinatus
Quote:

Originally Posted by girdav
Use the inequality $|\sin t|\leq |t|$ for $t\in\mathbb R$.

Also, dividing the polar coordinate limit's numerator and denominator by $(r\cos{\theta})^2 (r\sin{\theta})^2$ works fine. I'm not sure why I thought using l'hôpital three times was a good idea...