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I am new to multivariable calculus and my textbook doesn’t give out solutions so I’m just wondering how you go about proving something like this? I know that a function is differential at a point $a$ if it’s continous at $a$ and and the partial derivatives of $f$ exist near $a$ but I have never actually seen an example. Heres the question:

Assume $f(0, 0) = 0$, and determine whether $f(x, y) = \frac{xy^3}{x^2 + y^4}$ is differentiable at $(0, 0)$.

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Compute the partial derivative:

$${f_x’} (0,0)=\mathop {\lim }\limits_{x \to 0} {{f(x,0) – f(0,0)} \over x} = \mathop {\lim }\limits_{x \to 0} {{0 – 0} \over x} = 0$$

$${f_y’} (0,0)=\mathop {\lim }\limits_{y \to 0} {{f(0,y) – f(0,0)} \over y} = \mathop {\lim }\limits_{x \to 0} {{0 – 0} \over y} = 0$$

Compute the limit:

$$\mathop {\lim }\limits_{ x \to 0 \atop y \to 0} \frac {f(x,y) – {f_x’} (0,0)x – {f_y’} (0,0)y} {\sqrt {{x^2} + {y^2}} } $$

$$= \mathop {\lim }\limits_{ x \to 0 \atop y \to 0} \frac {xy^3} {(x^2+y^4)\sqrt {{x^2} + {y^2}} }$$

$$\xrightarrow{x=ky^2} \mathop {\lim }\limits_{y \to 0} {\frac{k{y^5}} {({k^2} + 1){y^4}\sqrt {{k^2}{y^4} + {y^2}} }}$$

$$=\mathop {\lim }\limits_{y \to 0} {\frac{k} {({k^2} + 1)\sqrt {k^2 y^2 + 1} }}=\frac{k}{k^2+1}$$

If $f(x,y) – {f_x’} (0,0)x – {f_y'(0,0)x} = o(\sqrt {{x^2} + {y^2}}) $ , which is equivalent to $ \mathop {\lim }\limits_{ x \to 0 \atop y \to 0} \cfrac {xy^3} {(x^2+y^4)\sqrt {{x^2} + {y^2}}}=0$ , we can say $f(x,y)$ is differentiable at $(0,0)$ , but in this case , the limit is relating to $k$ so that the limit doesn’t exist, that is to say $f(x,y)$ is non-differential at $(0,0)$.

Check the partial derivatives, if they are continuous at $(0,0)$. If they are then, $f$ is differentiable at $(0,0)$.

Using the following claim:

Partial derivatives exist and they are continuous at $(0,0)$, then it is differentiable at $(0,0)$.

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