Use the function to prove that (a) fx(0,0) and fy(0,0)...

Use the function to prove that (a) $f_{x}(0,0)$ and $f_{y}(0,0)$ exist, and (b) $f$ is not differentiable at $(0,0)$.
$$
f(x, y)=\left\{\begin{array}{ll}
\frac{3 x^{2} y}{x^{4}+y^{2}}, & (x, y) \neq(0,0) \\
0, & (x, y)=(0,0)
\end{array}\right.
$$

Key Concepts

Partial Derivatives

Partial derivatives measure the rate at which a multivariable function changes with respect to one of its variables, while holding the others fixed. They are defined using a limit of the difference quotient along the coordinate axes and are essential in analyzing the behavior of functions in several variables.

Differentiability in Multivariable Functions

Differentiability of a function in multiple dimensions means that the function can be well approximated by a linear function near the point of interest, which requires that the error between the function and its linear (first order) approximation becomes negligible relative to the distance from the point. The existence of partial derivatives alone does not guarantee differentiability; one must also verify that the remainder in the linear approximation vanishes appropriately, often by examining limits along various paths.

Transcript

00:01 In this problem of differential we have given that difference of f of x y say f of x and y is equals to this is 3 x square y divided with x to the power 4 plus y to the power 4 for x and y this is not equals to 0 and this is equal to 0 and this is equal to 0.

00:32 If ordered pair x and y is equals to 0.

00:41 And now we have to use the function to prove that, say, differentiation of the function at 00 with respect to x and differentiation of the function at 00 and differentiated with respect to y exist.

01:02 And also f is not differentiable at zero zero so here differentiation of the function at zero with respect to x is given by say limit and delta x tends to zero and this would be f of delta x and zero minus f of zero zero divided with delta x and here this would be limit delta x approaching towards 0 and this would be say this is delta x so 0 so 0 multiplied with any term would give you 0 so here in the numerator we have 0 and denominator is not equals to 0 so this is 0 minus 0 and this would be delta x so this term is equal to 0 now f say of 0 0 differentiated with respect to y so this would be limit delta y tends to 0 and this would be f of 0 and delta y minus f of 0 and divided with delta y similarly this would be again same so this is also equals to 0 so partial derivative exist at 0 0 so we conclude that partial derivative exist at 0 at 0 0 now along the line y is equals to x say this is the straight line this would be limit x and y approaches towards 0 0 say this would be function f of x y say this is the function f of x y and this would be say limit x approaches towards 0 say limit x approaches towards 0 and this would be 3x cube divided with x to the power 4 plus x square here y is equals to x we have already put here so this would be limit x approaches towards 0 and this is 3x 2 power can be cancelled from numerator and denominator and this would be x square plus 1 and if this is equals 0 so that means this is equals 0 now along the curve y is equals to x square so again this would be y is equals to x square that means we have to put y is equal to x square in the equation so limit this would be function f of x and y and limit x and y again approaches towards 0 0 now this would be 3 x to the power 4 divided with 2 x and to the power 4 which would give you 3 this is this is 2 x to the power 4 and this would give you 3 divided with 2 so we say that f is not continuous at 0 because both are giving you different value so we say that the function is not continuous at 0 0 so this is not continuous at 0 0 so f is not continuous at 0 0 so f is not continuous at 0 so f is not not differentiable at 0 because it is not continuous so this would be at 0 .0...

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