calculusimplicit-differentiationpartial derivative
I have the following:
$\frac{dy}{dx}=-\frac{x^2}{y^2}$
I would like to find $\frac{d^2y}{dx^2}$. One solution suggested was to take the partial derivative of the above function with respect to x, treating y as a constant. This gives the following solution:
$\frac{\partial \:}{\partial \:x}\left(-\frac{x^2}{y^2}\right)=-\frac{2x}{y^2}$
Which is correct. Why can y be treated as a constant here; why does taking this partial derivative give the second derivative of this implicit function?
Your approach is valid, but it's easy to lose ourselves in the notation. I will make the substitution $g = \frac{dy}{dx}$.
Equation $(A)$ becomes $4x^3 + 4y^3 g = 0$. Differentiating with respect to $y$, we have
$$12x^3 + 12y^2\frac{dy}{dx}g+4y^3\frac{dg}{dx}=0.$$
Notice that another $\frac{dy}{dx}$ popped up from $\frac{d}{dx}4y^3$. Now, $$12x^3 + 12y^2\left(\frac{dy}{dx}\right)^2+4y^3\frac{d^2y}{dx^2}=0.$$
Solving this for $\frac{d^2y}{dx^2}$ will yield the correct result after some manipulation.
With a bit of work, you can show that $R(x,y)$ is only defined for $y=x$ and $y=-x$ excluding the point $(0,0)$. Hence your function does not satisfy the conditions of the implicit function theorem, which states that your function should be at least continuously differentable to apply the method of implicit differentiation.
Note, you could extend your problem to complex numbers, in that case the domain of applicability of $R(x,y)$ would be extended. However, even in that case, it would still not be the entire plane because no definition of the square root or the inverse cosine is analytic on the entire complex plane. In particular, if you choose the most "natural" extension to the complex plane of the inverse cosine, it is not defined on the branch points $-1$ and $1$. Those correspond exactly to the cases $y=x$ and $y=-x$. In other words, naïve implicit differentiation does not work in that case either. You would therefore need analytic extensions of the square root and the inverse cosine such that they are valid in the case $y=x$. But, those extensions will not correspond to the definitions you adopted in the real case.
Best Answer
This is not the second total derivative of $y$ with respect to $x$. For instance, $y=−x$ satisfies the first equation, but the second derivative is zero and isn't given by the second equation.