How to take second derivative implicitly Let $$y^4 + 5x = 21.$$
What is the value of $d^2y/dx^2$ at the point $(2, 1)$?
I’m stuck at trying to work out the second implicit derivative of this function. As far as I can work out, the first derivative implicitly is
$$\dfrac {-5}{4y^{3}}$$
How do you take the second derivative implicitly with respect to $x$ when $x$ has vanished? There would be nowhere to plug in $x=2$. What am I missing here?

You got:
$$\frac{dy}{dx}=\frac{-5}{4y^3}$$
This is correct. Now notice the product rule:
$$\frac{d}{dx}(pq)=p'q+pq'$$
Here:
$$p=4y^3\to p'=12y^2\frac{dy}{dx}$$
$$q=\frac{dy}{dx}\to q'=\frac{d^2y}{dx^2}$$
Thus via implicit differentiation we have:
$$4y^3\frac{d^2y}{dx^2}+12y^2(\frac{dy}{dx})^2=0$$
$$4y^3\frac{d^2y}{dx^2}-\frac{15}{y}=0$$
and go from there
