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?

There's a typo in your problem statement.  The point $(2,1)$ is not on the curve.  They probably meant $(1,2)$.
I would not have solved for the first derivative.  Just differentiate the equation again, remembering that both $y$ and $y'$ are functions of $x$, and so the chain rule applies:
$$y^4+5x=21$$
$$4y^3y'+5=0$$
$$4y^3y'' + 12y^2y'y' = 0.$$
Plug in $y=2$ in the second line to get $4\cdot 8 y'+5=0$
and solve to get $y' = -5/32.$
Then plut $y=2$ and $y'=-5/32$ in the third line and solve for $y''$.  I get $-75/2048.$  (But I've had only one coffee this morning.)
