how do u find the second derivative of this
3y^2 - 4x = 16
In implicit differentiation, we take nothing for granted. we always identify what we are taking the derivative of, and with respect to what. That is, if we differentiate a y-term with respect to x, we attach the notation $\displaystyle \frac {dy}{dx}$ to it. This means we took the derivative of y with respect to x. We can simply call this y'. We don't attach anything if we differentiate a variable that is the same as the variable that we are differentiating with respect to. That is, if we differentiate x with respect to x, we don't attach anything. Why is that? Technically we do, if we differentiate an x-term with respect to x, we attach $\displaystyle \frac {dx}{dx}$, but since derivative notations can function as fractions, we simplify this to 1, so you don't see it. If we differentiate a term with both x's and y's in it, we use the product rule. and we attach the appropriate notation for each part. when we differentiate the x-term we attach nothing. when we differentiate the y-term, we attach $\displaystyle \frac {dy}{dx}$
Let's see how this works
$\displaystyle 3y^2 - 4x = 16$
We proceed by Implicit differentiation
$\displaystyle \Rightarrow 6y~y' - 4 = 0$ .......remember, the derivative of a constant is zero, so the 16 disappears. we attach y' to the derivative of the y-term, but nothing to the derivative of the x-term as explained above.
Now we simply solve for y'
$\displaystyle \Rightarrow y' = \frac {4}{6y}$
Now try the second derivative