I can't do this for the life of me, I should be able to. Argh.
For the relation 3x^2 - y^2 = 7, show that y'' = -21/y^3.
Do I use implicit differentiation after re-arranging for y?
I'm stuck. :|
Solving for y:
$\displaystyle y = (3x^2-7)^{\frac{1}{2}}$
Then, using the chain rule:
$\displaystyle y' = \frac{1}{2} (3x^2 - 7)^{- \frac{1}{2}}*(6x) = \frac {3x}{(3x^2-7)^{\frac{1}{2}}}$
Now, differentiating again, using the quotient rule:
$\displaystyle y'' = \frac{3(3x^2-7)^{\frac{1}{2}} - 3x* \frac{1}{2} (3x^2-7)^{- \frac{1}{2}}(6x)}{3x^2-7}$
which does work out to $\displaystyle \frac{-21}{y^3}$.
Let me know if you have trouble with the simplifying, but I assume your question was more about the differentiation.