Easy Relation Second Derivate Proof

**Jeavus** · March 30th 2008, 05:59 PM

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. :|

**Mathnasium** · March 30th 2008, 06:28 PM

Solving for y:

$y = (3x^2-7)^{\frac{1}{2}}$

Then, using the chain rule:

$y' = \frac{1}{2} (3x^2 - 7)^{- \frac{1}{2}}*(6x) = \frac {3x}{(3x^2-7)^{\frac{1}{2}}}$

Now, differentiating again, using the quotient rule:

$y'' = \frac{3(3x^2-7)^{\frac{1}{2}} - 3x* \frac{1}{2} (3x^2-7)^{- \frac{1}{2}}(6x)}{3x^2-7}$

which does work out to $\frac{-21}{y^3}$ .

Let me know if you have trouble with the simplifying, but I assume your question was more about the differentiation.

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