# Thread: Second Derivative in Terms of X and Y

1. ## Second Derivative in Terms of X and Y

So, i have no problem finding the first derivative. But then i cant get my answer to match the books answer for the second derivative. How are they getting the highlighted area? It doesnt look like they used the Quotient Rule. Have they moved the denominator up and then power rule?

Thanks for the help.

2. ## Re: Second Derivative in Terms of X and Y

Originally Posted by petenice
So, i have no problem finding the first derivative. But then i cant get my answer to match the books answer for the second derivative. How are they getting the highlighted area? It doesnt look like they used the Quotient Rule. Have they moved the denominator up and then power rule?

Thanks for the help.
They multiplied both sides by \displaystyle \begin{align*} x^2y \end{align*} and then moved all non-constant terms to the LHS before trying to evaluate the second derivative.

3. ## Re: Second Derivative in Terms of X and Y

Hey petenice.

They applied implicit differentiation to the expression in line 3. In other words they did d/dx [x^2*y*y' + xy^2 - 9] = d/dx(0)

4. ## Re: Second Derivative in Terms of X and Y

Originally Posted by chiro
Hey petenice.

They applied implicit differentiation to the expression in line 3. In other words they did d/dx [x^2*y*y' + xy^2 - 9] = d/dx(0)
Ok so now i can get past that part. However i get stuck at this next part. I keep going and keep getting the same answer but its incorrect. :/ Im pretty sure im doing everything right up to this point. But now what?

Thanks for any help!

5. ## Re: Second Derivative in Terms of X and Y

The differentiation of the first term is not correct: you differentiating x^2*y correctly but not the other one.

So d/dx[x^2*y*y']
= 2xyy' + x^2[d/dx](yy')
= 2xyy' + x^2[y'y' + y*y'']
= 2xyy' + x^2[y']^2 + x^2[yy'']

Try using this and see how you go.