OpenStudy (anonymous):

Compute d^2y/dx^2 at the point (3, 2). x^2-y^2 = 5

5 years ago
OpenStudy (zehanz):

You could try this: First, write y as function of x:\[y^2=x^2-5 \Leftrightarrow\]\[y=\pm \sqrt{x^2-5}\]Now, there are two options for y, apparently. We can do away with the negative one, however, because the question is about the point (3, 2). So y is positive. So we have:\[y=\sqrt{x^2-5}\]Now first calculate the derivative of y:\[\frac{ dy }{ dx }=\frac{ 1 }{ 2\sqrt{x^2-5} }\cdot 2x=\frac{ x }{ \sqrt{x^2-5} }\]If you differentiate again and set x=3, you're done...

5 years ago
OpenStudy (anonymous):

is the ans -5/8 ??

5 years ago
OpenStudy (zehanz):

It is! So did you get\[\frac{ d^2y }{ dx^2 }=-\frac{ 5 }{ (x^2-5)\sqrt{x^2-5} }\]This is rather tedious, because you have to use the Quotient Rule and then simplify... Setting x=3 then gives =-5/8

5 years ago