compute d^2/dx^2 at the point (4,2)
x^2+y^2=20

Question

amistre64 · Answer

is that our f(x) sitting there?

amistre64 · Answer

2x x' + 2y y' = 0

2x + 2y y' = 0

y' = -x/y

amistre64 · Answer

y'' = -1/y + y'x/y^2

amistre64 · Answer

$$y'' = -\frac{1}{y} + \frac{{-x^2}}{y^3}$$

amistre64 · Answer

(x=4, y=2)

y'' = -1/2 - 16/8

amistre64 · Answer

y'' = -32/8 = -4 perhaps?

amistre64 · Answer

thats wrong

anonymous · Answer

why not 5/2?

amistre64 · Answer

-4/8 - 16/8 = -20/8

y'' = -5/2

amistre64 · Answer

i was gettin there, my head on holds so much at once lol

anonymous · Answer

cool, thanks!

amistre64 · Answer

i was wondering if i could have done that from this step like this:

2x + 2y y' = 0

2 + 2y' + 2y y'' = 0

y'' = -2 -2(-x/y)     -1 -(-x/y)     -1 -(-2)
        ---------- = -------- =  ------- = doesnt look like it
                2y                 y                 2

anonymous · Answer

that looks more complicated, do you happen to also know about definite integrals?

amistre64 · Answer

i can do them with somewhat of skill and mostly luck ;)

amistre64 · Answer

definite integrals are integrating within an interval

anonymous · Answer

lol okay, let's see how lucky you are, I'm about to post one =D

amistre64 · Answer

yay!!

anonymous · Answer

okay, for some reason it won't let me ask a question like normal

amistre64 · Answer

refresh your browser

amistre64 · Answer

f5