Question

If x^2=25-y^2, what is the value of the second derivative at the point (3, 4)?

Answer 1

did you find dy/dx first ?

Answer 2

yeah, I got that as -x/y

Answer 3

seems correct,
now
dy/dx= -x/y

differentiate again both sides with respect to x..you will get a dy/dx in the RHS..substitute (-x/y) in its place, and voila, 99% of the question is complete! ;)

Answer 4

I'm confused, doesn't differentiating in the right get -x/dy?

Answer 5

ofcorse not,,how do you get that,,
d/dx(-x/y) => use product rule or quotient rule of differentiation..

Answer 6

still stuck ?

Answer 7

I think I'm closer. Using the quotient rule I got (x-dy-y)/y^2

Answer 8

seems you applied that wrong,,allow me to elaborate

d/dx(-x/y) = [ y*d(-x)/dx - (-x)*(dy/dx) ] /y^2

is that clear so far ?

Answer 9

yeah, that's what I got

Answer 10

so now you have
d2y/dx2 = [ y*d(-x)/dx - (-x)*(dy/dx) ] /y^2

just simplify RHS,,use dy/dx= -x/y

finally when all is done , substitute x=3 and y=4..

hope this helps..

Answer 11

This can actually get a little messy. First, I'd use y' for dy/dx.

Differentiating once you get

2x = 2yy'
Don't express y' yet, just differentiate again. This gives

0 = 2(y')^2 + 2y'' or y'' = - (y')^2

You don't actually need the whole function y'' just its value at (3,4). So use the first equation to calculate y' at (3,4), then plug it in as a number.

This is easier than getting y'' as a function of x and y.

Answer 12

yep..seconded!