Question

1) Let x^3 + y^3 = 28. Find y"(x) at the point (3, 1). y"(3)=

The correct answer is: -2*3*28 = -168

3x^2 + 3y^2(dx/dy) = 0

6x + 6y(dx/dy) = 0

dx/dy = -6(3)/6(1) = -3

That's as far as i get... i don't even know if that's started right...



You have
x^3 + + y^3 = 28
3x^2 + 3y^2(dy/dx) = 0 (that should be dy/dx) so
dy/dx = -x^2 / y^2
Now evaluate y=1 x=3 so y'(3)=-9
Now we want y"
d²y/dx² = [y^2(-2x) - -x^2*2y*y']/y^4
Now just evaluate
y"(3)= [-6 - 9*2*9]/1 = -168
Verify the algebra is correct.





good work.


Thnx, but in hindsight I'm wondering how much the students learn when I do the entire problem. I really need to practice giving just the set-up and letting them complete it. Gotta' admit, these problems are fun, now. I can recall struggling with this stuff just like some of them are, so I keep telling people to persevere until you 'breakthrough'.