I am finding the second derivative of (x^2)(y^3) = 1
2x(y^3) + (x^2)(y')(3y^2) = 0
And I simplified the answer to
y'= (-2y)/(3x)
so it would be...
[(-2y')(3x)-(-2y)(3)]/(9x^2) and it's simplified to (-6xy'+6y)/(9x^2)= y''
I could plug in (-2y)/(3x) in place of y'.
The correct answer is y''=(10y)/(9x^2), but I did not get this answer. Please help me!
This looks correct, but you dropped the y'
2x(y^3) + (x^2)(y')(3y^2)y' = 0 so
y'=-2x(y^3)/(x^2)(3y^2)=-2y/3x
If we now differentiate this again we get
y"=[3x*(-2y')-(-2y)*3]/(9x^2)=(-6xy'+6y)/(9x^2)
If we now substitute -2y/3x for y' we get
y"=[(-6x*-2y/3x)+6y]/9x^2=[4y+6y]/9x^2
Yes, you should make the substitution for y' and you'll get the correct answer in terms of x and y only.