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.
To find the second derivative of the given equation (x^2)(y^3) = 1, follow these steps:
1. Start with the first derivative using the product rule:
d/dx((x^2)(y^3)) = d/dx(x^2) * (y^3) + x^2 * d/dx(y^3) = 2x(y^3) + (x^2)(3y^2)y'
2. Simplify the obtained equation by substituting the value of y' from the previous calculation:
y' = -2y/(3x)
Therefore, replacing y' in the equation:
2x(y^3) + (x^2)(3y^2)(-2y/(3x)) = 0
3. Simplify this equation further by expanding and canceling out common terms:
2x(y^3) - 2xy^3 = 0
-2xy^3 + 2xy^3 = 0
4. Simplify the equation to get the expression for the second derivative:
(-6xy' + 6y)/(9x^2) = y''
Substitute the value of y' obtained previously (-2y/(3x)) into the equation:
(-6x(-2y/(3x)) + 6y)/(9x^2) = (-6xy'+6y)/(9x^2)
Simplify further by canceling out common terms:
= (4y + 6y)/(9x^2)
Finally, combine like terms to get the correct answer in terms of x and y:
y'' = (10y)/(9x^2)