Implicitly find the second derivative of x^2- y^3= 3

[BraedenP]

Homework Statement

Find \frac{d^2y}{dx^2} of the following equation:

x^2-y^3=3

Homework Equations

Quotient Rule: \frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)\cdot f^{'}(x)-f(x)\cdot g^{'}(x)}{g^2(x)}

The Attempt at a Solution

I solved for the first derivative: \frac{2x}{3y^2}

But I'm stuck as to how I'd find the second derivative. I can generate answers, but none of them are correct.

How would I go about doing this?

 

[Mark44]

Homework Statement

Find \frac{d^2y}{dx^2} of the following equation:

x^2-y^3=3

Homework Equations

Quotient Rule: \frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)\cdot f^{'}(x)-f(x)\cdot g^{'}(x)}{g^2(x)}

The Attempt at a Solution

I solved for the first derivative: \frac{2x}{3y^2}


But I'm stuck as to how I'd find the second derivative. I can generate answers, but none of them are correct.

How would I go about doing this?

You have
y' = \frac{2x}{3y^2}

Use the quotient rule (and chain rule) to find y''. Replace any occurrences of y' by what you already found so that y'' is in terms of x and y, but not y'.

 

[BraedenP]

You have
y' = \frac{2x}{3y^2}

Use the quotient rule (and chain rule) to find y''. Replace any occurrences of y' by what you already found so that y'' is in terms of x and y, but not y'.

I'm so retarded. I forgot to equate the whole thing to y', and instead I equated it to 0 and solved for y' again. Ugh.

Thanks for your help!

 

[HallsofIvy]

From x^2- y^3= 3, 2x- 3y^2 y'= 0.

I would NOT solve for y' here. Instead, just differentiate "implicitely" again:
2- (6yy')y'- 3y^2y''= 0 so
y''= \frac{6y(y')^2- 2}{3y^2}

That is, in my opinion, a perfectly good answer but you could now use y'= 2x/(3y^2) to get y" in terms of x and y.