Implicit Derivation: Finding 2nd Derivative of y^2 = x^3

[Bashyboy]

Homework Statement

I am suppose to find the second derivative implicitly of the function y^2 = x^3. I find the first derivative to be dy/dx = 3x^2/2y, but shortly find myself having difficulty in the second derivation. My steps for the second derivative is in the file attached; there are a few additionally steps, but, at the last step, I am in no way nearing the actual answer.

Homework Equations

The Attempt at a Solution

 

[SammyS]

Homework Statement

I am suppose to find the second derivative implicitly of the function y^2 = x^3. I find the first derivative to be dy/dx = 3x^2/2y, but shortly find myself having difficulty in the second derivation. My steps for the second derivative is in the file attached; there are a few additionally steps, but, at the last step, I am in no way nearing the actual answer.

I assume that you result for implicitly differentiating y^2 = x^3, the first time was something like:

\displaystyle 2y(dy/dx)=3x^2​

Differentiate that again before solving for dy/dx. It's easier to work with.

 

[Bashyboy]

Well, I took the derivative from where you advised me to; but I still feel I am getting it wrong. Here are some of the steps I took, though there are few because I had the intuition that I was getting them wrong.

 

[SammyS]

Well, I took the derivative from where you advised me to; but I still feel I am getting it wrong. Here are some of the steps I took, though there are few because I had the intuition that I was getting them wrong.

That looks OK so far.

 

[Bashyboy]

Well, the answer is 3x/4y, and I am still not getting this.

 

[ehild]

That looks OK so far.

You miss a square in the second step . d/dx(yy')=y'²+yy".

ehild

 

[Bashyboy]

ehild, I am terribly sorry: I don't really understand what you are saying; I can't not see what it is that I should be fixing in my second step.

 

[Bashyboy]

ehild, I believe you may have made a mistake when typing with latex.

 

[ehild]

Do again the derivative of y dy/dx. It is (dy/dx)²+yd²y/dx².

ehild

 

[SammyS]

[STRIKE]That looks OK so far[/STRIKE].

I was wrong. As ehild pointed out, the dy/dx should be squared.

\displaystyle \frac{d}{dx}(2y\frac{dy}{dx})=2\frac{dy}{dx}\frac{dy}{dx}+2y\frac{d^2y}{dx^2}