# Implicit Derivation

1. Mar 3, 2012

### Bashyboy

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution

File size:
25 KB
Views:
197
2. Mar 3, 2012

### SammyS

Staff Emeritus

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.

3. Mar 4, 2012

### 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.

File size:
12.7 KB
Views:
122
4. Mar 4, 2012

### SammyS

Staff Emeritus

That looks OK so far.

5. Mar 4, 2012

### Bashyboy

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

6. Mar 4, 2012

### ehild

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

ehild

7. Mar 4, 2012

### 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.

8. Mar 4, 2012

### Bashyboy

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

9. Mar 4, 2012

### ehild

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

ehild

10. Mar 4, 2012

### SammyS

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

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