# Derivatives, chain rule 5

1. Nov 10, 2013

### physics604

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

Find the second derivative of $$9x^2+y^2=9$$

2. Relevant equations

Chain rule

3. The attempt at a solution

I find the first derivative first.

$$18x+2y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=-9\frac{x}{y}$$

I then find the second derivative.

$$-9(\frac{y-x\frac{d'y}{dx''}}{y^2})$$ $$y^2=-9y+9x\frac{d'y}{dx''}$$ $$y^2+9y=9x\frac{d'y}{dx''}$$ $$\frac{d'y}{dx''}=\frac{y^2+9y}{9x}=\frac{y(y+9)}{9x}$$

My textbook says that the answer should be $$\frac{-81}{y^3}$$ What did I do wrong?

2. Nov 10, 2013

### physics604

Nevermind I got it! I should be plugging in y' into my equation for y''.

3. Nov 10, 2013

### physics604

I actually still need help.

When I plug in the y' into y'' I still don't get the right answer.

$$\frac{-9y-9x\frac{dy}{dx}}{y^2}$$ $$=\frac{-9y-9x(-\frac{9x}{y})}{y^2}$$ $$=\frac{-9y+\frac{81x^2}{y}}{y^2}$$

This is still not the answer. What did I do wrong now?

4. Nov 10, 2013

### Dick

First correct a sign error in your numerator. Then multiply numerator and denominator by y. Then use your original function to simplify the numerator.

5. Nov 10, 2013

### physics604

I'm sorry, but I don't get what you mean. When I multiply the numerator and denominator by y I get

$$\frac{-9y^2+81x^2}{y^3}$$

6. Nov 10, 2013

### Dick

I edited my post when I realized you had a sign error in the numerator. Find it and fix it.

7. Nov 10, 2013

### physics604

Okay, so I've fixed the sign error and multiplied the top and bottom by y. Now I have this

$$\frac{-9y^2-81x^2}{y^3}$$

The most I can do is factor out the -9, but I still don't get the correct answer.

8. Nov 10, 2013

### Dick

Your original function says y^2+9x^2=9. Use that to simplify the numerator.

9. Nov 10, 2013

### physics604

Okay I got it! Thanks so much!

$$\frac{-9(y^2+9x^2)}{y^3}=\frac{-9(9)}{y^3}=\frac{-81}{y^3}$$