How do you find the second derivative of a function using the chain rule?

[physics604]

Homework Statement

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

Homework Equations

Chain rule

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?

 

[physics604]

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

 

[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?

 

[Dick]

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?

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

 

[physics604]

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

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

 

[Dick]

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

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

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

 

[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.

 

[Dick]

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.

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

 

[physics604]

Okay I got it! Thanks so much!

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