# Math Help - 2nd implicit derivative

1. ## 2nd implicit derivative

I was just wondering if somebody could check over my math if they have a second.

Find $x^3+y^3 = 126$ and find y''(x) @ (5 , 1)

Took d/dx $3x^2+3y^2(dy/dx) = 0$

Solved for dy/dx by subtracting and dividing $(dy/dx) = (-3x^2)/(3y^2)$

Took $d^2y/dx^2$ using quotient rule $\frac {(-3x^2*6y(dy/dx)-3y^2(-6x))}{(3y^2)^2}$

Plugged in dy/dx to the second derivative $\frac {(-3x^2*6y(-3x^2/3y^2)-3y^2(-6x))}{(3y^2)^2}$ At this point does the procedure look correct? I plug in x = 5 and y = 1 and get 1260.

2. ## Re: 2nd implicit derivative

I would simplify the first derivative to:

$\frac{dy}{dx}=-\frac{x^2}{y^2}$

and so:

$\frac{d^2y}{dx^2}=-\frac{2xy^2-2x^2y\frac{dy}{dx}}{y^4}=-\frac{2xy^2+2x^4y^{-1}}{y^4}=-\frac{\frac{2x}{y}(x^3+y^3)}{y^4}$

thus:

$\frac{d^2y}{dx^2}\left|_{(5,1)}=-1260$

3. ## Re: 2nd implicit derivative

I would have avoided the quotient rule altogether.

$x^{3}+y^{3} = 126,$

$3x^{2} + 3y^{2}\frac{dy}{dx}=0,$ (so $\frac{dy}{dx}= -\frac{x^{2}}{y^{2}}),$

$6x + 6y\left(\frac{dy}{dx}\right)^{2} + 3y^{2}\frac{d^{2}y}{dx^{2}} = 0,$ etc.

4. ## Re: 2nd implicit derivative

Originally Posted by BobP
I would have avoided the quotient rule altogether.

$6x + 6y\left(\frac{dy}{dx}\right)^{2} + 3y^{2}\frac{d^{2}y}{dx^{2}} = 0,$ etc.
Why did you square $\frac {dy}{dx}$ and then add another $3y^2\frac{d^2y}{dx^2}$?

5. ## Re: 2nd implicit derivative

I didn't square it, I just differentiated $3y^{2}\frac{dy}{dx}$ as a product.

Differentiating $3y^{2}$ gets you $6y\frac{dy}{dx}$ and differentiating $\frac{dy}{dx}$ gets you $\frac{d^{\2}y}{dx^{2}}.$

Put them together using the product rule.

6. ## Re: 2nd implicit derivative

Interesting results!

Direct extraction of y(x):

$y''(x)=-\frac{2 x^4}{\left(126-x^3\right)^{5/3}}-\frac{2 x}{\left(126-x^3\right)^{2/3}}$

and:

$y''(5) = -1260$

7. ## Re: 2nd implicit derivative

Originally Posted by BobP
I didn't square it, I just differentiated $3y^{2}\frac{dy}{dx}$ as a product.

Differentiating $3y^{2}$ gets you $6y\frac{dy}{dx}$ and differentiating $\frac{dy}{dx}$ gets you $\frac{d^{\2}y}{dx^{2}}.$

Put them together using the product rule.
Ah, I see. I'm still kinda figuring out implicits.