# Second Derivative

• October 9th 2012, 06:15 PM
petenice
Second Derivative
Ok first post here so i apologize if i do something wrong.

So i know to use implicit differentiation to get the first derivative
5x^4/3x^2.
But then i dont understand two things...

1) Why do they multiply by xy/xy?
2) And in the last step, what happened to the y' ? How did they get rid of it? I hope my question makes sense.

• October 9th 2012, 06:54 PM
chiro
Re: Second Derivative
Hey petenice.

The reason why they multiply it is because they want to get it into the form of x^5/y^3 which is equal to 1 by the definition of the initial relationship so it cancels.

For your second question, this is just the normal quotient rule where d/dx(u/v) = [vu' - uv']/v^2.

u = 5y, v = 3x, u' = 5y', u' = 3 so [vu' - uv']/v^2 = [(3x)(5y') - (5y)(3)]/(9x^2) because y' = dy/dx and d/dx(3x) = 3.
• October 9th 2012, 07:08 PM
MarkFL
Re: Second Derivative
Another approach:

$y=x^{\frac{5}{3}}$

$\frac{dy}{dx}=\frac{5}{3}x^{\frac{2}{3}}$

$\frac{d^2y}{dx^2}=\frac{10}{9}x^{-\frac{1}{3}}=\frac{10x^{\frac{5}{3}}}{9x^2}=\frac{ 10y}{9x^2}$
• October 9th 2012, 07:41 PM
petenice
Re: Second Derivative
Im going to use this forum more often now. I've been staring at this question for hours and now I totally get it!

Chiro, Mark,

Thanks so much!