# implicit differenitaion

• Mar 11th 2010, 01:34 PM
ascendancy523
implicit differenitaion
so the problem asks us to solve for $\displaystyle \frac{d^2y}{dx^2}$ from the equation
$\displaystyle x^5+y^5=4$

I found $\displaystyle \frac{dy}{dx}=\frac{(-5x^4)}{(5y^4)}$, how would I go about solving for $\displaystyle \frac{d^2y}{dx^2}$?
• Mar 11th 2010, 01:40 PM
e^(i*pi)
Quote:

Originally Posted by ascendancy523
so the problem asks us to solve for $\displaystyle \frac{d^2y}{dx^2}$ from the equation
$\displaystyle x^5+y^5=4$

I found $\displaystyle \frac{dy}{dx}=\frac{(-5x^4)}{(5y^4)}$, how would I go about solving for $\displaystyle \frac{d^2y}{dx^2}$?

Cancel 5 to make things easier, then use the quotient rule

I get
$\displaystyle \frac{4x^3(y-x)}{y^5}$
• Mar 11th 2010, 02:30 PM
skeeter
Quote:

Originally Posted by ascendancy523
so the problem asks us to solve for $\displaystyle \frac{d^2y}{dx^2}$ from the equation
$\displaystyle x^5+y^5=4$

I found $\displaystyle \frac{dy}{dx}=\frac{(-5x^4)}{(5y^4)}$, how would I go about solving for $\displaystyle \frac{d^2y}{dx^2}$?

I get ...

$\displaystyle \frac{d^2y}{dx^2} = - \frac{16x^3}{y^9}$
• Mar 11th 2010, 03:43 PM
ascendancy523
Quote:

Originally Posted by skeeter
I get ...

$\displaystyle \frac{d^2y}{dx^2} = - \frac{16x^3}{y^9}$

I got a different answer from both of you but I typed in $\displaystyle \frac{d^2y}{dx^2} = - \frac{16x^3}{y^9}$ and it was right. How did you go about solving it?