# Thread: More Implicit Differentiation

1. ## More Implicit Differentiation

$\displaystyle x^2+y^2=1$
Find dy/dx and the second derivative.

The answer says both derivatives equal $\displaystyle \frac{-x}{y}$.
The first one does, but the second derivative I got as $\displaystyle \frac{y}{x}$.

2. Originally Posted by Truthbetold
$\displaystyle x^2+y^2=1$
Find dy/dx and the second derivative.

The answer says both derivatives equal $\displaystyle \frac{-x}{y}$.
The first one does, but the second derivative I got as $\displaystyle \frac{y}{x}$.
and how did you get that answer?

i got neither. my answer for the second derivative was more complicated. i differentiated -x/y with the quotient rule. but you can use the product rule also

3. aww, i got $\displaystyle \frac{-1}{y^3}$..

4. $\displaystyle x^2 + y^2 = 1$
$\displaystyle 2x\frac{dy}{dx} + 2y = 0$
$\displaystyle \frac{dy}{dx} = \frac{-y}{x}$

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

I got something different too. Hmm, did I miss something?

5. Originally Posted by colby2152
$\displaystyle x^2 + y^2 = 1$
$\displaystyle 2x\frac{dy}{dx} + 2y = 0$
$\displaystyle \frac{dy}{dx} = \frac{-y}{x}$

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

I got something different too. Hmm, did I miss something?

it's because, $\displaystyle \frac{dy}{dx} = \frac{-x}{y}$ and not $\displaystyle \frac{dy}{dx} = \frac{-y}{x}$.. Ü

6. Originally Posted by colby2152
$\displaystyle x^2 + y^2 = 1$
$\displaystyle 2x\frac{dy}{dx} + 2y = 0$
$\displaystyle \frac{dy}{dx} = \frac{-y}{x}$

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

I got something different too. Hmm, did I miss something?
You're OK, except it should be $\displaystyle y^{3}$ in the denominator.

7. Originally Posted by galactus
You're OK, except it should be $\displaystyle y^{3}$ in the denominator.
I see what I did, I flipped the derivative in the second line. It's easy to miss these things when typing them.