Math Help - Using implicit differentation for second deriviate

1. Using implicit differentation for second deriviate

Hello,

I am stuck on the following problem:

sorry i don't know how to write in math on here

i have

on the curve at (1, 1 ) using implicit differentiation for the eq
rootx + rooty + y = 3

the first question is find y' (1). I got the answer of -1/3 which is right. Now it asks for y''(1) and I can't figure it out for the life of me. The answer is 5/27 and I just can't get it.

2. The first derivative is

$|\frac{1}{2\sqrt{x} = \frac{1}{2\sqrt{y}\times y' = 0$

Thre second derivative is

$\frac{-1}{2x^3/2} + y'(\frac{-1}{2y^3/2}\timey' + ( 1 + \frac{1}{sqrt{y} = 0$

Now substitute the values and find the result.

3. The first derivative is

4. i can't see what you put down it says error

thanks for the response though!

5. The first derivative is

\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}\times y' = 0

The second derivative is

\frac{-1}{2x^3/2} + y'(\frac{-1}{2y^3/2}) + ( 1 + \frac{1}{sqrt{y})y" = 0

Now substitute the values and find the result.

6. whaaaaat. i can't really understand what you wrote

7. Originally Posted by Zippyz
whaaaaat. i can't really understand what you wrote
The first derivative is

1/(2x^1/2) + y'[1/(2y^1/2) + 1] = 0

The second derivative is

-1/2(2x^3/2) + y'[-1/2(2y^3/2)*y'] + y"( 1 + 1/2y^1/2) = 0

8. EDIT:

$\frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{dy}{dx} + \frac{dy}{dx} = 0$

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

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

so $\frac{d^{2}y}{dx^{2}} (1,1) = \frac{5}{27}$ using the fact that $\frac{dy}{dx} (1,1) = - \frac{1}{3}$

9. d/dx[1/(2y^1/2)*(dy/dx)] = (-1/(4y^3/2)(dy/dx)^2 + [1/(2y^1/2)]*d^2y/dx^2