Using implicit differentation for second deriviate

Oct 2009
6
0
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.

Thanks in advance.
 
Jun 2009
806
275
The first derivative is

\(\displaystyle |\frac{1}{2\sqrt{x} = \frac{1}{2\sqrt{y}\times y' = 0\)

Thre second derivative is

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

Now substitute the values and find the result.
 
Oct 2009
6
0
i can't see what you put down it says error

thanks for the response though!
 
Jun 2009
806
275
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.
 
Oct 2009
6
0
whaaaaat. i can't really understand what you wrote
 
Jun 2009
806
275
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
 
Last edited:
May 2009
959
362
EDIT:

\(\displaystyle \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{dy}{dx} + \frac{dy}{dx} = 0 \)

\(\displaystyle \frac{dy}{dx} = \frac{-\sqrt{y}}{\sqrt{x} + 2 \sqrt{x}\sqrt{y}} \)

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

so \(\displaystyle \frac{d^{2}y}{dx^{2}} (1,1) = \frac{5}{27} \) using the fact that \(\displaystyle \frac{dy}{dx} (1,1) = - \frac{1}{3} \)
 
Last edited:
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Jun 2009
806
275
d/dx[1/(2y^1/2)*(dy/dx)] = (-1/(4y^3/2)(dy/dx)^2 + [1/(2y^1/2)]*d^2y/dx^2