Thread: 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.

Thanks in advance.

The first derivative is



Thre second derivative is



Now substitute the values and find the result.

The first derivative is

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

thanks for the response though!

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.

whaaaaat. i can't really understand what you wrote

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

EDIT:







so using the fact that

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