# Thread: implicit differenciation

1. ## implicit differenciation

find y" of x+xy+y=2

so i did the first differncition as

1+y+x(dy/dx)+1(dy/dx) = 0
y= 1+y/x+1

y"= ?

somebody solve this for me ?

y" = ( 1+y)-(x+1)/ (x+1)^2

i think this is not right .
what is the answear. show me with working !

2. ## Re: implicit differenciation

So you've already taken the first differentiation which is,

$1+y+x\left[ \frac{dy}{dx}\right] +\frac{dy}{dx}=0\qquad \qquad (1)$

from the given equation we know that,

$y=\frac{(2-x)}{(x+1)}$

for later use, let's isolate $\frac{dy}{dx}$ from (1),

$\frac{dy}{dx}=\frac{-(1+y)}{(x+1)}=\frac{-\left( 1+\frac{(2-x)}{(x+1)}\right) }{(x+1)}=\frac{-3}{(x+1)^2}\qquad \qquad (2)$

to find $y''$, differentiate the equation (2) w.r.t x,

3. ## Re: implicit differenciation

why is ot

-(1+y) ?

i differentiated
-3/(x+1)^2

to get
6/(x+1)^3

what now . i dont get it

4. ## Re: implicit differenciation

Sorry, I think that I was not much clear on the previous post. So I did some editing. Do you understand it now?

5. ## Re: implicit differenciation

6/(x+1)^3

so the answear is that ?

6. ## Re: implicit differenciation

Yes, I think it is

7. ## Re: implicit differenciation

Hello, arsenal12345!

$\text{Differentiate implicitly. \;Find }y"\text{of }x+xy+y\:=\:2$

Differentiate: . $1 + y + xy' + y' \:=\:0$

n . . . . . . . . . . . . . . $(x+1)y' \:=\:-(y+1)$

. . . . . . . . . . . . . . . . . . . . $y' \:=\:-\frac{y+1}{x+1}$ .[1]

Differentiate: . . $y" \:=\:-\frac{(x+1)y' - (y+1)1}{(x+1)^2}$

Substitute [1]: . $y" \:=\:-\frac{(x+1)\left(-\frac{y+1}{x+1}\right) - (y+1)}{(x+1)^2}$

. . . . . . . . . . . . $y" \:=\:\frac{y+1 + y+1}{(x+1)^2} \:=\:\frac{2(y+1)}{x+1)^2}$