1. ## Implicit Differentiation ???

Ok... I have gotten this far with this problem:

xy+y^3=1

[x'*y+x*y'] +[3y^2]=0

My reasoning:
To get [x'*y+x+y'] I used the product rule
To get [3y^2] I used the product rule
to get "0" I used the constant rule

so... what's next ?

I know I am solving for y'

but can't figure out the next step

How to get y' alone one side or should it be taken away at this point ?

VERY VERY confused

2. Originally Posted by kbgemini16
Ok... I have gotten this far with this problem:

xy+y^3=1

[x'*y+x*y'] +[3y^2]=0

My reasoning:
To get [x'*y+x+y'] I used the product rule
To get [3y^2] I used the product rule
to get "0" I used the constant rule

so... what's next ?

I know I am solving for y'

but can't figure out the next step

How to get y' alone one side or should it be taken away at this point ?

VERY VERY confused
Sorry not that simple..will correct and edit in a minute.

First, you wanna move the $\displaystyle xy'$ and $\displaystyle 3y^2$ terms over.

Just to give you a tip, it's easier if you keep track of your differentials in this situation, because the $\displaystyle 3y^2$ needs a $\displaystyle y'$ term that is easily lost in this situation, if you don't do so.

So now that you have $\displaystyle y=-xy'-3y^2y'$

You can factor out the $\displaystyle y'$ and solve for it, to get the final answer of $\displaystyle y'=\frac{y}{-x-3y^2}$

3. ## Confused...what happened ?

Originally Posted by Kalter Tod
Sorry not that simple..will correct and edit in a minute.

First, you wanna move the $\displaystyle xy'$ and $\displaystyle 3y^2$ terms over.

Just to give you a tip, it's easier if you keep track of your differentials in this situation, because the $\displaystyle 3y^2$ needs a $\displaystyle y'$ term that is easily lost in this situation, if you don't do so.

So now that you have $\displaystyle y=-xy'-3y^2y'$

You can factor out the $\displaystyle y'$ and solve for it, to get the final answer of $\displaystyle y'=\frac{y}{-x-3y^2}$

ok..
So the step with [x'*y+x*y']+[3y62]=0 is wrong?

what step did I miss ?

4. Originally Posted by kbgemini16
ok..
So the step with [x'*y+x*y']+[3y62]=0 is wrong?

what step did I miss ?
Err...No...you didn't miss anything. I just worked from that step...

Edit..if it helps clear it up, I did not include the $\displaystyle x'$ term in the factor of $\displaystyle x'y$ because it's more or less irrelevant when solving for $\displaystyle y'$

5. ## Ok

Originally Posted by kbgemini16
Ok... I have gotten this far with this problem:

xy+y^3=1

[x'*y+x*y'] +[3y^2]=0

My reasoning:
To get [x'*y+x+y'] I used the product rule
To get [3y^2] I used the product rule
to get "0" I used the constant rule

so... what's next ?

I know I am solving for y'

but can't figure out the next step

How to get y' alone one side or should it be taken away at this point ?

VERY VERY confused
$\displaystyle xy+y^3=1$ differentiating implicity we get $\displaystyle xy'+y+3y^2\cdot{y'}=0$ now solve for $\displaystyle y'$

6. Originally Posted by Mathstud28
$\displaystyle xy+y^3=1$ differentiating implicity we get $\displaystyle xy'+y+3y^2\cdot{y'}=0$ now solve for $\displaystyle y'$
Yeah, Stud. The OP had already gotten to that point. They wanted help going from there.

7. ## Is it ???

Originally Posted by Mathstud28
$\displaystyle xy+y^3=1$ differentiating implicity we get $\displaystyle xy'+y+3y^2\cdot{y'}=0$ now solve for $\displaystyle y'$

y'= -3y^2
x

8. Originally Posted by kbgemini16
y'= -3y^2
x
Check the work I already posted. I ended up with the negative in the denominator because I did the problem slightly different, but it's the same answer.