Derivative of y with Respect to x: Solving Implicit Differentiation Problems

[Twinflower]

Homework Statement

Derivate y with regards to x, \frac{dy}{dx}

2xy+y^2-4x=10

Homework Equations

The Attempt at a Solution

I am not very good at differentiation, so I haven't got off to a good start, really.
I guess I can differentiate y^2 and -4x "the usual way", but I am kind of stuck with the 2xy-part.

I also recon I have to put the equation = 0

 

[SammyS]

Homework Statement

Derivate y with regards to x, \frac{dy}{dx}

2xy+y^2-4x=10

Homework Equations

The Attempt at a Solution

I am not very good at differentiation, so I haven't got off to a good start, really.
I guess I can differentiate y^2 and -4x "the usual way", but I am kind of stuck with the 2xy-part.

I also recon I have to put the equation = 0

To find the derivative of 2xy, use the product rule.

 

[I like Serena]

Hi Twinflower!

How did you differentiate y²?

 

[Twinflower]

So (2x)'(y) + (2x)(y)' ---> y + 2x ?

That would yeld something like: y + 2x +2y - 1, but the answer is supposed to be
\frac{dy}{dx} = \frac{2-y}{x+y}

 

[Twinflower]

Hi Twinflower!

How did you differentiate y²?

(y^2)' = 2y

I think :)

 

[SammyS]

So (2x)'(y) + (2x)(y)' ---> y + 2x ?

(2x)' = 2, because x' = 1.

Where did y' disappear to?

That would yeld something like: y + 2x +2y - 1, but the answer is supposed to be
\frac{dy}{dx} = \frac{2-y}{x+y}

 

[I like Serena]

So (2x)'(y) + (2x)(y)' ---> y + 2x ?

That would yeld something like: y + 2x +2y - 1, but the answer is supposed to be
\frac{dy}{dx} = \frac{2-y}{x+y}

I'm afraid you can't simplify (2x)(y)'.
y' is what you want to find.

(y^2)' = 2y

I think :)

Not quite.
You should apply the chain rule here, and leave the y'.

 

[Mark44]

(y^2)' = 2y

I think :)

No.
d/dx(y²) = 2y\cdotdy/dx = 2y\cdoty'

You didn't use the chain rule.

 

[Twinflower]

Ok guys, I am going to sit down and try to work this out.
Thanks for pushing me in the right direction.

I'll be back later :)

 

[Twinflower]

OK, I finally got it:
2xy+y^2-4x=10

Differentiating each term:

\frac{dy}{dx}2xy = (2x)' \times y + 2x \times y' = 2y +2x \frac{dy}{dx}
\frac{dy}{dx}y^2 = 2y \frac{dy}{dx}
\frac{dy}{dx}-4x = -4
\frac{dy}{dx} 10 = 0

Combining the terms yelds:

2y +2x \frac{dy}{dx}+2y \frac{dy}{dx}-4=0

Diving each term by 2 and combining both y'-terms:

\frac{dy}{dx}\times (x+y) + y-2 = 0

Subtracting (y-2) on both sides:

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

Dividing by (x+y) yelds the final answer:

\frac{dx}{dy}=\frac{2-y}{x+y}Edit: Alcubierre posted the answer while I was creating mine :)

 

[Alcubierre]

Haha we posted pretty much at the same time. Congratulations though, good job!

 

[I like Serena]

Good!

Btw, you did make a mistake here:
\frac{dy}{dx}y^2 = 2y + \frac{dy}{dx}
which should be:
\frac{dy}{dx}y^2 = 2y \cdot \frac{dy}{dx}
But apparently you fixed that while moving along.

 

[Twinflower]

Ah, thanks Serena. Just a typo, it's not in my book.
Gonna fix my post to avoid any confusion in case others are trying to use this thread :)

 

[I like Serena]

Hey Twinflower.

You still got 1 typo left, which is the one I actually mentioned. ;)

 

[Twinflower]

Done!