# Thread: Need help on Implicit Differentiation.

1. ## Need help on Implicit Differentiation.

The problem goes as such:

$\displaystyle 4xy-2x+4=0$

Evaluate y' at the point (-2,1)

What I did was (I use 'D' since I'm not sure on the proper notation at the moment, but I want to say it's d / dx)

$\displaystyle D(4xy)-D(2x)+D(4)=D(0)$

$\displaystyle 4y'-2+0=0$

$\displaystyle \frac{4y'}{4}-2+2+0=\frac{0+2}{4}$

$\displaystyle y'= \frac{2}{4}$

2. Originally Posted by Zanderist
The problem goes as such:

$\displaystyle 4xy-2x+4=0$

Evaluate y' at the point (-2,1)

What I did was (I use 'D' since I'm not sure on the proper notation at the moment, but I want to say it's d / dx)

$\displaystyle D(4xy)-D(2x)+D(4)=D(0)$

$\displaystyle 4y'-2+0=0$

$\displaystyle \frac{4y'}{4}-2+2+0=\frac{0+2}{4}$

$\displaystyle y'= \frac{2}{4}$

you have to use the product rule for the first term of the equation ...

$\displaystyle \frac{d}{dx}(4xy) = 4xy' + 4y$

3. What in this case is what is:
g(x)
g'(x)
f(x)
f'(x)

I assume f(x) is the first term.

4xy-2x+4=0

but I'm kinda lost after that.

4. Originally Posted by Zanderist
What in this case is what is:
g(x)
g'(x)
f(x)
f'(x)

I assume f(x) is the first term.

4xy-2x+4=0

but I'm kinda lost after that.

$\displaystyle \frac{d}{dx}(4xy - 2x + 4 = 0)$

$\displaystyle 4x \cdot y' + 4y - 2 = 0$

$\displaystyle 4x \cdot y' = 2 - 4y$

$\displaystyle 2x \cdot y' = 1 - 2y$

$\displaystyle y' = \frac{1-2y}{2x}$

5. How is 4y created?

6. Originally Posted by Zanderist
How is 4y created?
I told you in my first reply ... the product rule.

(4x) is the first function, (y) is the second.

$\displaystyle \frac{d}{dx} (4x)(y) = (4x)(y') + (4)(y)$

7. Okay... just a few more things:

$\displaystyle \frac{d}{dx}$

Means in respect to 'x' right?

And Implicit Differentiation is used to find a Z-axis? What is it's applications?

8. Originally Posted by Zanderist
Okay... just a few more things:

$\displaystyle \frac{d}{dx}$

Means in respect to 'x' right? yes

And Implicit Differentiation is used to find a Z-axis? no, its used to find a derivative.

What is it's applications? besides more advanced problems, your first application of implicit differentiation will most likely be solving related rates problems.
...

9. Originally Posted by skeeter

$\displaystyle 4x \cdot y' = 2 - 4y$

$\displaystyle 2x \cdot y' = 1 - 2y$
How does that top step become the bottom step?

4x becoming 2x, -4y becoming -2y, and finally 2 becoming a 1.

10. Originally Posted by Zanderist
How does that top step become the bottom step?

4x becoming 2x, -4y becoming -2y, and finally 2 becoming a 1.
$\displaystyle 4x \cdot y' = 2 - 4y$

$\displaystyle 4x \cdot y' = 2(1 - 2y)$

Now divide both sides of the equation by 2

11. Skeeter, I think I got it now! (for this problem type)

I was looking at what was going on in this post.

http://www.mathhelpforum.com/math-he...161-post5.html

$\displaystyle 4xy-2x+4=0$

Divide everything by 2:
$\displaystyle 2xy-x+2=0$
(I did this step because I thought it would be easier to get my end result )

Differentiate: $\displaystyle \frac {dy}{dx}(2xy-x+2)=0$
(To respect of 'x' of course)

Becoming:
$\displaystyle 2x \frac {dy}{dx} +2y - 1 = 0$

Next: $\displaystyle 2x \frac {dy}{dx} = 1-2y$

Final: $\displaystyle \frac {dy}{dx}=\frac {1-2y}{2x}$

Just tell me, did I do anything illegal.