# Thread: Implicit differentiation - where'd the x' go?

1. ## Implicit differentiation - where'd the x' go?

Book example (and everywhere else online):
Differentiate the folium of descartes:

x^3 + y^3 = 6xy

Left side: 3x^2 + 3y^2 y'

Right side using product rule: 6xy' + 6y

Where did the x' go? (Ignoring the 6 for now):

The product rule is (fg)' = fg' + gf'
or:
(xy)' = xy' + yx'

"xy' + yx'" is not "xy' + y" -- Where the heck does the x' go?

2. Originally Posted by cnb
Book example (and everywhere else online):
Differentiate the folium of descartes:

x^3 + y^3 = 6xy

Left side: 3x^2 + 3y^2 y'

Right side using product rule: 6xy' + 6y

Where did the x' go? (Ignoring the 6 for now):

The product rule is (fg)' = fg' + gf'
or:
(xy)' = xy' + yx'

"xy' + yx'" is not "xy' + y" -- Where the heck does the x' go?
What is the first derivative of x with respect to x?

Or
$\displaystyle \left ( \frac{d}{dx} \right) x =$?

Or if that is still confusing, let p = x. Then what is p' ?

-Dan

3. From the book example, it would have to be 1
Also, treating d/dx (x) as a fraction would be dx/dx = 1

I am not all that clear on why this is though. It seems like it should be terribly obvious to me, but I am unclear on just what "the derivative of x with respect to x" really means, particularly the "with respect to" part.
Is there a way to say "d(x) with respect to q" in Newton's prime notation?

4. Originally Posted by cnb
From the book example, it would have to be 1
Also, treating d/dx (x) as a fraction would be dx/dx = 1

I am not all that clear on why this is though. It seems like it should be terribly obvious to me, but I am unclear on just what "the derivative of x with respect to x" really means, particularly the "with respect to" part.
Is there a way to say "d(x) with respect to q" in Newton's prime notation?

$\displaystyle x^{\prime} = \lim_{h \to 0}\frac{(x + h) - x}{h}$
$\displaystyle = \lim_{h \to 0}\frac{h}{h} = \lim_{h \to 0} 1 = 1$