1. ## implicit differentiation con't

Problem: $\displaystyle x^4(x+y)=y^2(3x-y)$

Solution attempt:

$\displaystyle \frac{d}{dx}[x^4(x+y)]-\frac{d}{dx}[y^2(3x-y)]=0$

$\displaystyle [4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3y-y')]=0$

upon distribution of coefficients,

$\displaystyle 4x^4+4x^3y+x^4+x^4y'-6xyy'+2y^2y'-3y^2+y^2y'=0$

If I didn't subtract $\displaystyle \frac{d}{dx}[y^2(3x-y)]$ as I did in step 1 I end up with,

$\displaystyle 4x^4+4x^3y+x^4+x^4y'=6xyy'+2y^2y'-3y^2+y^2y'$

This is where I am confronted with many mathematically valid yet counter productive steps as I try to solve for $\displaystyle y'$

combining like terms I have,

$\displaystyle 5x^4+4x^3y+x^4y'=6xyy'+3y^2y'-3y^2$

$\displaystyle 5x^4+4x^3y+x^4y'=y'(6xy+3y^2)-3y^2$

Answer in book: $\displaystyle y'=\frac{3y^2-5x^4-4x^3y}{x^4+3y^2-6xy}$

notes: what gets me is the fact that I have two $\displaystyle y'$ and $\displaystyle \frac{y'}{y'}=1$

2. i'm going to combine like termms and see where that gets me.

3. Originally Posted by Foxlion
i'm going to combine like termms and see where that gets me.
First, you have a sign error.

4. Originally Posted by Foxlion
Problem: $\displaystyle x^4(x+y)=y^2(3x-y)$

Solution attempt:

$\displaystyle \frac{d}{dx}[x^4(x+y)]-\frac{d}{dx}[y^2(3x-y)]=0$

$\displaystyle [4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3y-y')]=0$
$\displaystyle [4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3-y')]=0$

-Dan

5. $\displaystyle [4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3-y')]=0$

-Dan
is typo, following lines do not take it into account

6. I'll just springboard off of Dan's post (and yours...):
yes, you want to combine like terms. You have, among other things,
$\displaystyle x^4y' - 6xyy' + 2y^2y' +y^2y' = y'(x^4 - 6xy + 3y^2)$

After you move the rest of the terms to the other side and divide by
$\displaystyle (x^4 - 6xy + 3y^2)$, you'll have the desired result.

7. <sigh>...I got it...

$\displaystyle x^4(x+y) = y^2(3x-y)$
differentiating both sides...
$\displaystyle 4x^3(x+y)+x^4(1+y') = 2yy'(3x-y+y^2(3-y')$
distributing coefficients...
$\displaystyle 4x^4+4x^3y+x^4+x^4y' = 6xyy'-2y^2y'+3y^2-y^2y'$
combining like terms...
$\displaystyle 5x^4+4x^3y+x^4y' = 6xyy'-3y^2y'+3y^2$
subtracting $\displaystyle 3y^2$from both sides...
$\displaystyle 5x^4+4x^3y+x^4y'-3y^2 = 6xyy'-3y^2y'$
subtracting $\displaystyle x^4y'$ from both sides...
$\displaystyle 5x^4+4x^3y-3y^2 = 6xyy'-3y^2y'-x^4y'$
factoring out $\displaystyle y'$ from right...
$\displaystyle 5x^4+4x^3y-3y^2 = y'(6xy-3y^2-x^4)$
dividing both sides by $\displaystyle (6xy-3y^2-x^4)$...
$\displaystyle \frac{5x^4+4x^3y-3y^2}{6xy-3y^2-x^4} = y'$
now for some rearrangement as well as multiplying both sides by $\displaystyle \frac{-1}{-1}$...
$\displaystyle y' = \frac{3y^2-5x^4-4x^3y}{x^4+3y^2-6xy}$

Thank you both