1. ## implicit differentiation con't

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

Solution attempt:

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

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

upon distribution of coefficients,

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

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

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

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

combining like terms I have,

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

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

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

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

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

Solution attempt:

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

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

-Dan

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

-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,
$x^4y' - 6xyy' + 2y^2y' +y^2y' = y'(x^4 - 6xy + 3y^2)$

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

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

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

Thank you both