implicit differentiation con't

**Foxlion** · March 10th 2011, 12:35 PM

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 $\frac{y'}{y'}=1$

**Foxlion** · March 10th 2011, 01:17 PM

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

**TheChaz** · March 10th 2011, 01:21 PM

Originally Posted by Foxlion

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

First, you have a sign error.

**topsquark** · March 10th 2011, 01:30 PM

Originally Posted by Foxlion

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

Solution attempt:

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

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

The last line should read
$[4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3-y')]=0$

-Dan

**Foxlion** · March 10th 2011, 01:34 PM

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

-Dan

is typo, following lines do not take it into account

**TheChaz** · March 10th 2011, 01:37 PM

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.

**Foxlion** · March 10th 2011, 02:24 PM

<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

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