1. ## Implicit differentiation

using the implicit differentiation, find y' (or dy/dx) of y^4 - 3xy + x = y^3

4y^3 - 3y - 3x + 1 = 3y^2

dy/dx(4y^3 - 3y - 3x) + 1 = 3y^2

dy/dx(4y^3 - 3y - 3x) = 3y^2 - 1

dy/dx = (3y^2 - 1)/(4y^3 - 3y - 3x)

what did i do wrong? =/

the answer is supposed to be (3y - 1)/(4y^3 - 3y^2 - 3x)

2. Originally Posted by jeph
using the implicit differentiation, find y' (or dy/dx) of y^4 - 3xy + x = y^3

4y^3 - 3y - 3x + 1 = 3y^2

dy/dx(4y^3 - 3y - 3x) + 1 = 3y^2

dy/dx(4y^3 - 3y - 3x) = 3y^2 - 1

dy/dx = (3y^2 - 1)/(4y^3 - 3y - 3x)

what did i do wrong? =/

the answer is supposed to be (3y - 1)/(4y^3 - 3y^2 - 3x)
Hello,

First you have to rearrange your equation:
$\displaystyle f(x,y)=y^4-y^3-3xy+x$
Then I use this formula:

$\displaystyle y'=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$

$\displaystyle \frac{\partial f}{\partial x}=-3y+1$. And
$\displaystyle \frac{\partial f}{\partial y}=4y^3-3y^2-3x$

Plug in these results in the formula mentioned above and you'll get the given result.

EB

3. Originally Posted by jeph
using the implicit differentiation, find y' (or dy/dx) of y^4 - 3xy + x = y^3

4y^3 - 3y - 3x + 1 = 3y^2

dy/dx(4y^3 - 3y - 3x) + 1 = 3y^2

dy/dx(4y^3 - 3y - 3x) = 3y^2 - 1

dy/dx = (3y^2 - 1)/(4y^3 - 3y - 3x)

what did i do wrong? =/

the answer is supposed to be (3y - 1)/(4y^3 - 3y^2 - 3x)
An alternative to Earboth's post is:
$\displaystyle y^4 - 3xy + x = y^3$

$\displaystyle 4y^3y' - 3y - 3xy' + 1 = 3y^2y'$

$\displaystyle 4y^3y' - 3y^2y' -3xy' = 3y - 1$

$\displaystyle (4y^3 - 3y^2 -3x)y' = 3y - 1$

$\displaystyle y' = \frac{3y - 1}{4y^3 - 3y^2 -3x}$

Remember, when you take differentiate a function (actually take the derivative) of a function f(y) of y you need to use the chain rule:
$\displaystyle f'(y) \cdot y'$

-Dan

4. so when you differentiate 3xy you get 3y+3xy', but you don't change the y' to a 1? I think I got it...

how do you gets write the math equations like that? It's so much easier to look at. Is it easier to type in the math as well that way?

5. Originally Posted by jeph
...
how do you gets write the math equations like that? It's so much easier to look at. Is it easier to type in the math as well that way?
Hello, Jeph,

have a look here: http://www.mathhelpforum.com/math-he...rial-latex.pdf

EB