# Thread: Find dy/dx by implicit differentiation

1. ## Find dy/dx by implicit differentiation

Hello,

I have

2x^3 + x^2 Y - xy^3 = 2

I got
6x^2 + x^2 y' +2yx - x^3 yy' + y^3 = 0

6x^2 + 2yx + y^3 = x^3 yy' - x^2 y'

6x^2 + 2yx + y^3 = y'(x^3 y - x^2)

y' = (6x^2 + 2yx + y^3) / (x^3 y - x^2)

y' = (6x^2 + 2yx + y^3) / x^2(xy-1)

My question, is that right and can I go further, especially with the numerator? I can not check my answer, since I did not have the solution.

Thanks

2. Originally Posted by DBA
Hello,

I have

2x^3 + x^2 Y - xy^3 = 2

I got
6x^2 + x^2 y' +2yx - x^3 yy' + y^3 = 0 (e^(i*pi)) - you should still differentiate x terms. This requires the product rule)

6x^2 + 2yx + y^3 = x^3 yy' - x^2 y'

6x^2 + 2yx + y^3 = y'(x^3 y - x^2)

y' = (6x^2 + 2yx + y^3) / (x^3 y - x^2)

y' = (6x^2 + 2yx + y^3) / x^2(xy-1)

My question, is that right and can I go further, especially with the numerator? I can not check my answer, since I did not have the solution.

Thanks
$\displaystyle 2x^3 + x^2y - xy^3 = 2$

$\displaystyle 2x^3$ is obvious

$\displaystyle x^2y$ and $\displaystyle xy^3$ require the product rule:

Spoiler:
For $\displaystyle x^2y$

$\displaystyle u = x^2 \: \longrightarrow \: u' = 2x$

$\displaystyle v = y \: \longrightarrow \: v' = \frac{dy}{dx}$

$\displaystyle \frac{d}{dx}(x^2y) = u'v + v'u = 2xy + x^2\frac{dy}{dx}$

For $\displaystyle xy^3$

$\displaystyle u = x \: \longrightarrow \: u' = 1$

$\displaystyle v = y^3 \: \longrightarrow \: v' = 3y^2 \frac{dy}{dx}$

$\displaystyle \frac{d}{dx}(xy^3) = y^3 + 3xy^2 \frac{dy}{dx}$

Therefore differentiating it all gives:

$\displaystyle 6x^2 + 2xy + x^2\frac{dy}{dx} - (y^3 + 3xy^2 \frac{dy}{dx}) = 0$

Collect any $\displaystyle \frac{dy}{dx}$ terms onto one side

$\displaystyle x^2 \frac{dy}{dx} - 3xy^2 \frac{dy}{dx} = y^3-2xy-6x^2$

Factor:

$\displaystyle \frac{dy}{dx}(x^2-3xy^2) = y^3-2xy-6x^2$

Isolate $\displaystyle \frac{dy}{dx}$:

$\displaystyle \frac{dy}{dx} = \frac{y^3-2xy-6x^2}{x^2-3xy^2}$

Which to my knowledge cannot be simplified.

3. Great, thank you so much, I found what I did wrong.