Use implicit differentiation to find dy/dx [Answer check]

[lamerali]

Use implicit differentiation to find [tex]\frac{dy}{dx}[/tex] for xy[tex]^{2}[/tex] – yx[tex]^{2}[/tex] = 3xy

i've answered the question but i think I'm doing it wrong
any help is appreciated!

x(2y)[tex]\frac{dy}{dx}[/tex] – y(2x) = 3xy
2xy [tex]\frac{dy}{dx}[/tex] – 2yx = 3xy
2xy[tex]\frac{dy}{dx}[/tex] = 5xy
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{5xy}{2xy}[/tex]
[tex]\frac{dy}{dx}[/tex] = 3xy

Thank you!

 

[Defennder]

Your first line is the mistake. You forgot to differentiate the other term on the left.

 

[lamerali]

Ok I'm not sure if I've done any better this time but here it goes:

x(2y)[tex]\frac{dy}{dx}[/tex] - y(2x)[tex]\frac{dy}{dx}[/tex] = 3xy

2xy[tex]\frac{dy}{dx}[/tex] - 2yx[tex]\frac{dy}{dx}[/tex] = 3xy

[tex]\frac{dy}{dx}[/tex] (2xy - 2yx) = 3xy

[tex]\frac{dy}{dx}[/tex] = [tex]\frac{3xy}{2xy - 2yx}[/tex]

i'm not sure if the 2xy and 2yx would cancel each other out but it made more sense then dividing by a zero!
thank you

 

[danago]

Ok I'm not sure if I've done any better this time but here it goes:

x(2y)[tex]\frac{dy}{dx}[/tex] - y(2x)[tex]\frac{dy}{dx}[/tex] = 3xy

2xy[tex]\frac{dy}{dx}[/tex] - 2yx[tex]\frac{dy}{dx}[/tex] = 3xy

[tex]\frac{dy}{dx}[/tex] (2xy - 2yx) = 3xy

[tex]\frac{dy}{dx}[/tex] = [tex]\frac{3xy}{2xy - 2yx}[/tex]

i'm not sure if the 2xy and 2yx would cancel each other out but it made more sense then dividing by a zero!
thank you

You didnt take Defenner's advice; When you differentiate one side of the equals sign, you need to also differentiate the other side. So you need to differentiate the 3xy part too.

Also, be careful how you are differentiating xy² for example. You need to remember that xy² is the product of two functions, x and y². You need to use the product rule.

 

[lamerali]

Alright...i THINK i got it...lets see

y[tex]^{2}[/tex] + x(2y) [tex]\frac{dy}{dx}[/tex] - x[tex]^{2}[/tex] [tex]\frac{dy}{dx}[/tex] - y(2x) = 3y + 3x [tex]\frac{dy}{dx}[/tex]

x(2y)[tex]\frac{dy}{dx}[/tex] - x[tex]^{2}[/tex] [tex]\frac{dy}{dx}[/tex] - 3x [tex]\frac{dy}{dx}[/tex] = 3y - y[tex]^{2}[/tex] + y(2x)

[tex]\frac{dy}{dx}[/tex] (2yx - x[tex]^{2}[/tex] - 3x) = 3y - y[tex]^{2}[/tex] + 2xy

[tex]\frac{dy}{dx}[/tex] = (3y - y^2) / (-x^2 - 3x)
am I getting any closer to a correct answer?

 

[Dick]

You are getting there. But you can't 'cancel' the 2xy from numerator and denominator, can you? Back up one line and try again.

 

[danago]

Alright...i THINK i got it...lets see

y[tex]^{2}[/tex] + x(2y) [tex]\frac{dy}{dx}[/tex] - x[tex]^{2}[/tex] [tex]\frac{dy}{dx}[/tex] - y(2x) = 3y + 3x [tex]\frac{dy}{dx}[/tex]

x(2y)[tex]\frac{dy}{dx}[/tex] - x[tex]^{2}[/tex] [tex]\frac{dy}{dx}[/tex] - 3x [tex]\frac{dy}{dx}[/tex] = 3y - y[tex]^{2}[/tex] + y(2x)

[tex]\frac{dy}{dx}[/tex] (2yx - x[tex]^{2}[/tex] - 3x) = 3y - y[tex]^{2}[/tex] + 2xy

[tex]\frac{dy}{dx}[/tex] = (3y - y^2) / (-x^2 - 3x)
am I getting any closer to a correct answer?

Where did the 2xy from both the numerator and denominator go in the last step? Other than that, looks alright to me.

Edit: Looks like Dick beat me to it :p

 

[lamerali]

Ok great! i'll pop them back into the equation!
Thank you guys!

 

[Dick]

Ok great! i'll pop them back into the equation!
Thank you guys!

Fine. As long as you understand why you can't cancel that way and promise not to do it again. (1+2)/(3+2)=3/5, but it turns into 1/3 if you 'cancel' the 2.

 

[Almanzo]

I could not read your picture, so perhaps this is superfluous.

xy² - x²y - 3xy = 0
dx/dx = 1
d/dx(xy² - x²y - 3xy) = d0/dx = 0
y² + 2xy(dy/dx) - 2xy - x²(dy/dx) - 3y - 3x(dy/dx)=0
y² - 2xy - 3y = x²(dy/dx) + 3x(dy/dx) - 2xy(dy/dx)
dy/dx = (y² - 2xy - 3y)/(x² + 3x - 2xy)

 

[lamerali]

As long as you understand why you can't cancel that way and promise not to do it again.

I understand Thank you I promise I will try really hard not to do it again!
Take care =)

 

[Arpit]

use implicit differentiation to find [tex]\frac{dy}{dx}[/tex] for xy[tex]^{2}[/tex] – yx[tex]^{2}[/tex] = 3xy

i've answered the question but i think I'm doing it wrong
any help is appreciated!

X(2y)[tex]\frac{dy}{dx}[/tex] – y(2x) = 3xy
2xy [tex]\frac{dy}{dx}[/tex] – 2yx = 3xy
2xy[tex]\frac{dy}{dx}[/tex] = 5xy
[tex]\frac{dy}{dx}[/tex] = [tex]\frac{5xy}{2xy}[/tex]
[tex]\frac{dy}{dx}[/tex] = 3xy

thank you!

u forgot the use of product rule here...and therefore...your answer is wrong...!