# Homework Help: Use implicit differentiation to find dy/dx [Answer check]

1. Sep 13, 2008

### lamerali

Use implicit differentiation to find $$\frac{dy}{dx}$$ for xy$$^{2}$$ – yx$$^{2}$$ = 3xy

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

x(2y)$$\frac{dy}{dx}$$ – y(2x) = 3xy
2xy $$\frac{dy}{dx}$$ – 2yx = 3xy
2xy$$\frac{dy}{dx}$$ = 5xy
$$\frac{dy}{dx}$$ = $$\frac{5xy}{2xy}$$
$$\frac{dy}{dx}$$ = 3xy

Thank you!

2. Sep 13, 2008

### Defennder

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

3. Sep 13, 2008

### lamerali

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

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

2xy$$\frac{dy}{dx}$$ - 2yx$$\frac{dy}{dx}$$ = 3xy

$$\frac{dy}{dx}$$ (2xy - 2yx) = 3xy

$$\frac{dy}{dx}$$ = $$\frac{3xy}{2xy - 2yx}$$

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

4. Sep 13, 2008

### danago

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.

5. Sep 13, 2008

### lamerali

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

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

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

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

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

Last edited: Sep 13, 2008
6. Sep 13, 2008

### 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.

7. Sep 13, 2008

### danago

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

8. Sep 13, 2008

### lamerali

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

9. Sep 13, 2008

### Dick

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.

10. Sep 13, 2008

### Almanzo

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)

11. Sep 13, 2008

### lamerali

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

12. Jun 24, 2010

### Arpit

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