I am a 16 year old student who is due to start college. I need help with this question!

Find an expression in terms of x and y for dy/dx, given that:

y^3+(3x^2)y-4x=0

Any help would be greatly appreciated.

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- Sep 6th 2006, 11:23 AMShowcase_22Differentiation
I am a 16 year old student who is due to start college. I need help with this question!

Find an expression in terms of x and y for dy/dx, given that:

y^3+(3x^2)y-4x=0

Any help would be greatly appreciated. - Sep 6th 2006, 11:34 AMTD!
If it's hard to solve to the form y = f(x), you can use implicit differentiation and solve for y'(x).

In general, this will depend on both x and y, but that's allowed here :)

To check your answer, I find:

$\displaystyle

y' = \frac{{dy}}{{dx}} = - \frac{{2\left( {3xy - 2} \right)}}{{3\left( {x^2 + y^2 } \right)}}

$ - Sep 6th 2006, 11:43 AMSoroban
Hello, Showcase_22!

This requires Implicit Differentiation.

I hope you're familiar with it.

Quote:

Find an expression in terms of $\displaystyle x$ and $\displaystyle y$ for $\displaystyle \frac{dy}{dx}$

given that: .$\displaystyle y^3 + 3x^2y - 4x\:=\:0$

We have: .$\displaystyle y^3 \quad+ \quad3x^2y \;\;- \;4x\;=\;0$

. . . . . . . .$\displaystyle \downarrow$ . . . . . $\displaystyle \downarrow$ . . . . .$\displaystyle \downarrow$ . . $\displaystyle \downarrow$

Then: .$\displaystyle 3y^2\frac{dy}{dx} + \overbrace{3x^2\frac{dy}{dx} + 6xy} - 4 \;= \;0$

. . . . . Chain Rule . . Product Rule

Now solve for $\displaystyle \frac{dy}{dx}$ . . .

We have: .$\displaystyle 3y^2\frac{dy}{dx} + 3x^2\frac{dy}{dx} \:=\:4 - 6xy$

Factor: .$\displaystyle (3y^2 + 3x^2)\frac{dy}{dx}\:=\:4-6xy$

Therefore: .$\displaystyle \frac{dy}{dx}\:=\:\frac{4-6xy}{3y^2 + 3x^2} $

- Sep 6th 2006, 11:44 AMShowcase_22
How do you differentiate like that?

this is what i'm doing:

y^3+ (3x^2)y-4x=0

3y^2(dy/dx)+(3x^2)(2x(dy/dx))+3y-4=0

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

(dy/dx)((3y^2)+6x^3)=4-3y

dy/dx=(4-3y)/((3y^2)+6x^3)

It looks a little confusing like this, but the answer I get is a little wrong. How would you solve this? - Sep 6th 2006, 11:48 AMShowcase_22
Wow, thanks for your reply!

I worked on that question for an hour this morning whilst on the bus, I thought it was impossible!