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:
Hello, Showcase_22!
This requires Implicit Differentiation.
I hope you're familiar with it.
Find an expression in terms of and for
given that: .
We have: .
. . . . . . . . . . . . . . . . . . . .
Then: .
. . . . . Chain Rule . . Product Rule
Now solve for . . .
We have: .
Factor: .
Therefore: .
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?