[RESOLVED] Implicit Differentiation

This is a problem out of Shaum's Outlines on Numerical Methods. Just some spare time fun.
Differentiate the following function wrt x (i.e. find y'):

x³y⁴ + 2y = 3x

The solution from the back of the book is y' = y(1-x²y⁴) / x(1+x²y⁴))

I know the solution is right because I checked it using RK4. I didn't think my differentiation skills were THAT rusty, but I come up with something different.

Differentiating both sides I get:

(1) 3x²y⁴ + 4x³y³y' + 2y' = 3

(2) y' (4x³y³ + 2) = (3 - 3x²y⁴)

(3) y' = (3 - 3x²y⁴) / (4x³y³ + 2)

Can anyone point out my error(s)? Thanks. :sick:

Interesting problem you have there!

When I differentiated I got exactly the same answer as you.

You could also:
- rearrange (the original equation) to x^3.y^3 = 3x-2y
- (in your expression for y') factor out a y in the numerator and an x in the denominator
- Substitute x^3.y^3 = 3x-2y

This yeilds

y' = (6y - 6x)/(12x - 6y)

But I can see no way of getting the other solution as yet.

All the best, Matt

Turns out both expressions for y' are equivalent! :eek: I verified that RK4 solutions for both are the same. Also, if the 2 expressions for y' are the same, then setting them equal to each other ought to yield something:

If x³y⁴ + 2y = 3x yields the following for y':

y' = y(1 - x²y⁴) / x(1 + x²y⁴)) or y' = (3 - 3x²y⁴) / (4x³y³ + 2)

then

(1)... y(1 - x²y⁴) / x(1 + x²y⁴)) = (3 - 3x²y⁴) / (4x³y³ + 2) ----- set derivatives equal to each other

(2)... y(1 - x²y⁴) / x(1 + x²y⁴)) = 3(1 - x²y⁴) / 2(2x³y³ + 1) ----- factor right side

(3)... y(1 - x²y⁴) / x(1 + x²y⁴)) = y(1 - x²y⁴) / (2/3)y(1 + 2x³y⁴) ----- multiple right side by y/y

(4)... x(1 + x²y⁴) = (2/3)y(1 + 2x³y⁴) ----- since numerators are the same, denominators equal each other

(5)... x + x³y⁴ = (2/3)y + (4/3)x³y⁴ = (2/3)y + x³y⁴ + (1/3)x³y⁴ ----- multiply out

(6)... 3x = 2y + x³y⁴ ----- subtract x³y⁴ from both sides and multiply by 3

qed :cool:

Nice!!