1. ## derivative

find all the points on the graph x^4+y^4+2=4xy^3

2. Originally Posted by eayettey
find all the points on the graph x^4+y^4+2=4xy^3

Do you mean graph it?

Since the title mentioned a derivative we can use implicit differentation to find it.

$\displaystyle 4x^3+4y^3\frac{dy}{dx}=4y^3+12xy^2\frac{dy}{dx}$

From here you can solve for $\displaystyle \frac{dy}{dx}$

3. i have this same problem and it says "Find all points on the graph x^4+y^2+2=4xy^3 at which the tangent line is horizontal" He forgot to add "the tangent line is horizontal". This same problem is getting me too

4. Originally Posted by gtgamer140@yahoo.com
i have this same problem and it says "Find all points on the graph x^4+y^2+2=4xy^3 at which the tangent line is horizontal" He forgot to add "the tangent line is horizontal". This same problem is getting me too
Ahhhh so now we can finish

So we know at the horizontal tangents that the derivative is zero

$\displaystyle \frac{dy}{dx}=0$

Now using the equation from above we get

$\displaystyle 4x^3+4y^3\frac{dy}{dx}=4y^3+12xy^2\frac{dy}{dx}$

$\displaystyle 4x^3=4y^3 \iff y=x$

Now if we plug this back into the original equation we get

$\displaystyle x^4+y^4+2=4xy^3$ setting $\displaystyle y=x$ we get

$\displaystyle x^4+x^4+2=4x^4 \iff 0=2x^4-2 \iff 0=2(x^4-1)$

$\displaystyle 0=2(x^2-1)(x^2+1) \iff 0 =2(x-1)(x+1)(x^2+1)$

so it happens when $\displaystyle x=\pm 1$

So when x=1 we get

$\displaystyle 1+y^4+2=4y^3 \iff y^4-4y^3+3=0$

Now by the rational roots theorem we know the only possible rational roots are $\displaystyle \pm1, \pm 3$

If you check on y=1 works so (1,1) is a horizontal tangent

I will leave the other cases for you to check.

Note there may be other Real roots