Find normal and tangent vector to a curve Find a normal vector and a tangent vector to the curve given by the equation: $x^5 + y ^5 =2x^3$ at the point $P(1, 1)$. Find the equation of the tangent line. 
Edit: The notes I have:
Taking $f(x, y) = x^5 - 2x^3 + y^5 = 0$, I got $m_n = (-1,5), m_t = (5, 1)$ and $x = 5y-4$ for the tangent line. 
I'm very much unsure on if this formula is what I should/can use and if I've used it correctly. 
 A: You can find the slope of the tangent line by differentiating. If you don't want to do implicit differentiation (which may be simpler in this case), you can just do some algebra beforehand:
$$x^5+y^5=2x^3 \to y^5=2x^3-x^5 \to y=\sqrt[5]{2x^3-x^5}$$
and differentiate according to the power rule. You have the slope of your tangent line; knowing that it goes through $(1,1)$, you should have enough information to solve for that.
The tangent vector will have a slope exactly the same as that of the tangent line. The normal vector will have a slope that is the negative inverse of that of the tangent vector. If $m_t$ is the slope of the tangent vector, the slope $m_n$ of the normal vector will be $-\frac{1}{m_t}$.
In the method you mentioned, you really just have to find $c$ here to match it up with that equation. You would just re-arrange the equation so that all the constants are on one side of the equation. Here, we find that
$$F(x,y)=x^5-2x^3+y^5=c$$
and $c=0$.
A: Differentiate explicitly:
$$5x^4 + 5y^4y' = 6x^2$$
Solve for $y'$
