How to find the normal to a curve at a specific point? Can someone walk me through how to find the normal to a curve at a specific point? Taking one example from my Calculus Textbook, it states Find the equation of the normal to the curve $sqrt(y)+xy^2=5$ at the point (4,1)?

Now, I know a Normal is perpendicular to a tangent line, and has a reverse-signed reciprocal slope to our tangent. But beyond the general knowledge there, how do I go about solving this as asked?
There are two basic approaches here:


Rewrite the defining equation for your curve to make one coordinate a function of the other. In this case it is easy to reach $$ x = \frac{5-\sqrt y}{y^2}$$ which you can differentiate symbolically to find a tangent vector and then produce the equation for the normal (goes through the specified point, and perpendicular to the tangent).
Know that the normal to a level curve $f(x,y)=c$ is always in the direction of the gradient of $f$, so compute $(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ at $(4,1)$ construct a line through $(4,1)$ in that direction.
We call $ \mathbf {r} $ the curve. To find the normal vector in a point $ p $ of the curve you derive the vector$ \mathbf {r'}/||\mathbf {r'}||$  and calculate in $ p $ the derivate. In your case first find a parametrization of the curve.