Find an equation for the tangent and normal to the curve x^2+y^2=9 at x=1
Do you have to simplify the problem to get y on one side.
or
is your u=x^2 v=y^2
du=2x dv=2y
if so then what next?
Differentiate implicitly both sides of the equation with respect to x, to get dy/dx. Treat y as a function of x.
2x + 2y*(dy/dx) = 0
When x=1, y^2 = 8 y = 2.828
2 + 5.856*(dy/dx)= 0
dy/dx = -0.342
That is the slope of the tangent line. The normal at that point has a slope that is -1 divided by that.