Find the equation of the normal to the curve y= x(1-2x)^{1/2 }at the point (-4,-12)
y= x(1-2x)^{1/2 }dy/dx = 1/2 (1-2x)^{-1/2}
do i sub in -4 now? If so, am i getting the gradient of the normal or the curve itself?
Your derivation is wrong. Use product rule
d(uv)/dx = u dv/dx + v du/dx.
the derivative comes out to be dy / dx = (1-3x)/(1-2x)^1/2
Now get dy/dx at x = -4
that is the slope of tangent at x = -4.
Negative reciprocal of this slope will give us slope of normal to the curve at x = -4.
Thus slope of normal would be -1/[dy/dx] = - (1-2x)^1/2 / (1-3x)
Now you can proceed further