Just realized that differentiating gives -1, not 1 as I thought. This eliminates the x term in the numerator. The littlest things get you sometimes!
This question, in a list of past examination questions, has me stuck. It says:
"For the curve ,
Obtain an expression for , by first differentiating .
Thus, show that the gradient of the normal to the curve at the point is given by ."
Differentiating as told, I get .
Using this and the chain rule to differentiate y, I get .
Some algebraic gymnastics makes this simplify to , for the gradient of the tangent to the curve...
But I cannot eliminate the x in the numerator to arrive at the given expression for the gradient of the normal to the curve, even after checking all working carefully. It is produced by the differential to and that seems correct. Several friends have tried and failed as well. Is there something very obvious that we're missing?