Thread: Find the gradient of the normal

1. Find the gradient of the normal

This question, in a list of past examination questions, has me stuck. It says:

"For the curve $y= \sqrt{\frac{1-x}{1+x } }$,

Obtain an expression for $\frac{dy}{dx }$, by first differentiating $\frac{1-x}{1+x }$.

Thus, show that the gradient of the normal to the curve at the point $(x,y)$ is given by $(1+x)\sqrt{(1-x^2)}$."

Differentiating $\frac{1-x}{1+x }$ as told, I get $\frac{2x}{(1+x)^2}$.

Using this and the chain rule to differentiate y, I get $\frac{1}{2}\sqrt{(\frac{1-x}{1+x})}\frac{2x}{(1+x)^2}$.

Some algebraic gymnastics makes this simplify to $\frac{x}{(1+x)\sqrt{1-x^2}}$, 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 $\frac{1-x}{1+x }$ and that seems correct. Several friends have tried and failed as well. Is there something very obvious that we're missing?

2. Just realized that differentiating $(1-x)$ gives -1, not 1 as I thought. This eliminates the x term in the numerator. The littlest things get you sometimes!