This is what my book says:

If C is given by the vector equation $\displaystyle \vec{r}(t)=x(t)\vec{i}+y(t)\vec{j} $ $\displaystyle a \leq t \leq b$

then the unit tangent vector is $\displaystyle \vec{T}(t)= \frac{x'(t)}{|\vec{r}'(t)|}\vec{i} + \frac{y'(t)}{|\vec{r}'(t)|}\vec{j}$

You can verify that the outward unit normal vector to C is given by $\displaystyle \vec{n}(t) = \frac{y'(t)}{|\vec{r}'(t)|}\vec{i} - \frac{x'(t)}{|\vec{r}'(t)|}\vec{j}$

I don't get how that's true. I tried proving it using the fact that $\displaystyle \vec{n}(t)=\frac{\vec{T}'(t)}{|\vec{T}'(t)|}$ and all I got was this:

$\displaystyle \vec{n}(t)=\frac{\frac{d}{dt}\frac{x'(t)}{|\vec{r} '(t)|} + \frac{d}{dt}\frac{y'(t)}{|\vec{r}'(t)|}}{\sqrt{(\f rac{d}{dt}\frac{x'(t)}{|\vec{r}'(t)|})^2+ \frac{d}{dt}\frac{y'(t)}{|\vec{r}'(t)|}^2}}$

and I don't know how to simplify that to $\displaystyle \vec{n}(t) = \frac{y'(t)}{|\vec{r}'(t)|}\vec{i} - \frac{x'(t)}{|\vec{r}'(t)|}\vec{j}$

Thanks in advance!