Given the vector function $\displaystyle r(t)=<\sqrt{2}t,e^t,e^{-t}>$ I need to find the tangent unit vector $\displaystyle T(t)=\frac{r'(t)}{\sqrt{x'^2+y'^2+z'^2}}$. The denominator is the absolute value of the vector derivative $\displaystyle r'(t)=<\sqrt{2},e^t,-e^{-t}>$. I write an expression for $\displaystyle T(t)$:

$\displaystyle T(t)=\frac{1}{\sqrt{2+e^{2t}+e^{-2t}}}<\sqrt{2},e^t,-e^{-t}>$

$\displaystyle =\frac{1}{\sqrt{2(1+\frac{e^{2t}+e^{-2t}}{2}})}<\sqrt{2},e^t,-e^{-t}>$

$\displaystyle =\frac{1}{\sqrt{(1+cosh2t)}}(i+\frac{e^t}{\sqrt{2} }j-\frac{e^{-t}}{\sqrt{2}}k)$

I don't think this could be wrong since it's pretty straight forward. It requires zero intelligence, you just apply the definition directly. However, I can't seem to obtain the same expression given in the book:

$\displaystyle T(t)=\frac{1}{e^{2t}+1}<\sqrt{2}e^t,e^{2t},-1>$

I can see that $\displaystyle e^tr'(t)=<\sqrt{2}e^t,e^{2t},-1>$, which suggests that they just multiplied the numerator and denominator of $\displaystyle T(t)$ by $\displaystyle e^t$. However I can't get the demonator in the form $\displaystyle e^{2t}+1$. I'm confused by the algebra, I can't show that $\displaystyle e^t\sqrt{2+e^{2t}+e^{-2t}}=e^{2t}+1$.

????