Answer
$t^{\prime}(x)=\displaystyle \frac{|x|}{x}-\frac{1}{x^{2}}$
Work Step by Step
SUMMARY:
The Power Rule$:\ \ \ [x^{n}]^{\prime}=nx^{n-1 } $
Sum Rule: $\ \ \ \ \ \ [f\pm g]^{\prime}(x)=f^{\prime}(x)\pm g^{\prime}(x) $
Constant Multiple Rule:$\ \ \ [cf]^{\prime}(x)=cf^{\prime}(x) $
Constant times x:$\ \ \ \displaystyle \frac{d}{dx}(cx)=c $
Constant:$\displaystyle \ \ \ \ \ \frac{d}{dx}(c)=0 $
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$|x|=\left\{\begin{array}{lll}
x, & x>0\\
-x, & x<0 &
\end{array}\right.$,
t(x) is not defined for x=0, so
$t(x)=\left\{\begin{array}{lll}
x+\frac{1}{x}, & x>0\\
-x+\frac{1}{x}, & x<0 &
\end{array}\right.$
For $x>0,\ t^{\prime}(x)=[x+\displaystyle \frac{1}{x}]^{\prime}=... $Sum Rule,
$=[x^{1}]^{\prime}+[x^{-1}]^{\prime}=$...Power Rule...$=1-x^{-2}=1-\displaystyle \frac{1}{x^{2}}$
For $x<0,\ t^{\prime}(x)=[-x+\displaystyle \frac{1}{x}]^{\prime}=... $Sum Rule,
$=[-x^{1}]^{\prime}+[x^{-1}]^{\prime}=$...Power Rule...$=-1-x^{-2}=-1-\displaystyle \frac{1}{x^{2}}$
So
$t^{\prime}(x)=\left\{\begin{array}{lll}
1-\frac{1}{x^{2}}, & x>0\\
-1-\frac{1}{x^{2}}, & x<0 &
\end{array}\right.$
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another way is to use $[|x|]^{\prime}=\displaystyle \frac{|x|}{x}$ (from sec.10.6):
$t^{\prime}(x)=[|x|+x^{-1}]^{\prime}=\displaystyle \frac{|x|}{x}-\frac{1}{x^{2}}$
which is equivalent to our first result
($\displaystyle \frac{|x|}{x}=+1, $when $x>0$, and (-1) when $x<0$)
