Answer
$$y' = - 10{\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)^9}\left( {\frac{1}{{\sqrt {1 + {x^2}} }} - {{\operatorname{csch} }^{ - 1}}x} \right)$$
Work Step by Step
$$\eqalign{
& y = {\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)^{10}} \cr
& {\text{find the derivatvive}} \cr
& y' = \left( {{{\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)}^{10}}} \right)' \cr
& {\text{chain rule}} \cr
& y' = 10{\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)^{10 - 1}}\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)' \cr
& y' = 10{\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)^9}\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)' \cr
& {\text{product rule for }}x{\operatorname{csch} ^{ - 1}}x \cr
& y' = 10{\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)^9}\left( {x\left( {{{\operatorname{csch} }^{ - 1}}x} \right)' + {{\operatorname{csch} }^{ - 1}}x\left( x \right)'} \right) \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.5 }}\left( {{\text{ see page 479}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {{{\operatorname{csch} }^{ - 1}}u} \right] = - \frac{1}{{\left| u \right|\sqrt {1 + {u^2}} }}\frac{{du}}{{dx}} \cr
& y' = 10{\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)^9}\left( {x\left( { - \frac{1}{{\left| x \right|\sqrt {1 + {x^2}} }}} \right) + {{\operatorname{csch} }^{ - 1}}x\left( 1 \right)} \right) \cr
& {\text{simplifying}} \cr
& y' = 10{\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)^9}\left( { - \frac{1}{{\sqrt {1 + {x^2}} }} + {{\operatorname{csch} }^{ - 1}}x} \right) \cr
& y' = - 10{\left( {1 + x{{\operatorname{csch} }^{ - 1}}x} \right)^9}\left( {\frac{1}{{\sqrt {1 + {x^2}} }} - {{\operatorname{csch} }^{ - 1}}x} \right) \cr} $$
