Answer
$$\frac{{dy}}{{dx}} = 2{\sinh ^{ - 1}}\left( {2x} \right)$$
Work Step by Step
$$\eqalign{
& y = 2x{\sinh ^{ - 1}}\left( {2x} \right) - \sqrt {1 + 4{x^2}} \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {2x{{\sinh }^{ - 1}}\left( {2x} \right)} \right] - \frac{d}{{dx}}\left[ {\sqrt {1 + 4{x^2}} } \right] \cr
& {\text{Use product rule}} \cr
& \frac{{dy}}{{dx}} = 2x\frac{d}{{dx}}\left[ {{{\sinh }^{ - 1}}\left( {2x} \right)} \right] + 2{\sinh ^{ - 1}}\left( {2x} \right)\frac{d}{{dx}}\left[ x \right] - \frac{d}{{dx}}\left[ {\sqrt {1 + 4{x^2}} } \right] \cr
& {\text{Use Theorem 5}}{\text{.20 }}\frac{d}{{dx}}\left[ {{{\sinh }^{ - 1}}u} \right] = \frac{{u'}}{{\sqrt {{u^2} + 1} }} \cr
& \frac{{dy}}{{dx}} = 2x\left( {\frac{2}{{\sqrt {{{\left( {2x} \right)}^2} + 1} }}} \right) + 2{\sinh ^{ - 1}}\left( {2x} \right)\left( 1 \right) - \frac{{8x}}{{2\sqrt {1 + 4{x^2}} }} \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{4x}}{{\sqrt {4{x^2} + 1} }} + 2{\sinh ^{ - 1}}\left( {2x} \right) - \frac{{4x}}{{2\sqrt {1 - 4{x^2}} }} \cr
& \frac{{dy}}{{dx}} = 2{\sinh ^{ - 1}}\left( {2x} \right) \cr} $$
