## Calculus: Early Transcendentals (2nd Edition)

$$f'\left( x \right) = {\sinh ^{ - 1}}x$$
\eqalign{ & f\left( x \right) = x{\sinh ^{ - 1}}x - \sqrt {{x^2} + 1} \cr & {\text{find the derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {x{{\sinh }^{ - 1}}x - \sqrt {{x^2} + 1} } \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {x{{\sinh }^{ - 1}}x} \right) - \frac{d}{{dx}}\left( {\sqrt {{x^2} + 1} } \right) \cr & {\text{by the product rule}} \cr & f'\left( x \right) = x\frac{d}{{dx}}\left( {{{\sinh }^{ - 1}}x} \right) + {\sinh ^{ - 1}}x\frac{d}{{dx}}\left( x \right) - \frac{d}{{dx}}\left( {\sqrt {{x^2} + 1} } \right) \cr & {\text{use derivatives of the inverse hyperbolic functions and the chain rule}} \cr & f'\left( x \right) = x\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right) + {\sinh ^{ - 1}}x\left( 1 \right) - \frac{{2x}}{{2\sqrt {{x^2} + 1} }} \cr & {\text{simplify}} \cr & f'\left( x \right) = \frac{x}{{\sqrt {{x^2} + 1} }} + {\sinh ^{ - 1}}x - \frac{x}{{\sqrt {{x^2} + 1} }} \cr & f'\left( x \right) = {\sinh ^{ - 1}}x \cr}