Show that if $y=\sinh ^{-1}x$, then $$\boxed{ {y^\prime =\dfrac{d}{dx}\sinh ^{-1}x=\dfrac{1}{\ \sqrt{x^{2}+1}}.}}$$

Solution Since $y=\sinh ^{-1}x=\ln \big( x+\ \sqrt{x^{2}+1}\big)$, we have $$\begin{eqnarray*} y^\prime &=&\dfrac{d}{dx}\sinh ^{-1}x=\dfrac{d}{dx}\left[ \ln \big( x+\ \sqrt{x^{2}+1}\big) \right] \underset{\underset{\color{#0066A7}{\hbox{\(\tfrac{d}{dx}\ln (u)=\tfrac{u^{\,\prime} ( x) }{u( x) }\)}}}{\color{#0066A7}{\uparrow }}}{=} \dfrac{\dfrac{d}{dx}\big( x+\ \sqrt{x^{2}+1}\big) }{x+\ \sqrt{x^{2}+1}}\\ &=&\dfrac{1+\dfrac{1}{2}(x^{2}+1) ^{-1/2}\cdot 2x}{x+\ \sqrt{x^{2}+1}}=\dfrac{1+\dfrac{x}{\ \sqrt{x^{2}+1}}}{x+\ \sqrt{x^{2}+1}}=\dfrac{\dfrac{\ \sqrt{ x^{2}+1}+x}{\ \sqrt{x^{2}+1}}}{x+\ \sqrt{x^{2}+1}}\\ &=&\dfrac{x+\ \sqrt{x^{2}+1}}{ ( x+\ \sqrt{x^{2}+1}) \ \sqrt{x^{2}+1}}=\dfrac{1}{\ \sqrt{x^{2}+1}} \end{eqnarray*}$$