## Calculus: Early Transcendentals (2nd Edition)

$$\frac{{dy}}{{dx}} = - \frac{{3{{\operatorname{csch} }^2}3x}}{{2\sqrt {\coth 3x} }}$$
\eqalign{ & y = \sqrt {\coth 3x} \cr & {\text{rewriting}} \cr & y = {\left( {\coth 3x} \right)^{1/2}} \cr & {\text{computing }}dy/dx \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}{\left( {\coth 3x} \right)^{1/2}} \cr & {\text{by the chain rule}} \cr & \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\coth 3x} \right)^{1/2 - 1}}\frac{d}{{dx}}\left( {\coth 3x} \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\coth 3x} \right)^{1/2 - 1}}\left( { - {{\operatorname{csch} }^2}3x} \right)\frac{d}{{dx}}\left( {3x} \right) \cr & \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {\coth 3x} \right)^{ - 1/2}}\left( { - {{\operatorname{csch} }^2}3x} \right)\left( 3 \right) \cr & {\text{multiplying}} \cr & \frac{{dy}}{{dx}} = - \frac{{3{{\operatorname{csch} }^2}3x{{\left( {\coth 3x} \right)}^{ - 1/2}}}}{2} \cr & \frac{{dy}}{{dx}} = - \frac{{3{{\operatorname{csch} }^2}3x}}{{2{{\left( {\coth 3x} \right)}^{ - 1/2}}}} \cr & \frac{{dy}}{{dx}} = - \frac{{3{{\operatorname{csch} }^2}3x}}{{2\sqrt {\coth 3x} }} \cr}