Answer
$\frac{{dy}}{{dx}} = 2x^2\cosh \left( {{x^2}} \right) + \sinh \left( {{x^2}} \right)$
Work Step by Step
$$\eqalign{
& y = x\sinh \left( {{x^2}} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x\sinh \left( {{x^2}} \right)} \right] \cr
& {\text{Use the product rule}} \cr
& \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left[ {\sinh \left( {{x^2}} \right)} \right] + \sinh \left( {{x^2}} \right)\frac{d}{{dx}}\left[ x \right] \cr
& {\text{Use }}\frac{d}{{dx }}\left[ {\sinh u} \right] = \cosh u\frac{{du}}{{dx}},{\text{ let }}u = {x^2} \cr
& \frac{{dy}}{{dx}} = x\cosh \left( {{x^2}} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] + \sinh \left( {{x^2}} \right)\frac{d}{{dx}}\left[ x \right] \cr
& {\text{Compute derivatives and simplify}} \cr
& \frac{{dy}}{{dx}} = \cosh \left( {{x^2}} \right)\left( {2x^2} \right) + \sinh \left( {{x^2}} \right)\left( 1 \right) \cr
& \frac{{dy}}{{dx}} = 2x^2\cosh \left( {{x^2}} \right) + \sinh \left( {{x^2}} \right) \cr} $$
