Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = 4{x^2}{\sin ^3}x\cos x + 2x{\sin ^4}x + 2x{\cos ^{ - 3}}x\sin x + {\cos ^{ - 2}}x$$
\eqalign{ & y = {x^2}{\sin ^4}x + x{\cos ^{ - 2}}x \cr & y = {x^2}{\left( {\sin x} \right)^4} + x{\left( {\cos x} \right)^{ - 2}} \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^2}{{\left( {\sin x} \right)}^4}} \right] + \frac{d}{{dx}}\left[ {x{{\left( {\cos x} \right)}^{ - 2}}} \right] \cr & {\text{use the product rule for each term}} \cr & \frac{{dy}}{{dx}} = {x^2}\frac{d}{{dx}}\left[ {{{\left( {\sin x} \right)}^4}} \right] + {\left( {\sin x} \right)^4}\frac{d}{{dx}}\left[ {{x^2}} \right] + x\frac{d}{{dx}}\left[ {{{\left( {\cos x} \right)}^{ - 2}}} \right] + {\left( {\cos x} \right)^{ - 2}}\frac{d}{{dx}}\left[ x \right] \cr & {\text{solve derivatives using the chain rule for }} \cr &\frac{d}{{dx}}\left[ {{{\left( {\sin x} \right)}^4}} \right]{\text{ and }}\frac{d}{{dx}}\left[ {{{\left( {\cos x} \right)}^{ - 2}}} \right] \cr & \frac{{dy}}{{dx}} = {x^2}\left( 4 \right){\left( {\sin x} \right)^3}\frac{d}{{dx}}\left[ {\sin x} \right] + {\left( {\sin x} \right)^4}\left( {2x} \right) + x\left( { - 2} \right){\left( {\cos x} \right)^{ - 3}}\frac{d}{{dx}}\left[ {\cos x} \right] + {\left( {\cos x} \right)^{ - 2}}\left( 1 \right) \cr & \frac{{dy}}{{dx}} = {x^2}\left( 4 \right){\left( {\sin x} \right)^3}\left( {\cos x} \right) + {\left( {\sin x} \right)^4}\left( {2x} \right) + x\left( { - 2} \right){\left( {\cos x} \right)^{ - 3}}\left( { - \sin x} \right) + {\left( {\cos x} \right)^{ - 2}} \cr & {\text{simplify}} \cr & \frac{{dy}}{{dx}} = 4{x^2}{\sin ^3}x\cos x + 2x{\sin ^4}x + 2x{\cos ^{ - 3}}x\sin x + {\cos ^{ - 2}}x \cr}