## Calculus: Early Transcendentals (2nd Edition)

$$- 2\cos \left( {{x^{2\sin x}}} \right)\sin \left( {{x^{2\sin x}}} \right)\left( {x\cos x + \sin x} \right)$$
\eqalign{ & \frac{d}{{dx}}\left( {\cos \left( {{x^{2\sin x}}} \right)} \right) \cr & {\text{evaluate the derivative}} \cr & = - \sin \left( {{x^{2\sin x}}} \right)\frac{d}{{dx}}\left( {{x^{2\sin x}}} \right) \cr & {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr & = - \sin \left( {{x^{2\sin x}}} \right)\frac{d}{{dx}}\left( {{e^{{x^{2\sin x}}}}} \right) \cr & = - \sin \left( {{x^{2\sin x}}} \right)\frac{d}{{dx}}\left( {{e^{2x\sin x}}} \right) \cr & {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr & = - 2\sin \left( {{x^{2\sin x}}} \right){e^{2x\sin x}}\frac{d}{{dx}}\left( {x\sin x} \right) \cr & {\text{product rule}} \cr & = - 2\sin \left( {{x^{2\sin x}}} \right){e^{2x\sin x}}\left( {x\cos x + \sin x} \right) \cr & {\text{simplify}} \cr & = - 2\cos \left( {{x^{2\sin x}}} \right)\sin \left( {{x^{2\sin x}}} \right)\left( {x\cos x + \sin x} \right) \cr}