Answer
$$\frac{{dy}}{{dx}} = {\left( {1 + {x^2}} \right)^{\sin x}}\left[ {\frac{{2x\sin x}}{{1 + {x^2}}} - \ln \left( {1 + {x^2}} \right)\left( {\cos x} \right)} \right]$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}{\left( {1 + {x^2}} \right)^{\sin x}} \cr
& {\text{let }}y = {\left( {1 + {x^2}} \right)^{\sin x}} \cr
& {\text{taking the natural logarithm of both sides of the equation}} \cr
& \,\,\,\ln y = \ln {\left( {1 + {x^2}} \right)^{\sin x}} \cr
& {\text{using logarithmic properties}} \cr
& \,\,\,\ln y = \sin x\ln \left( {1 + {x^2}} \right) \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {\sin x\ln \left( {1 + {x^2}} \right)} \right] \cr
& {\text{product rule}} \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \sin x\frac{d}{{dx}}\left[ {\ln \left( {1 + {x^2}} \right)} \right] - \ln \left( {1 + {x^2}} \right)\frac{d}{{dx}}\left[ {\sin x} \right] \cr
& {\text{compute derivatives}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \sin x\left( {\frac{{2x}}{{1 + {x^2}}}} \right) - \ln \left( {1 + {x^2}} \right)\left( {\cos x} \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{{2x\sin x}}{{1 + {x^2}}} - \ln \left( {1 + {x^2}} \right)\left( {\cos x} \right) \cr
& {\text{solving for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left[ {\frac{{2x\sin x}}{{1 + {x^2}}} - \ln \left( {1 + {x^2}} \right)\left( {\cos x} \right)} \right] \cr
& {\text{replace }}y{\text{ with }}{\left( {1 + {x^2}} \right)^{\sin x}} \cr
& \frac{{dy}}{{dx}} = {\left( {1 + {x^2}} \right)^{\sin x}}\left[ {\frac{{2x\sin x}}{{1 + {x^2}}} - \ln \left( {1 + {x^2}} \right)\left( {\cos x} \right)} \right] \cr} $$
