## anonymous 5 years ago i mean sin (x)^y

$dy/dx= y * \sin^{y-1}(x)*d(\sin(x))/dx + \sin^y(x)*\ln(y)*dy/dx$ - that second term comes from d(u^x)/dx = u^x * ln(u). Therefore, $dy/dx = y * \sin^{y-1}(x)*\cos(x)+\sin^y(x)*\ln(y)*dy/dx.$