The quotient #\frac[f(x)}{g(x)}# of two derivable functions #f(x)# and #g(x)# is derivable for all #x# in which #g(x)\ne 0#. And its derivative is equal to:
#frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{(g(x))^2}#
So, going back to the exercise we have:
#f(x)=\sin(x)#
#g(x)=x^8#
Thus:
#frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{x^8\cdot (-\sin(x)) - \cos(x)\cdot 8x^7}{(x^8)^2}#
#frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{-x^8\sin(x)-8x^7\cos(x)}{x^16}#
Factoring:
#frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{-x^7(x\sin(x)+8\cos(x))}{x^16}#
Reducing #frac{x^7}{x^16}# we finally get:
#frac{d}{dx}(\frac[f(x)}{g(x)})=-\frac{x\sin(x)+8\cos(x)}{x^9}#
For the second derivative your functions #f(x)# and #g(x)# will be:
#f(x)=-(x\sin(x)+8\cos(x))#
#g(x)=x^9#
Try doing it!
