How do you find the derivative of #y= cos(x)/x^8#?

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Answer 1 · 2016-02-13T20:01:15

#y'=(-(xsin(x)+8cos(x)))/x^9#

Use the quotient rule, which states that, for this problem,

#y'=(x^8d/dx(cos(x))-cos(x)d/dx(x^8))/(x^8)^2#

The two derivatives are:

#d/dx(x^8)=8x^7" "" "#(Through the power rule.)

#d/dx(cos(x))=-sin(x)#

Plugging these both back in, we see that

#y'=(-x^8sin(x)-8x^7cos(x))/x^16#

Factor out a #-x^7# term.

#y'=(-x^7(xsin(x)+8cos(x)))/x^16#

#y'=(-(xsin(x)+8cos(x)))/x^9#