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

Answer 1

Ezra Adams

#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#

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Answer 2

Adam Adkins

To find the derivative of ( y = \frac{\cos(x)}{x^8} ), you can use the quotient rule, which states that for functions ( u(x) ) and ( v(x) ), the derivative of ( \frac{u(x)}{v(x)} ) is given by ( \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ). Applying this rule to ( y = \frac{\cos(x)}{x^8} ), the derivative is:

[ y' = \frac{-x^8 \sin(x) - 8x^7 \cos(x)}{(x^8)^2} ]

Simplifying further:

[ y' = \frac{-x^8 \sin(x) - 8x^7 \cos(x)}{x^{16}} ]

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