You actually have two ways of approaching this derivative.

First, notice that your function #y# can be written as a *quotient* of two other functions

#y = f(x)/g(x)#. where

#{(f(x) = cosx), (g(x) = x^8) :}#

Likewsie, you can write #y# as a *product* of two functions

#y= f(x) * h(x)#, where

#{(f(x) = cosx), (h(x) = x^(-8)) :}#

You can thus differentiate this function by using

The quotient rule allows you to calculate the derivative of a function expressed as a quotient of two other functions by using the

#color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2)#, where #g(x)!=0#

In your case, the derivative of #y# will be

#d/dx(y) = ([d/dx(cosx)] * x^8 - cosx * d/dx(x^8))/(x^8)^2#

#d/dx(y) = (-sinx * x^8 - cosx * 8x^7)/x^16#

#d/dx(y) = -(color(red)(cancel(color(black)(x^7))) ( x * sinx + 8cosx))/x^(color(red)(cancel(color(black)(16)))color(blue)(9))#

#d/dx(y) = color(green)(-(x * sinx + 8 * cosx)/x^9)#

This time ,the derivative of #y# can be determined using the formula

#color(blue)(d/dx(y) = f^'(x) * h(x) + f(x) * h^'(x)#

This will get you

#d/dx(y) = [d/dx(cosx)] * x^(-8) + cosx * d/dx(x^(-8))#

#d/dx(y) = -sinx * x^(-8) + cosx * (-8) * x^(-9)#

#d/dx(y) = -(sinx/x^8 + (8cosx)/x^(9))#

This is equivalent to

#d/dx(y) = -( (x * sinx)/x^9 + (8 * cosx)/x^9) = color(green)(-(x * sinx + 8 * cosx)/x^9)#