How do you find the derivative of #1/x^8#?

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Answer 1 · 2016-11-02T08:29:04

#(dy)/(dx)=-8/x^9#

Recall that #1/x^n=x^-n#

#:.1/x^8=x^-8#

#(dy)/(dx)=-8x^-9->#power rule

#=-8/x^9#

OR

Using quotient rule, if #y=1/x^8#, let #u=1# and #v=x^8#

#(dy)/(dx)=(v(du)/(dx)-u(dv)/(dx))/v^2#

#(du)/(dx)=0#

#(dv)/(dx)=8x^7#

#:. (dy)/(dx)=(x^8*0-8x^7)/x^16#

#=(-8x^7)/x^16#

#=-8/x^9#

#=-8x^-9#

Answer 2 · 2016-11-02T08:31:46

#d/dx(x^8) = -8/x^9 #

# y = 1/x^8 #
# implies y= x^-8 #
# implies dy/dx= -8 x^(-8-1) #
# implies dy/dx= -8 x^-9 #
# implies dy/dx = -8/x^9 #