What is the derivative of #5/x^7#?

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Answer 1 · 2016-05-25T06:22:23

#(d y)/(d x)=-35x^-8#

#y=5/x^7#

#y=a/b" "y^'=(a^'*b-b^'*a)/b^2#

#(d y)/(d x)=(0*x^7-7*x^6*5)/x^(14)#

#(d y)/(d x)=-(35x^6)/x^14#

#(d y)/(d x)=-(35)/x^8" "=-35*x^-8#

#" or :"#

#y=5*x^-7#

#y=a*b#

#y^'=a^'*b+b^'*a#

#(d y)/(d x)=-7*0*x^-8-7*x^-8*5#

#(d y)/(d x)=0-35x^-8#

#(d y)/(d x)=-35x^-8#