How do you differentiate #y=lnx/x^20#?

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Answer 1 · 2016-10-31T17:52:06

#dy/dx=(x^19-20x^19 ln x)/x^40#

#y=lnx/x^20#

Use quotient rule #(f/g)^' = (gf'-fg')/g^2#

#f=ln x, g= x^20#

#f'=1/x, g'=20x^19#

#dy/dx=(gf'-fg')/g^2#

#dy/dx=(x^20*1/x-lnx *20x^19)/(x^20)^2#

#dy/dx=(x^19-20x^19 ln x)/x^40#