What is the slope of the tangent line to the equation #y=x^2(3x+1/x^3)# at #x=1/3#?

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Answer 1 · 2016-11-19T01:01:02

Slope of tangent to #y# at #x=1/3# is #-8#

#y=x^2(3x+1/x^3)#

#= x^2(3x+x^(-3))#

#dy/dx= x^2(3-3x^(-4))+2x(3x+x^(-3))# Product Rule

#= 3x^2-3x^(-2)+6x^2+2x^(-2)#

#= 9x^2-x^(-2)#

The slope #(m)# of the tangent to #y# at #x=1/3# is #dy/dx# at #x=1/3#

Thus: #m=9*(1/3)^2 - (1/3)^(-2)#

#m= 1-9=8#