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

$m= 1-9=8$

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