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

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Answer 1 · 2016-11-19T01:01:02

Slope of tangent to $y$ $y$ at $x = \frac{1}{3}$ $x=1/3$ is $- 8$ $-8$

$y = x^{2} (3 x + \frac{1}{x^{3}})$ $y=x^2(3x+1/x^3)$

$= x^{2} (3 x + x^{- 3})$ $= x^2(3x+x^(-3))$

$\frac{d y}{d x} = x^{2} (3 - 3 x^{- 4}) + 2 x (3 x + x^{- 3})$ $dy/dx= x^2(3-3x^(-4))+2x(3x+x^(-3))$ Product Rule

$= 3 x^{2} - 3 x^{- 2} + 6 x^{2} + 2 x^{- 2}$ $= 3x^2-3x^(-2)+6x^2+2x^(-2)$

$= 9 x^{2} - x^{- 2}$ $= 9x^2-x^(-2)$

The slope $(m)$ $(m)$ of the tangent to $y$ $y$ at $x = \frac{1}{3}$ $x=1/3$ is $\frac{d y}{d x}$ $dy/dx$ at $x = \frac{1}{3}$ $x=1/3$

Thus: $m = 9 \cdot {(\frac{1}{3})}^{2} - {(\frac{1}{3})}^{- 2}$ $m=9*(1/3)^2 - (1/3)^(-2)$

$m = 1 - 9 = 8$ $m= 1-9=8$

What is the slope of the tangent line to the equation y=x^2(3x+1/x^3)y=x2(3x+1x3)y=x^2(3x+1/x^3) at x=1/3x=13x=1/3?