What is the tangent of y= 1/(x^3),(2, 1/8) ?

The tangent of y= 1/(x^3),(2, 1/8) is y=-3/16 x+1/2

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tangent of y= 1/(x^3),(2, 1/8)

tangent of

y = \frac{1}{x ^{3}}, (2, \frac{1}{8})

y = - \frac{3}{16} x + \frac{1}{2}

Solution steps

Find the slope of

y = \frac{1}{x ^{3}} : \frac{d y}{d x} = - \frac{3}{x ^{4}}

Compute the slope of

y = \frac{1}{x ^{3}}

(2, \frac{1}{8}) : m = - \frac{3}{16}

Find the line with slope m=

- \frac{3}{16}

and passing through

(2, \frac{1}{8}) : y = - \frac{3}{16} x + \frac{1}{2}

y = - \frac{3}{16} x + \frac{1}{2}

\frac{d}{d x} (5 + 4 x)

\frac{\partial}{\partial y} (cos (2 x) sin (3 y))

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (4) = 2, y^{^{'}} (4) = 2

\frac{d}{d x} (tan (x - 5))

\int xg (x^{2}) d x

What is the tangent of y= 1/(x^3),(2, 1/8) ?
The tangent of y= 1/(x^3),(2, 1/8) is y=-3/16 x+1/2