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}

x \to 3 + lim (\frac{1 - 2 x}{x - 3})

\frac{d}{d x} (\frac{14}{ln ( x )})

\frac{d}{d x} (cot (\frac{x}{2}))

\frac{\partial}{\partial x} (8 y - (x + y)^{3})

\frac{d}{d x} (5 cos^{2} (π 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