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}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

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