What is the tangent of 1/(x^3) ?

The tangent of 1/(x^3) is y=-3/(a_{0)^4}x+4/(a_{0)^3}

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tangent of 1/(x^3)

tangent of

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

y = - \frac{3}{a _{0}^{4}} x + \frac{4}{a _{0}^{3}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0}^{3}})

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}{a _{0}^{4}}

Find the line with slope m=

- \frac{3}{a _{0}^{4}}

and passing through

(a_{0}, \frac{1}{a _{0}^{3}}) : y = - \frac{3}{a _{0}^{4}} x + \frac{4}{a _{0}^{3}}

y = - \frac{3}{a _{0}^{4}} x + \frac{4}{a _{0}^{3}}

\int_{0}^{2} - 5 x^{3} x^{2} + 4 d x

\int 5 θ cos (θ) d θ

a x + 1

\int (6 x^{3} + 7 3 x^{2}) d x

y = sec^{2} (x), y = sin (x), [0, \frac{π}{4}]

What is the tangent of 1/(x^3) ?
The tangent of 1/(x^3) is y=-3/(a_{0)^4}x+4/(a_{0)^3}