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

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

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tangent of y= 1/(x^3)

tangent of

y = \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}}

Compute the slope of

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

x = a_{0} : 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_{- 2}^{2} x^{3} - 1 d x

z (v) = v^{\frac{3}{2}} (v + c e^{v})

y^{^{''}} + y = (cot (x))^{2}

\frac{\partial}{\partial x} (\frac{6 x}{x ^{2} + 5})

\int 4 ln (x) d x

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